Introductory notes in topology

Stephen SemmesRice University

Contents

1 Topological spaces 41.1 Neighborhoods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Other topologies on R 5

3 Closed sets 53.1 Interiors of sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 Metric spaces 7

4.1 Other metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

5 The real numbers 8

5.1 Additional properties . . . . . . . . . . . . . . . . . . . . . . . . . 95.2 Diameters of bounded subsets of metric spaces . . . . . . . . . . 9

6 The extended real numbers 10

7 Relatively open sets 117.1 Additional remarks . . . . . . . . . . . . . . . . . . . . . . . . . . 11

8 Convergent sequences 12

8.1 Monotone sequences . . . . . . . . . . . . . . . . . . . . . . . . . 128.2 Cauchy sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

9 The local countability condition 13

9.1 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149.2 Sequentially closed sets . . . . . . . . . . . . . . . . . . . . . . . 14

10 Local bases 15

11 Nets 15

11.1 Sub-limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

12 Uniqueness of limits 16

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13 Regularity 17

14 An example 1714.1 Topologies and subspaces . . . . . . . . . . . . . . . . . . . . . . 19

15 Countable sets 19

15.1 The axiom of choice . . . . . . . . . . . . . . . . . . . . . . . . . 2015.2 Condensation points . . . . . . . . . . . . . . . . . . . . . . . . . 21

16 Bases 2116.1 Totally bounded sets . . . . . . . . . . . . . . . . . . . . . . . . . 22

17 More examples 22

18 Stronger topologies 23

19 Normality 23

20 Continuous mappings 24

20.1 Sequential continuity . . . . . . . . . . . . . . . . . . . . . . . . . 2520.2 Simple examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

21 The product topology 25

21.1 Countable products . . . . . . . . . . . . . . . . . . . . . . . . . . 2621.2 Arbitrary products . . . . . . . . . . . . . . . . . . . . . . . . . . 27

22 Subsets of metric spaces 27

22.1 The Baire category theorem . . . . . . . . . . . . . . . . . . . . . 2822.2 Sequences of open sets . . . . . . . . . . . . . . . . . . . . . . . . 28

23 Open sets in R 2823.1 Collections of open sets . . . . . . . . . . . . . . . . . . . . . . . 29

24 Compactness 30

24.1 Compactness and convergent sequences . . . . . . . . . . . . . . 30

25 Properties of compact sets 30

25.1 The limit point property . . . . . . . . . . . . . . . . . . . . . . . 31

26 Countable compactness 3226.1 Related results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3326.2 Additional related results . . . . . . . . . . . . . . . . . . . . . . 33

27 Continuity and compactness 34

28 Characterizations of compactness 34

28.1 Sequential compactness . . . . . . . . . . . . . . . . . . . . . . . 3528.2 Some variants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

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29 Products of compact sets 36

29.1 Sequences of subsequences . . . . . . . . . . . . . . . . . . . . . . 3729.2 Compactness of closed intervals . . . . . . . . . . . . . . . . . . . 37

30 Filters 38

31 Mappings and filters 39

32 Ultrafilters 39

33 Continuous real-valued functions 40

34 Compositions and inverses 41

35 Disjoint compact sets 41

36 Local compactness 42

37 Localized separation conditions 42

38 !-Compactness 43

38.1 Subsets of R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

39 Topological manifolds 44

40 !-Compactness and normality 44

41 Unions of bases 45

42 Separating points 46

43 Urysohn’s lemma 46

43.1 R\{0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

44 One-point compactification 47

45 Supports of continuous functions 48

46 Connectedness 4846.1 Connected topological spaces . . . . . . . . . . . . . . . . . . . . 4846.2 Connected sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4946.3 Other properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 5046.4 Pathwise connectedness . . . . . . . . . . . . . . . . . . . . . . . 5046.5 Additional properties and examples . . . . . . . . . . . . . . . . . 5146.6 Local connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 5146.7 Local pathwise connectedness . . . . . . . . . . . . . . . . . . . . 52

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47 A little set theory 52

47.1 Mappings and their properties . . . . . . . . . . . . . . . . . . . . 5247.2 One-to-one correspondences . . . . . . . . . . . . . . . . . . . . . 5347.3 One-to-one mappings . . . . . . . . . . . . . . . . . . . . . . . . . 5347.4 Some basic properties . . . . . . . . . . . . . . . . . . . . . . . . 5447.5 Bases of topological spaces . . . . . . . . . . . . . . . . . . . . . . 5547.6 Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5547.7 Properties of exponentials . . . . . . . . . . . . . . . . . . . . . . 5547.8 Countable dense sets . . . . . . . . . . . . . . . . . . . . . . . . . 5647.9 Additional properties of exponentials . . . . . . . . . . . . . . . . 56

Appendix

A Additional homework assignments 56

A.1 Limit points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57A.2 Dense open sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57A.3 Combining topologies . . . . . . . . . . . . . . . . . . . . . . . . 57A.4 Complements of countable sets . . . . . . . . . . . . . . . . . . . 57A.5 Combining topologies again . . . . . . . . . . . . . . . . . . . . . 57A.6 Combining topologies on R . . . . . . . . . . . . . . . . . . . . . 58

References 58

1 Topological spaces

A topology on a set X is a collection " of subsets of X , the open subsets of Xwith respect to the topology, such that the empty set ! and X itself are opensets, the intersection of finitely many open sets is an open set, and the union ofany family of open sets is an open set. For any set X , the indiscrete topologyhas !, X as the only open sets, and the discrete topology is defined by sayingthat all subsets of X are open. On the real line R, a set U " R is open inthe standard topology if for every x # U there are real numbers a, b such thata < x < b and (a, b) " U , where (a, b) is the open interval consisting of all realnumbers r such that a < r < b.

Let (X, ") be a topological space. We say that X satisfies the first separationcondition if for every x, y # X with x $= y there is an open set V " X such thaty # V and x $# V . To be more precise, x and y may be chosen independentlyof each other, and so the first separation condition actually means that thereare a pair of open subsets V and !V of X such that y # V and x $# V as well asx # !V and y $# !V . We say that X satisfies the second separation condition, orequivalently that X is a Hausdor! topological space, if for every x, y # X withx $= y there are open subsets U , V of X such that x # U , y # V , and U , V aredisjoint, i.e., U % V = !. For the first homework assignment, discuss the casewhere X is a topological space with only finitely many elements which satisfiesthe first separation condition, and give an example of a topological space thatsatisfies the first separation condition but not the second.

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1.1 Neighborhoods

The definition of a topological space may seem a bit strange at first, and cer-tainly quite abstract. It is sometimes helpful to look at a topological spaceX in terms of neighborhoods, where a neighborhood of a point x # X can bedefined as any open set U " X such that x # U . Alternatively, one may takethe neighborhoods of x in X to be the subsets of X that contain an open setcontaining x. At any rate, the neighborhoods of x in X are supposed to containall elements of X which are su!ciently close to x. The definition of a topologyis a convenient way to make this precise, which can then be used for relatednotions of convergence, continuity, etc. In practice, one might start with a classof neighborhoods of elements of a set X , and use this to get a topology on X .Specifically, a set U " X would be defined to be an open set in this situationif for every x # U there is a neighborhood of x in X which is contained in U .For example, the standard topology on the real line is essentially defined in thisway, using open intervals as a basic class of neighborhoods in R.

2 Other topologies on R

In addition to the standard topology on the real line R, let us consider a coupleof “exotic topologies” "!, "+, defined as follows. We say that U " R is an openset with respect to the topology "! if for every x # U there is a real numbera < x such that (a, x] " U , where the interval (a, x] consists of the real numberst which satisfy a < t & x. Similarly, U " R is an open set with respect to thetopology "+ if for every x # U there is a real number b > x such that [x, b) " R,where [x, b) consists of the real numbers t which satisfy x & t < b. One cancheck that these are topologies on R, and that U " R is an open set in thestandard topology if and only if it is open with respect to "! and "+.

If (X, ") is a topological space and E " X , then we say that E is dense inX if for every nonempty open set U " X , E % U $= !. Equivalently, E is densein X if for every x # X and every open set U " X such that x # U , there is apoint y # E which approximates x in the sense that y # U . Observe that X isautomatically dense in itself, and if X is equipped with the discrete topology,then X is the only dense set in X . One can check that the set Q of rationalnumbers is dense in the real line with respect to the standard topology, andalso with respect to the topologies "!, "+ described in the previous paragraph.This uses the fact that for every pair of real numbers a, b with a < b there is arational number r such that a < r < b.

3 Closed sets

Let (X, ") be a topological space, and let E " X and p # X be given. We saythat p is an accumulation point of E in X if for every open set U " X suchthat p # U there is an x # E with x # U . We say that p is a limit point of Ein X if for every open set U " X such that p # U there is a y # E with y $= p

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and y # U . Every element of E is automatically an accumulation point of E,and every limit point of E is automatically an accumulation point of E. Anaccumulation point of E in X which is not an element of E is automatically alimit point of E, and in general a limit point of E may or may not be an elementof E.

For any A " X , the complement of A in X is denoted X\A and defined tobe the set of z # X such that z $# A. Hence the complement of the complementof A is equal to A. The complement of a union of subsets of X is equal to theintersection of the complements of the sets, and similarly the complement of anintersection of subsets of X is equal to the union of the complements of the sets.We say that A " X is closed if the complement of A in X is an open set. Itfollows from the definition of a topology that !, X are closed subsets of X , theunion of finitely many closed subsets of X is a closed set, and the intersectionof any family of closed subsets of X is a closed set.

The closure of E " X is denoted E and defined to be the set of accumulationpoints of E in X , which is the same as the union of E and the set of limit pointsof E in X . Equivalently, z # X is not in the closure of E if there is an open setU " X such that z # U and E % U = !, in which event U is contained in thecomplement of the closure of E. It follows that X\E is an open set, which is tosay that E is closed. Conversely, if E " X is closed, then E = E, because everyelement of the complement of E is contained in the open set X\E which doesnot intersect E by definition. Observe that E " X is dense in X if and only ifE = X .

3.1 Interiors of sets

Let X be a topological space. The interior of a set A " X is denoted IntA andis the set of x # A for which there is an open set U " X such that x # U andU " A. Hence

IntA " A

automatically. If U " X is an open set and U " A, then U " IntA, andin particular IntA is automatically an open set too. It is easy to check thatthe complement of the interior of A " X in X is equal to the closure of thecomplement of A in X , i.e.,

X\ IntA = X\A.

Equivalently,X\E = Int(X\E)

for every E " X . This extends the fact that the complement of an open set isa closed set and vice-versa.

If A1, A2 are subsets of X , then

Int(A1 % A2) = (IntA1) % (IntA2).

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This follows from the definition of the interior of a set. Similarly,

(IntA1) ' (IntA2) " Int(A1 ' A2).

Thus,E1 ' E2 = E1 ' E2

andE1 % E2 " E1 % E2

for every E1, E2 " X .

4 Metric spaces

By a metric space we mean a set M together with a function d(x, y) defined forx, y # M such that (i) d(x, y) is a nonnegative real number for every x, y # Mwhich is equal to 0 if and only if x = y, (ii) d(x, y) = d(y, x) for every x, y, z # M ,and (iii) the triangle inequality holds, which is to say that

d(x, z) & d(x, y) + d(y, z)

for every x, y, z # M . In this event we say that d(x, y) defines a metric on M .For every x # M and positive real number r, we put

B(x, r) = {y # M : d(x, y) < r},

the open ball centered at x with radius r in M . We say that U " M is an openset if for every x # U there is an r > 0 such that B(x, r) " U . One can checkthat this defines a topology on M . Specifically, if U1, . . . , Un are open subsetsof M in this sense and x is an element of the intersection U1 % · · · % Un, thenfor each i = 1, . . . , n, x is an element of Ui, and there is an ri > 0 such thatB(x, ri) " Ui. If r is equal to the minimum of r1, . . . , rn, then r > 0 and B(x, r)is contained in U1 % · · ·%Un. This shows that U1 % · · ·%Un is an open set in M ,and the fact that arbitrary unions of open subsets of M are open sets is evensimpler. For every x # M and r > 0, one can check that the open ball B(x, r)is an open set, because for every y # B(x, r) we have that

B(y, t) " B(x, r)

with t = r ( d(x, y) > 0, by the triangle inequality.On any set M , the discrete metric is defined by putting d(x, y) equal to 0

when x = y and to 1 when x $= y. One can check that this is a metric, and thatthe corresponding topology on M is the discrete topology.

4.1 Other metrics

Let (M, d(x, y)) be a metric space. For every positive real number #, # d(x, y)defines a metric on M which determines the same topology on M as d(x, y)

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does, i.e., the same class of open subsets of M . To be more precise, the openball in M centered at a point w # M and with radius t > 0 with respect to theinitial metric d(x, y) is the same as the open ball in M centered at w with radius# t with respect to the new metric # d(x, y). Similarly, for every positive realnumber " , d! (x, y) = min(d(x, y), ") defines a metric on M which determinesthe same topology as d(x, y). In order to check that d! (x, y) satisfied the triangleinequality, observe that

min(a + b, ") & min(a, ") + min(b, ")

for every a, b ) 0. The open ball in M centered at a point z # M and withradius r > 0 with respect to d(x, y) is the same as the open ball in M centeredat z with radius r with respect to d! (x, y) when r & " . When r > " , any openball in M with radius r with respect to d! (x, y) contains every element of M .However, the equivalence of balls with small radius is su!cient to imply thatthe two metrics determine the same class of open subsets of M , i.e., the sametopology. As another example along these lines, consider d"(x, y) =

"d(x, y).

For any real numbers a, b ) 0,

a2 + b2 & a2 + 2 a b + b2 = (a + b)2,

and one can use this to show that d"(x, y) satisfies the triangle inequality andhence is a metric on M . In this case, the open ball centered at a point p # Mand with radius r > 0 with respect to d(x, y) is the same as the open ballcentered at p with radius

*r with respect to d"(x, y), and this implies that the

associated topologies are the same.

5 The real numbers

If x is a real number, then the absolute value |x| of x is equal to x when x ) 0and to (x when x & 0. Hence |x| ) 0 for every x # R, and |x| = 0 if and onlyif x = 0. One can check that

|x + y| & |x| + |y|

for every x, y # R, and therefore that |x ( y| defines a metric on R, known asthe standard metric. The topology determined by the standard metric is thestandard topology. To be more precise, the open ball in R centered at x # Rwith radius r > 0 is the same as the open interval (x ( r, x + r).

If A " R and b # R, then we say that b is an upper bound for A if a & b forevery a # A. We say that c # R is the least upper bound or supremum of A,denoted supA, if c is an upper bound for A and if c & b for every upper boundb of A. It follows directly from the definition that the supremum of A is uniqueif it exists, i.e., if c1, c2 # R both satisfy the requirements of suprema of A, thenc1 & c2, c2 & c1, and thus c1 = c2. The completeness property of the real linestates that every nonempty set of real numbers which has an upper bound hasa supremum. The rational numbers are not complete in this sense, because the

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set of t # Q such that t ) 0 and t2 & 2 is a nonempty subset of Q with anupper bound that does not have a supremum in Q, and in fact the supremumin the real line is equal to

*2.

Similarly, w # R is a lower bound for B " R if w & x for every x # B,and z # R is the greatest lower bound or infimum of B, denoted inf B, if z isa lower bound of B and every lower bound w # R of B satisfies w & z. Asin the previous situation, the infimum is automatically unique if it exists. If Bis a nonempty set of real numbers with a lower bound, then the completenessproperty for the real numbers implies that B has an infimum, which can beobtained as the supremum of the set of lower bounds of B. Alternatively, theinfimum of B is equal to the negative of the supremum of

(B = {x # R : (x # B}.

5.1 Additional properties

The completeness property of the real numbers can be used to show that everya # R with a > 0 has a positive square root. Specifically, let A be the set ofr # R such that r ) 0 and r2 & a. Clearly 0 # A, and thus A $= !. If a & 1,then r & 1 for every r # A, while if a ) 1, then r & a for every a # A. HenceA is a nonempty set of real numbers with an upper bound, and thus there is ab # R which is the least upper bound of A. If b2 < a, then one can show thatthere is an $ > 0 such that (b+$)2 = b2 +2 b $+$2 < a. Similarly, if b2 > a, thenone can show that there is a % > 0, % < b, such that (b(%)2 = b2(2 b %+%2 > a.In both cases one can get a contradiction with the fact that b = supA, and itfollows that b2 = a. It is easy to show directly that a positive square root of ais unique.

If r, t # R and r & t, then the closed interval [r, t] consists of the x # R

such that r & x & t. Suppose that Il = [rl, tl], l = 1, 2, . . ., is a sequence ofclosed intervals in the real line such that Il " Il+1 for every l ) 1. Equivalently,rl & rl+1 & tl+1 & tl for every l ) 1, which implies that rl & rn & tn & tl when1 & l & n. Using completeness one can show that there is an x # R such thatx # Il for every l ) 1. To be more precise, the intersection

##

l=1 Il of all of theIl’s is the same as the closed interval [r, t], where r is the supremum of the setof rl’s and t is the infimum of the set of tl’s, l ) 1.

5.2 Diameters of bounded subsets of metric spaces

Let (M, d(x, y)) be a metric space. A set E " M is said to be bounded if thereis a point p # M and a positive real number r such that d(x, p) & r for everyx # E. In this case, d(x, q) & d(p, q) + r for every q # M and x # E, by thetriangle inequality. Equivalently, E " M is bounded if the set of real numbersof the form d(x, y), x, y # E, is bounded. If E " M is bounded and nonempty,then the diameter of E is denoted diamE and defined by

diamE = sup{d(x, y) : x, y # E}.

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Suppose that E1, E2 are nonempty subsets of M such that E1 " E2. If E2 isbounded, then E1 is bounded too, and

diamE1 & diamE2.

If E " M is bounded and nonempty, then the closure E of E is bounded, and

diamE = diamE.

For if x, y # E and $ > 0 is arbitrary, then there are x", y" # E such thatd(x, x"), d(y, y") < $. Hence d(x, y) & d(x", y") + 2 $, by the triangle inequality,which implies that d(x, y) & diamE +2 $. Because this holds for every x, y # E,we may conclude that E is bounded and diamE & diamE + 2 $, which impliesin turn that diamE & diamE since $ > 0 is arbitrary. The opposite inequalityholds automatically, and therefore the diameters of E and E are equal.

6 The extended real numbers

By definition, the extended real numbers consist of the real numbers togetherwith two additional elements, (+ and ++, such that

(+ < x < ++

for every x # R. The notions of upper and lower bounds make sense for setsof extended real numbers, with ++ as an upper bound and (+ as a lowebound for any set of extended real numbers. Using the completeness propertyfor the real numbers, one can check that every nonempty set of extended realnumbers has a unique supremum and infimum, defined in the same way as forsets of real numbers. In particular, if E is a nonempty set of real numberswhich does not have an upper bound in R, then the supremum of E is equal to++ in the extended real numbers. Similarly, the infimum of a nonempty set ofreal numbers without a finite lower bound is equal to (+ in the extended realnumbers.

A set U of extended real numbers is an open set if for every x # U eitherx # R and there are real numbers a, b such that a < x < b and (a, b) " U ,or x = (+ and there is a real number b such that the set [(+, b) of extendedreal numbers y < b is contained in U , or x = ++ and there is a real numbera such that the set (a, ++] of extended real numbers z > a is contained in U .This defines a topology on the set of extended real numbers, for which the realline R is a dense open set. This topology agrees with the one on R in the sensethat E " R is open in R with the standard topology if and only if E is openin the extended real numbers. If E " R, then ++ is a limit point of E in theextended real numbers if and only if E does not have an upper bound in R, andsimilarly (+ is a limit point of E in the extended real numbers if and only ifE does not have a lower bound in R.

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7 Relatively open sets

Suppose that (X, ") is a topological space and that Y " X . There is a naturaltopology on Y induced by the one on X , in which E " Y is open in Y whenthere is an open set U " X such that E = U % Y . In this event we say that Eis relatively open in Y . If Y happens to be an open set in X , then E " Y isrelatively open in Y if and only if E is an open set in X , but otherwise relativelyopen subsets of Y may not be open in X . Following the general definition ofclosed subsets of a topological space, E " Y is relatively closed in Y if Y \E isrelatively open in Y . Hence E is relatively closed in Y when Y \E = U % Y forsome open set U in X , which implies that E = (X\U)%Y and X\U is closed inX . Conversely, if E = A%Y for some closed set A in X , then Y \E = (X\A)%Yand X\A is open in X , and E is relatively closed in Y . In particular, if Y isclosed in X , then E " Y is relatively closed in Y if and only if E is closed inX . For any Y , the relative closure of E " Y , which is the closure of E in Y asa topological space itself, is equal to the intersection of Y with the closure of Ein X .

If (M, d(x, y)) is a metric space and Y " M , then the restriction of d(x, y) tox, y # Y defines a metric on Y . The topology on Y associated to the restrictionof the metric to Y is the same as the topology on Y induced from the topologyon X determined by the metric. Indeed, if U " X is open in X , then U % Yis open in Y with respect to the restriction of the metric to Y . Conversely, theopen ball in Y with center y # Y and radius r > 0 is automatically equal to theintersection of Y with the open ball in X with center y and radius r. Therefore,every open ball in Y is relatively open with respect to the topology induced fromthe one on X . If E " Y is an open set with respect to the topology determinedby the restriction of the metric to Y , then E is equal to a union of open balls inY , the open balls in Y that are contained in E. Each of these open balls is theintersection of Y with an open ball in X , and hence their union is the same asthe intersection of Y with the union of the corresponding open balls in X , anopen set in X .

7.1 Additional remarks

Let X be a topological space. If U " X is an open set, then for every Y " X ,U %Y is relatively open in Y . Conversely, if U " X and U %Y is relatively openin Y for every Y " X , then U is an open set in X , since we can take Y = X .Similarly, E " X is a closed set if and only if E % Y is relatively closed in Y forevery Y " X . As a refinement of this type of question, one can look for classesof subsets Y of X such that open and closed subsets of X are characterized byintersections with Y being relatively open or closed in Y , respectively, for everyY in the particular class of subsets of X . For example, the class of open subsetsY of X has this property. Actually, any class of open subsets of X whose unionis equal to X has this property. Any class of subsets of X for which the unionof their interiors is equal to X has this property for the same reasons.

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8 Convergent sequences

A sequence {xj}#j=1 of elements of a topological space X converges to x # X iffor every open set U " X with x # U there is an L ) 1 such that xj # U forevery j ) L. If X is equipped with the indiscrete topology, then every sequence{xj}#j=1 of elements of X converges to every x # X . If X is equipped with thediscrete topology and {xj}#j=1 is a sequence of elements of X that converges tox # X , then there is an L ) 1 such that xj = x for every j ) L, because {x}is an open set in X . For any topological space, if xj = x for each j, or if thereis an L ) 1 such that xj = x when j ) L, then {xj}#j=1 converges to x. In thereal line, xj = 1/j converges to 0 in the standard topology.

Let X be a topological space, let x be an element of X , and consider thesequence {xj}#j=1 such that xj = x for every j. As mentioned in the previousparagraph, this sequence converges to x in X . If X satisfies the first separationcondition, y # X , and y $= x, then there is an open set V " X such that y # Vand x $# V , and it follows that {xj}#j=1 does not converge to y in X . Conversely,if y # X and every open set V " X with y # V also contains x, then {xj}#j=1

converges to y.Suppose that X is a topological space, E " X , and {xj}#j=1 is a sequence

of elements of E which converges to x # X . Under these conditions, x is anaccumulation point of E, and hence an element of the closure E of E, and anelement of E if E is a closed set. If xj $= x for every j, then x is a limit point ofE. The first separation condition is equivalent to the statement that subsets ofX with exactly one element are closed. For if x # X , then the first separationcondition implies that the set of w # X with w $= x is a union of open sets.

8.1 Monotone sequences

A sequence {xl}#l=1 of real numbers is said to be monotone increasing if xl & xn

for every n ) l ) 1. Similarly, a sequence {yl}#l=1 of real numbers is said to bemonotone decreasing if yl & yn when n ) l ) 1. Every monotone increasing ordecreasing sequence of real numbers which is bounded in the sense that the termsof the sequence are contained in a bounded set in R converges. Specifically, onecan check that a monotone increasing sequence of real numbers which is boundedfrom above converges to the supremum of the set of terms in the sequence, anda monotone decreasing sequence of real numbers which is bounded from belowconverges to the infimum of the terms in the sequence. In particular, this usesthe completeness property of the real numbers.

Now consider the real line equipped with the exotic topologies "+, "!, whereU " R is an open set with respect to "+ if and only if for every x # U there isa t # R with t > x such that [x, t) " U , and V " R is an open set with respectto "! if and only if for every y # V there is an r # R with r < y such that(r, y] " V . A monotone increasing sequence of real numbers which is boundedfrom above converges to the supremum of the terms in the sequence with respectto the topology "!, but a monotone decreasing sequence converges with respectto "! if and only if it is eventually constant. In the same way, a monotone

12

decreasing sequence of real numbers which is bounded from below converges tothe infimum of the terms in the sequence with respect to the topology "+, buta monotone increasing sequence converges with respect to "+ if and only if itis eventually constant. To be more precise, a sequence is said to be eventuallyconstant if all but finitely many terms in the sequence are the same.

8.2 Cauchy sequences

Let (M, d(x, y)) be a metric space. A sequence {xl}#l=1 of elements of M is saidto be a Cauchy sequence if for every $ > 0 there is an L ) 1 such that

d(xl, xn) < $ for every l, n ) L.

If {xl}#l=1 converges to a point x # M , then {xl}#l=1 is a Cauchy sequence,because for every $ > 0 there is an L such that

d(xl, x) <$

2when l ) L,

and henced(xl, xn) & d(xl, x) + d(xn, x) <

$

2+$

2= $

when l, n ) L. A metric space in which every Cauchy sequence converges issaid to be complete.

One can use the completeness property of the real numbers in terms ofordering to show that the real line equipped with the standard metric |x ( y|is complete as a metric space. For suppose that {xl}#l=1 is a Cauchy sequenceof real numbers, and consider El = {xn : n ) l} for each l. Because {xl}#l=1 isa Cauchy sequence, El is a bounded set of real numbers for every l ) 1, andhence has an infimum rl and a supremum tl. The fact that {xl}#l=1 is a Cauchysequence implies that tl ( rl converges to 0 as a sequence of real numbers. Onecan check that the supremum of the rl’s is equal to the infimum of the tl’s, andthat {xl}#l=1 converges to their common value.

9 The local countability condition

A topological space X satisfies the local countability condition at a point p # Xif there is a sequence {U"(p)}#"=1 of open subsets of X such that p # U"(p) andU"+1(p) " U"(p) for every & ) 1, and for every open set U " X with p # Uthere is an & ) 1 such that U"(p) " U . It is easy to arrange for the U"(p)’sto be decreasing, by replacing U"(p) with U1(p) % · · · % U"(p) if necessary, andthis ensures that U"(p) " U implies Uk(p) " U when k ) &. The topology ofa metric space is locally countable at every point, because one can take U"(p)to be the open ball centered at p with radius 1/&. The real line equipped withthe topology "+, in which U " R is open if and only if for every p # U thereis a t # R with t > p such that [p, t) " U , is locally countable at every point,because one can take U"(p) = [p, p+1/&). Similarly, the real line equipped with

13

the topology "!, in which U " R is open if and only if for every p # U there isan r # R, r < t, such that (r, t] " U , is locally countable at every point, withU"(p) = (p ( 1/&, p].

Suppose that X is a topological space which is locally countable at a pointp # X , and let {U"(p)}#"=1 be a sequence of open sets as in the previous para-graph. If p is an accumulation point of E " X , then for every & ) 1 there is anx" # E such that x" # U"(p), and if p is a limit point of E, then one can choosex" such that x" $= p too. Under these conditions, {x"}#"=1 converges to p in X .If X is locally countable at every point, then it follows that E " X is closed ifand only if for every sequence {xj}#j=1 of elements of E which converges to apoint x # X we have that x # E. In particular the topology of a space which islocally countable at every point is determined by the convergent sequences andtheir limits.

9.1 Subsequences

Let {xi}#i=1 be a sequence of elements of some set X . A subsequence of {xi}#i=1

is a sequence of the form {xin}#n=1, where i1 < i2 < · · · is a strictly increasing

sequence of positive integers. Suppose now that X is a topological space. LetE1 be the set of x # X for which there is a subsequence {xin

}#n=1 of {xi}#i=1

such that {xin}#n=1 converges to x in X . Let E2 be the set of x # X with the

property that for every open set U " X with x # U there are infinitely manypositive integers l such that xl # U . It follows directly from the definitionsthat E1 " E2. Conversely, if X satisfies the local countability condition at eachpoint, then one can check that E2 " E1 and hence E1 = E2. In any topologicalspace X , it is easy to see that E2 is a closed set. Therefore, if X satisfies thefirst countability condition, then E1 is a closed set in X too.

9.2 Sequentially closed sets

Let (X, ") be a topological space. A set E " X is said to be sequentially closedif for every sequence {xi}#i=1 of elements of E and every x # X such that {xi}#i=1

converges to x in X we have that x # E. Closed subsets of X in the usual senseare automatically sequentially closed. Conversely, sequentially closed subsets ofX are closed in the usual sense when X satisfies the local countability conditionat each point. More generally, if Y " X satisfies the local countability conditionat each of its elements with respect to the topology induced by " on X and ifE is sequentially closed in X , then E % Y is relatively closed in Y .

It is easy to see that the intersection of any family of sequentially closedsubsets of X is sequentially closed. If E1, E2 are sequentially closed subsetsof X , then their union E1 ' E2 is sequentially closed, because any sequence ofelements of E1'E2 has a subsequence whose terms are contained in E1 or in E2.Of course a subsequence of a convergent sequence converges to the same limit.It follows that the collection !" of complements of sequentially closed subsetsof X defines a topology on X that contains " . However, it is not clear that a

14

sequence in X which converges with respect to " also converges with respect to!" .

10 Local bases

A local base for the topology of a topological space X at a point p # X is acollection B(p) of open subsets of X such that p # V for every V # B(p), andfor every open set U " X with p # U there is a V # B(p) which satisfies V " U .The collection of all open subsets of X that contain p as an element is a localbase for the topology at p. In a metric space, the open balls centered at p forma local base for the topology at p. The exotic topologies "+, "! on the realline are characterized by the local bases B+(p), B!(p), where B+(p) consists ofthe intervals [p, t), t > p, and B!(p) consists of the intervals (r, p], r < p. Inany topological space, the topology is determined by choices of local bases atall of the elements of the space. A topological space X is locally countable at apoint p # X if and only if there is a local base B(p) for the topology at p whoseelements can be enumerated by a sequence. As noted previously, a sequenceof open subsets of X that form a local base for the topology of X at p can bemodified to get a decreasing sequence of open sets which form a local base forthe topology at p using the intersection of the first n open sets in the originalsequence for each n.

Let B(p) be a local base for the topology of X at p, and let E " X be given.It is easy to see that p is an accumulation point of E if and only if for everyU # B(p) there is a q # E such that q # U . Similarly, p is a limit point of E ifand only if for every U # B(p) there is a q # E such that q # U and q $= p.

11 Nets

A partial order on a set A is a binary relation , such that (i) x , x for everyx # A, (ii) x , y and y , x imply x = y for every x, y # A, and (iii) x , yand y , z imply x , z for every x, y, z # A. A partially-ordered set (A,,) isa directed system if for every collection a1, . . . , an of finitely many elements ofA there is an a # A such that ai , a for i = 1, . . . , n. For example, the set ofpositive integers is a directed system with respect to the usual ordering. If Xis a topological space, p # X , and B(p) is a local base for the topology of X atp, then B(p) becomes a directed system with the ordering , defined by U , Vwhen V " U for every U, V # B(p). The ordering is the reverse of inclusion,because smaller open sets represent greater precision at p. If U1, . . . , Un # B(p),then their intersection U1 % · · · % Un is an open set in X that contains p, andhence there is a V # B(p) such that V " U1 % · · ·%Un. This shows that B(p) isa directed system with respect to the ordering ,.

If (A,,) is a directed system and X is any set, then a net {xa}a$A indexedby A with values in X assigns to every a # A an element xa of X . If A isthe set of positive integers with the usual ordering, then this is the same as

15

a sequence of elements of X . If X is a topological space, {xa}a$A is a net ofelements of X along the directed system A, and x # X , then {xa}a$A convergesto x if for every open set U " X there is a b # A such that xa # U for everya # A which satisfies b , a. This is the same as the definition of convergenceof a sequence when A is the set of positive integers. If E " X and {xa}a$A isa net of elements of E which converges to x # X , then x is an accumulationpoint of E, and conversely every accumulation point of E is the limit of a net ofelements of E. Similarly, if {xa}a$A is a net of elements of E which converges tox # X and xa $= x for every a # A, then x is a limit point of E, and every limitpoint of E is the limit of a convergent net of this type. It follows that E " Xis closed if and only if for every net {xa}a$A of elements of E which convergesto x # X we have that x # E. Therefore the topology on X is determined bythe convergence of nets.

11.1 Sub-limits

Let X be a topological space, let (A,,) be a directed system, and let {xa}a$A

be a net of elements of X indexed by A. Let E be the set of x # X with theproperty that for every a # A and open set U " X with x # U there is a b # Asuch that a , b and xb # U . It is easy to see from the definition that E is aclosed set in X . Fix an x # E. Let Ax be the set of ordered pairs (a, U) wherea # A and U " X is an open set with x # U . It would be su!cient for thepresent purposes to use only open sets U in a local base for the topology of Xat x. If (a, U), (b, V ) # Ax, then put (a, U) ,x (b, V ) when a , b and V " U .One can check that this defines a partial ordering on Ax which makes Ax intoa directed system. For (a, U) # Ax, let x(a,U) be an element of X of the formxb, where b # A, a , b, and xb # U . By construction, {x(a,U)}(a,U)$Ax

is a netof elements of X indexed by Ax that converges to x.

12 Uniqueness of limits

Let X be a Hausdor" topological space, let (A,,) be a directed system, andlet {xa}a$A be a net of elements of X which converges to x, y # X . If x $= y,then the Hausdor" property implies that there are open subsets U , V of X suchthat x # U , y # V , and U % V = !. Because of convergence, there are b, b" # Asuch that xa # U when b , a and xa # V when b" , a. If c # A and b, b" , c,then xa # U % V when c , a, a contradiction. This shows that the limit of aconvergent net in a Hausdor" space is unique, and in particular the limit of aconvergent sequence is unique.

Conversely, suppose that X is a topological space which is not Hausdor".Hence there are x, y # X such that x $= y and for every pair of open subsets U ,V of X with x # U and y # V , we have that U % V $= !. Let A be the set ofordered pairs (U, V ) of open subsets of X with x # U and y # V . There is anatural partial ordering , on A, where (U, V ) , (U ", V ") for (U, V ), (U ", V ") # Awhen U " " U and V " " V . If (U1, V1), . . . , (Un, Vn) are finitely many elements

16

of A, and if we put U = U1 % · · · % Un, V = V1 % · · · % Vn, then (U, V ) # A and(Ui, Vi) , (U, V ) for i = 1, . . . , n. This shows that (A,,) is a directed system.For each (U, V ) # A, let x(U,V ) be an element of U %V . This defines a net alongA with values in X . By construction, this net converges to x and to y. It followsthat a topological space is Hausdor" if every convergent net in the space has aunique limit.

13 Regularity

Let (M, d(x, y)) be a metric space, and let x, y be elements of M , with x $= y.If r = d(x, y), then B(y, r) is an open set in M that contains y but not x, andhence the topology on M determined by the metric satisfies the first separationcondition. If t = d(x, y)/2, then B(x, t), B(y, t) are disjoint open sets in Mwhich contain x, y, respectively, and it follows that M is a Hausdor" topologicalspace. For every x # M and r ) 0, let Vr(x) be the set of w # M such thatd(w, x) > r. If z # Vr(x) and t = d(z, x) ( r, then t is a positive real number,and one can use the triangle inequality to check that B(z, t) is contained inVr(x). It follows that Vr(x) is an open set in M . For x # M and r ) 0, theclosed ball with center x and radius r is denoted B(x, r) and defined to be theset of w # M with d(w, x) & r. This is the same as the complement of Vr(x)in M , and hence B(x, r) is a closed set in M . One can also show that B(x, r)is a closed set by verifying directly that every limit point of B(x, r) in M is anelement of B(x, r).

A topological space X satisfies the third separation condition, or is regular,if it satisfies the first separation condition and if for every x # X and closed setE " X such that x $# E there are open subsets U , V of X such that x # U ,E " V , and U % V = !. The first separation condition is equivalent to thestatement that for every y # X , the set consisting of y and no other elementsis closed in X , and therefore a regular topological space is Hausdor" becauseone can take E = {y} when y $= x. Suppose that (M, d(x, y)) is a metric space,x # M , and E " M is a closed set such that x # M\E. Because M\E is anopen set, there is an r > 0 such that B(x, r) " M\E. It follows that B(x, r/2),Vr/2(x) are disjoint open subsets of M that contain x, E, respectively, and thusM is a regular topological space.

14 An example

Let V be the space of real-valued functions on the set Z+ of positive integers.This is basically the same as the space of sequences of real numbers, but forthe sake of notational simplicity it is convenient to express the elements of V asfunctions. If # is a positive real-valued function on Z+ and f # V , then put

N#(f) = {g # V : |f(l) ( g(l)| < #(l) for every l # Z+},

the neighborhood of f associated to # in V . Let us say that U " V is an openset if for every f # U there is a positive real-valued function # on Z+ such that

17

N#(f) " U . Clearly the empty set and V are open subsets of V in this sense.Suppose that U1, . . . , Un are open subsets of V , and that f # U1 % · · ·%Un. Foreach i = 1, . . . , n, there is a positive real-valued function #i on Z+ such thatN#i

(f) " Ui. If#(l) = min(#1(l), . . . , #n(l))

for l # Z+, then # is a positive real-valued function on Z+, and

N#(f) = N#1(f) % · · · % N#n

(f) " U1 % · · · % Un.

Hence U1 % · · · % Un is an open set in V . It is easy to see from the definitionthat a union of open subsets of V is an open set, and therefore that we have atopology on V .

If f # V , # is a positive real-valued function on Z+, and g # N#(f), then

#"(l) = #(l) ( |f(l) ( g(l)| > 0

for every l # Z+ and N#!(g) " N#(f), which implies that N#(f) is an open setin V . If f1, f2 are distinct elements of V , which means that f1(l) $= f2(l) forsome l, then one can choose a positive real-valued function # on Z+ such that2 #(l) < |f1(l) ( f2(l)| for some l, and hence N#(f1) % N#(f2) = !. Thus V isHausdor", and one can also show that V is regular. However, V does not satisfythe local countability condition at any point. For suppose that f # V and thatU1, U2, . . . is a sequence of open sets in V that contain f and form a local basisfor the topology of V at f . Let #1, #2, . . . be a sequence of positive real-valuedfunctions on Z+ such that N#i

(f) " Ui for each i. Put

#(n) =min(#1(n), . . . , #n(n))

n.

Thus # is a positive real-valued function on Z+, #(l) < #i(l) when l ) i, andN#(f) does not contain N#i

(f) " Ui for any i, a contradiction.If {fj}#j=1 is a sequence of elements of V , f # V , {fj(l)}#j=1 converges to f(l)

as a sequence of real numbers in the standard topology on R for every l # Z+,and there is an n ) 1 such that fj(l) = f(l) when j, l ) n, then it is easy to seethat {fj}#j=1 converges to f in the topology defined on V . Conversely, supposethat {fj}#j=1 is a sequence of elements of V which converges to f # V in thistopology. Let # be the nonnegative real-valued function on Z+ defined by

#(n) =sup{min(|fj(l) ( f(l)|, 1) : j, l ) n}

2.

By construction, # is monotone decreasing on Z+. If #(n) > 0 for every n # Z+,then there is an n0 ) 1 such that fj # N#(f) when j ) n0, by convergence inV . This implies that

|fj(l) ( f(l)| < #(l) & #(n0)

when j, l ) n0, contradicting the definition of #(n0). Therefore #(n) = 0 forsome n, which implies that fj(l) = f(l) when j, l ) n. Also, {fj(l)}#j=1 converges

18

to f(l) as a sequence of real numbers in the standard topology on R for everyl # Z+, and so we are back in the situation described at the beginning of theparagraph. If V0 is the space of f # V such that f(l) $= 0 for only finitely manyl, then V0 is closed in V . Indeed, if f(l) $= 0 for infinitely many l, then we canchoose a positive function # on Z+ such that #(l) & |f(l)| when f(l) $= 0, andevery h # N#(f) satisfies h(l) $= 0 when f(l) $= 0, which is to say that N#(f) iscontained in the complement of V0 in V . The functions on Z+ with values inthe rational numbers Q are dense in V , and similarly the Q-valued functions inV0 are dense in V0.

14.1 Topologies and subspaces

Let X be a set, and suppose that X1, X2, . . . is a sequence of subsets of X suchthat Xi " Xi+1 for each i and X =

$#

i=1 Xi. Suppose too that each Xi isequipped with a topology "i, and that these topologies are compatible in thesense that the topology on Xi induced by "i+1 on Xi+1 is the same as "i. Itfollows that the topology on Xi induced by "l on Xl is the same as "i whenl > i. Let us say that U " X is an open set in X if U %Xi is an open set in Xi

with respect to "i for every i ) 1. It is easy to check that this defines a topology" on X , using the fact that each "i is a topology on Xi. If Ui " Xi is an openset with respect to "i, then the compatibility condition implies that there is anopen set Ui+1 " Xi+1 with respect to ti+1 such that Ui = Ui+1%Xi. Repeatingthe process, for every l > i there is an open set Ul " Xl with respect to "l suchthat Ul!1 = Ul % Xl!1. Hence Ul = Un % Xl when n ) l ) i. If U =

$#

l=i Ul,then Un = U % Xn for every n ) i. In particular, U is an open set in X , and itfollows that "i is the same as the topology induced on Xi by " .

As a variant of this construction, suppose also that X is a vector space overthe real numbers, and that each Xi is a linear subspace of X . Under suitableconditions, one might first say that a convex set U " X is an open set in X ifU % Xi is an open set in Xi for every i ) 1. One can then use the convex opensets to generate a topology on X in the usual way, i.e., where V " X is an openset if for every x # V there is a convex open set U " X such that x # U andU " V .

15 Countable sets

Let A be a set, and suppose that there is a sequence {xj}#j=1 of elements of Ain which every element of A appears at least once. It may be that A has onlyfinitely many elements, and that some elements are repeated infinitely often inthe sequence. At any rate, one can systematically go through the sequence andskip terms that are repeated. If A has infinitely many elements, then this willleave a sequence {yl}#l=1 of elements of A in which every element of A occursexactly once. In this event we say that A is countably infinite. If B " A hasinfinitely elements, then we can also go through the sequence and just keep theterms in B, to get a sequence {zn}#n=1 of elements of B in which every element

19

of B appears exactly once. It follows that B is countably infinite too.If one adds an element to a countably-infinite set, then the resulting set is

countably-infinite, because one can get an enumeration of the elements in thenew set by putting the new element in the first term and sliding the other termsover one step. Similarly, the union of a countably-infinite set with a set with onlyfinitely many elements is countably infinite. One can enumerate the elements ofa union of two countably-infinite sets by a sequence by using the terms indexedby even integers for one set and the terms indexed by odd integers for the otherset. If A1, A2, . . ., is a sequence of sets with only finitely many elements, thenone can make a sequence listing the elements of the union A =

$#

j=1 Aj by firstlisting the elements of A1, then the elements of A2, etc., and hence A has onlyfinitely many elements or is countably infinite. The Cartesian product Z+-Z+,consisting of ordered pairs of positive integers, can be expressed as a union of asequence of finite subsets, and is therefore countable infinite. Using this one cancheck that the union of a sequence of countably-infinite sets is countably-infinite.Basically, one can list the elements of the union by a doubly-indexed sequence,and convert that into a sequence of the usual type using an enumeration ofZ+ - Z+.

The set Z+ of positive integers can be enumerated trivially with the sequencexj = j and hence is countably infinite. The set Z of all integers can be enu-merated by the sequence y2l = l, y2l+1 = (l and is also countably infinite. Forevery positive integer n, the set of integer multiples of 1/n is basically the sameas the set of all integers and therefore countably infinite, and consequently theset Q of rational numbers is countably infinite, being the union of these setsover all n # Z+. Let B be the set of functions f on Z+ such that f(l) = 0 or1 for every l # Z+. Suppose that {fj}#j=1 is a sequence of functions in B. Iff is the function on Z+ such that f(l) = 1 when fl(l) = 0 and f(l) = 0 whenfl(l) = 1, then f # B and f is di"erent from fj for every positive integer j. Itfollows that B is uncountable, which means that it is neither finite nor countablyinfinite. One can use this to show that the real line is uncountable. Specifically,every element of B corresponds to a real number in the interval [0, 1], using theelement of B as a binary expansion. Every element of [0, 1] occurs in this wayat least once and at most twice. If there were a sequence listing all of the realnumbers in [0, 1], then B would be countably infinite as well, a contradiction.

15.1 The axiom of choice

Let A be a set, and suppose that f is a function on A which assigns to everya # A a set f(a) " A. Let B be the set of a # A such that a # B when a $# f(a)and a $# B when a # f(a). By construction, B $= f(a) for every a # A, sincea is an element of exactly one of B, f(a) for every a # A. This generalizes theuncountability of the set of all binary sequences.

The axiom of choice can be formulated as the statement that for any setA there is a mapping from the collection of all nonempty subsets of A into Awhich associates to every B " A with B $= ! an element b of A such that b # B.As another common formulation, suppose that I is a nonempty set and that for

20

every i # I, Ai is a nonempty set. That is to say, {Ai}i$I is a nonempty familyof nonempty sets. The axiom of choice asserts that there is a function f on Iwith values in

$i$I Ai such that f(i) # Ai for every i # I. The generalized

Cartesian product%

i$I Ai of the Ai’s is customarily defined as the set of suchfunctions f , and the axiom of choice is then the claim that this generalizedCartesian product is nonempty when I and every Ai, i # I, are nonempty.

15.2 Condensation points

Let X be a topological space, and suppose that E " X and p # X . Let ussay that p is a strong limit point of E if for every open set U " X with p # Uthere are infinitely many elements of E in U . In particular, there is a q # E %Usuch that q $= p under these conditions, which implies that a strong limit pointis automatically a limit point in the usual sense. Conversely, if X satisfies thefirst separation condition, then one can check that an ordinary limit point isautomatically a strong limit point. In any topological space X , the set of stronglimit points of a set E " X is a closed set.

Similarly, a point p # X is a condensation point of a set E " X if for everyopen set U " X with p # U , E %U is uncountable. Hence a condensation pointof E is a strong limit point. Again one can check that the set of condensationpoints of E is a closed set in X .

16 Bases

Let X be a topological space. A collection B of open subsets of X is said tobe a base for the topology of X if every open set V " X can be expressed as aunion of open subsets of X in B. Equivalently, B is a base for the topology of Xif for every open set V " X and every x # V there is a U # B such that x # Uand U " V . This is the same as saying that for every p # X , the collectionB(p) of U # B such that p # U is a local base for the topology of X at p. Forexample, the collection of open intervals (a, b) with a, b # R, a < b, is a basefor the standard topology of the real line. If X is any set, then the collection ofone-element subsets {x}, x # X , is a base for the discrete topology on X . If Mis a metric space, then the collection of open balls in M with arbitrary centersand radii form a base for the topology associated to the metric.

The existence of a base for the topology with only finitely or countably manyelements is a very interesting condition on a topological space. For instance,the collection of open intervals (a, b) with a, b # Q, a < b, forms a countablebase for the standard topology on the real line. More generally, suppose that(M, d(x, y)) is a metric space and that {xj}#j=1 is a sequence of points in M suchthat the set E consisting of all of the xj ’s, j # Z+, is dense in M . In this eventthe family B(xj , 1/l), j, l # Z+, of open balls forms a base for the topologyof M , which can be rearranged into a ordinary sequence using the countabilityof Z+ - Z+. Conversely, let X be any topological space with a base B for itstopology. For every U # B, pick a point x(U) # U , and let E be the set of x(U),

21

U # B. Because B is a base for the topology of X , E is dense in X . If B hasonly finitely or countably many elements, then E has only finitely or countablymany elements.

16.1 Totally bounded sets

Let (M, d(x, y)) be a metric space. A set E " M is totally bounded if for every$ > 0 there is a finite set A$ " M such that

E "&

x$A!

B(x, $).(16.1)

Thus a totally bounded set is bounded, since the union of finitely many boundedsubsets of M is bounded. Suppose that A$ " M satisfies (16.1). For everyx # A$ with B(x, $) % E $= !, choose y(x) # E such that d(y(x), x) < $. Let !A$

be the set of y(x)’s obtained in this way. By construction, !A$ " E and

E "&

y$!A!

B(y, 2 $).

Clearly the number of elements of !A$ is less than or equal to the number ofelements of A$ when A$ has only finitely many elements. Therefore, withoutloss of generality, we may suppose that A$ " E in the definition of a totallybounded set. If M is totally bounded, then M is separable, in the sense thatthere is a dense set A " M with only finitely or countably many elements, i.e.,A =

$#

n=1 A1/n.

17 More examples

The exotic topologies "+, "! on the real line can be characterized by the basesB+, B! for their topologies, respectively, where B+ consists of the intervals ofthe form [a, b) and B! consists of the intervals of the form (a, b], with a, b # R

and a < b. It has been mentioned previously that the rational numbers aredense in the real line with respect to each of these exotic topologies, and thateach of these topologies has a countable local base for the topology at everypoint. If B"

+ is any base for the topology of R equipped with the topology "+,then for every x # R there is an element of B"

+ which contains x and is containedin [x,++). If x, y # R and x $= y, then it is easy to see that any two elementsof B"

+ corresponding to x, y in this way have to be di"erent. It follows that anybase for the topology of R equipped with "+ has to be uncountable, using thefact that the real line is uncountable. Similarly, any base for the R equippedwith "! has to be uncountable. Consequently, there are no metrics on R forwhich the associated topologies are "+ or "!, since a metric space that containsa countable dense set has a countable base for the topology.

Each interval [a, b) in R is an open set with respect to the topology "+, andit is also closed, because its complement ((+, a) ' [b, ++) in the real line is

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an open set in the topology "+. If x # R, E " R is closed with respect to thetopology "+, and x $# E, then there is a t # R such that [x, t) " R\E. Hence(R, "+) is a regular topological space, since [x, t), ((+, x)' [t, ++) are disjointopen sets in the topology that contain x, E, respectively. For the same reasons,(R, "!) is a regular topological space.

18 Stronger topologies

Suppose that "1, "2 are topologies on a set X such that "2 contains "1. Thus theopen subsets of X with respect to "1 are also open subsets of X with respectto "2. In this event one may say that "2 is a stronger topology on X than"1. If p # X is a limit point of E " X with respect to "2, then p is also alimit point of E with respect to "1. A sequence or net of elements of X thatconverges with respect to "2 automatically converges with respect to "1, andwith the same limit. If X satisfies the first separation condition with respect to"1, then X automatically satisfies the first separation condition with respect to"2. The analogous statement works for the Hausdor" property too. The sametype of argument does not work for regularity, since a stronger topology hasmore closed sets as well.

19 Normality

A topological space X is said to normal if it satisfies the first separation condi-tion and for every pair E1, E2 of disjoint closed subsets of X there are disjointopen subsets U1, U2 of X such that E1 " U1 and E2 " U2. Because of thefirst separation condition, which implies that subsets of X with one element areclosed, a normal topological space is Hausdor" and regular. If (M, d(x, y)) is ametric space, then M is normal with respect to the topology associated to themetric. For let a pair E1, E2 of disjoint closed subsets of M be given. For everyx # E1, choose a positive real number r1(x) such that B(x, r1(x)) " M\E2,which we can do because x # M\E2 and M\E2 is an open set. Similarly, forevery y # E2, choose an r2(y) > 0 such that B(y, r2(y)) " M\E1. Let U1 bethe union of the open balls B(x, r1(x)/2), x # E1, and let U2 be the union ofthe open balls B(y, r2(y)/2), y # E2. Thus U1, U2 are open subsets of M whichcontain E1, E2, respectively, by construction. Suppose for the sake of a contra-diction that there is a point z # U1 % U2. Hence there are x # E1 and y # E2

such that d(x, z) < r1(x)/2 and d(y, z) < r2(y)/2. By the triangle inequality,

d(x, y) <r1(x) + r2(y)

2.(19.1)

However, r1(x) and r2(y) are each less than or equal to d(x, y), because of theway that they were chosen and the fact that x # E1, y # E2. This contradicts(19.1), and it follows that U1, U2 are disjoint, as desired.

Now consider the real line equipped with the exotic topology "+, in whichU " R is an open set if and only if for every x # U there is a t # R, t > x, such

23

that [x, t) " U . Let E1, E2 be disjoint closed subsets of R with respect to thetopology "+. For every x # E1, there is a t1(x) > x such that [x, t1(x)) " R\E2,since R\E2 is an open set with respect to "+ which contains x. Similarly, forevery y # E2 there is a t2(y) > y such that [y, t2(y)) " R\E1. Let U1 be theunion of the intervals [x, t1(x)), x # E1, and let U2 be the union of the intervals[y, t2(y)), y # E2. By construction, U1, U2 are open subsets of R with respect to"+ which contain E1, E2, respectively. Suppose for the sake of a contradictionthat there is a point z # U1 % U2. Hence there are x # E1 and y # E2 suchthat z is an element of both [x, t1(x)) and [y, t2(y)). Equivalently, x, y & zand z < t1(x), t2(y). If x & y, then y # [x, t1(x)), contradicting the way thatt1(x) was chosen, since y # E2. For the same reasons, y & x is impossible,which implies that U1%U2 = !. This shows that (R, "+) is a normal topologicalspace, and (R, "!) is a normal topological space as well, where "! is the exotictopology generated by intervals of the form (a, b].

For many properties of a topological space, if the property holds for a spaceX , then it also holds for any subspace Y " X equipped with the topologyinduced from the one on X . This is not so clear for normality.

20 Continuous mappings

Suppose that X , Y are topological spaces and that f : X . Y is a functiondefined on X with values in Y , which is the same as a mapping from X to Y .We say that f is continuous at a point p # X if for every open set V " Ywith f(p) # V there is an open set U " X such that p # U and f(x) # V forevery x # U . If f is continuous at p, (A,,) is a directed system, and {xa}a$A

is a net that converges to p in X , then the net {f(x)}a$A converges to f(p) inY . Conversely, if f is not continuous at p, then there is an open set V " Ysuch that f(p) # V and for every open set U " X with p # U there is a pointx(U) # U such that f(x(U)) $# V . The x(U)’s form a net indexed by the opensubsets of X that contain p ordered by reverse inclusion, and this net convergesto p by construction. The f(x(U))’s form a net in Y indexed in the same waythat does not converge to f(p). Therefore f is continuous at p if f maps everyconvergent net in X with limit p to a convergent net in Y with limit f(p).

We say that f is a continuous mapping from X to Y if f is continuous atevery point p # X . If f : X . Y is continuous and W " Y is an open set, then

f!1(W ) = {x # X : f(x) # W}

is an open set in X , because for every x # f!1(W ) there is an open set U inX such that x # U and U " f!1(W ). It is easy to see from the definitionof continuity that every mapping with this property is continuous. For anymapping f : X . Y , one can check that

X\f!1(A) = f!1(Y \A)

for every A " Y , and hence that f is continuous if and only if f!1(A) is closedin X for every closed set A " Y .

24

20.1 Sequential continuity

Let X , Y be topological spaces, and let f be a mapping from X to Y . We saythat f is sequentially continuous at a point p # X if for every sequence {xi}#i=1

of elements of X which converges to p in X , {f(xi)}#i=1 converges to f(p) as asequence in Y . If f is continuous at p in the usual sense, then f is sequentiallycontinuous at p, because sequences are special cases of nets. Conversely, sequen-tial continuity at p implies continuity at p when the domain satisfies the localcountability condition at p. Similarly, f : X . Y is sequentially continuous if itis sequentially continuous at every point p # X . Hence continuous mappings aresequentially continuous. If X satisfies the local countability condition at eachpoint, then every sequentially continuous mapping f : X . Y is continuous.

20.2 Simple examples

Here are some simple examples of continuous mappings f from a topologicalspace X into a topological space Y .

(1) If X is equipped with the discrete topology, then any mapping f fromX into any topological space Y is continuous.

(2) If Y is equipped with the indiscrete topology, then any mapping f fromany topological space X into Y is continuous.

(3) If X = Y is the real line with the standard topology, then continuity isequivalent to the familiar definition in terms of $’s and %’s. Similarly, if X andY are metric spaces, then continuity is equivalent to an analogous definition interms of $’s and %’s.

(4) Constant mappings between arbitrary topological spaces are continuous.(5) If X = Y with the same topology, then the identity mapping f(x) = x

is continuous.(6) If X = Y as sets, and "1, "2 are topologies on X such that "2 " "1, then

the identity mapping is continuous as a mapping from (X, "1) to (X, "2).(7) If X " Y and X is equipped with the topology induced from the one on

Y , then the standard embedding f(x) = x is continuous as a mapping from Xinto Y .

(8) If X is equipped with the indiscrete topology and Y satisfies the firstseparation condition, then every continuous mapping from X into Y is constant.

(9) A mapping f : X . Y is said to be locally constant if for every p # Xthere is an open set U " X such that p # U and f is constant on U . Locallyconstant mappings are always continuous.

(10) If Y is equipped with the discrete topology, then every continuous map-ping from X into Y is locally constant.

21 The product topology

Suppose that (X1, "1), (X2, "2) are topological spaces. Consider the Cartesianproduct X1 - X2, consisting of all ordered pairs (x1, x2) with x1 # X1 andx2 # X2. The product topology "1 - "2 on X1 - X2 can be defined by saying

25

that W " X1 -X2 is an open set if and only if for every (x1, x2) # W there areopen sets U1 " X1, U2 " X2 such that x1 # U1, x2 # U2, and U1 - U2 " W .Equivalently, If U " X1 and V " X2 are open sets, then U - V is an open setin X1 - X2, and products of open sets like these form a basis for the producttopology on X1-X2. If (A,,) is a directed system and {(x1,a, x2,a)}a$A is a netin X1-X2, then this net converges to (x1, x2) # X1-X2 in the product topologyif and only if the nets {x1,a}a$A, {x2,a}a$A converge to x1, x2, respectively, inX1, X2.

Let p1 : X1 - X2 . X1, p2 : X1 - X2 be the two coordinate projectionsassociated to the Cartesian product, which is to say that

p1(x1, x2) = x1, p2(x1, x2) = x2

for every (x1, x2) # X1 -X2. It is easy to see from the definition of the producttopology that p1, p2 are automatically continuous as mappings from X1 - X2

with the product topology to X1, X2 with their given topologies, respectively.In general, if X , Y are topological spaces and f is a mapping from X to Y , thenf : X . Y is an open mapping if

f(U) = {y # Y : y = f(x) for some x # U}

is an open set in Y for every open set U " X . One can check that the coordinatemappings p1, p2 are automatically open mappings too.

21.1 Countable products

If X1, X2, . . ., is a sequence of sets, then their Cartesian product X =%#

i=1 Xi

can be defined as the set of sequences x = {xi}#i=1 such that xi # Xi forevery i ) 1. Suppose that each Xi is equipped with a topology "i too. Thecorresponding product topology " on X is defined by saying that U " X is anopen set if for every x # U there are open sets Ui " Xi, i ) 1, such that xi # Ui

for every i ) 1,%#

i=1 Ui " U , and Ui = Xi for all but finitely many i. Anothertopology " " on X can be defined in the same way but without the restrictionthat Ui = Xi for all but finitely many i, which means that the open subsets ofX with respect to " " are much smaller in general than for the product topology.Observe for instance that if each Xi is a Hausdor" topological space, then theCartesian product X is Hausdor" with respect to either of these topologies.

If Xi satisfies the local countability condition at each of its elements forevery i ) 1, then one can check that the Cartesian product X equipped withthe product topology also satisfies the local countability condition at every point.Similarly, if there is a countable base for the topology of each Xi, then there isa countable base for the Cartesian product equipped with the product topology.Suppose that each Xi is equipped with a metric di(·, ·) which determines thetopology "i. Without loss of generality, we may suppose also that di(p, q) & 1/ifor every p, q # Xi and i ) 1, since otherwise we can replace di(p, q) withmin(di(p, q), 1/i). In this case, one can check that

d(x, y) = sup{di(xi, yi) : i ) 1}

26

is a metric on X for which the corresponding topology is the product topology.

21.2 Arbitrary products

Let I be a set, and suppose that for every i # I we have a set Xi. Let Xbe the generalized Cartesian product

%i$I Xi of the Xi’s, which is to say the

set of functions f defined in I with values in$

i$I Xi such that f(i) # Xi forevery i # I. Suppose that each Xi is equipped with a topology "i. The producttopology on X is defined by saying that U " X is an open set if for everyf # U there are open sets Ui " Xi, i # I, such that f(i) # Ui for every i # I,%

i$I Ui " U , and Ui = X for all but finitely many i # I. The condition thatf(i) # Ui for every i # I is equivalent to f #

%i$I Ui. One can get another

topology with much smaller open subsets in general by allowing the Ui’s to bearbitrary open subsets of the Xi’s, without the requirement that Ui = Xi for allbut finitely many i. As an intermediate definition, one might ask that Ui = Xi

for all but finitely or countably many i # I. For any of these topologies, notefor instance that X is a Hausdor" topological space when Xi is Hausdor" forevery i # I.

22 Subsets of metric spaces

Let (M, d(x, y)) be a metric space. If A " M and r > 0, then put

Ar =&

x$A

B(x, r),

which is the same as the set of y # M for which there is an x # A such thatd(x, y) < r. By construction,

A " Ar

and Ar is an open set in M for every r > 0, being a union of open sets. Observethat

Ar " At

when r < t. Furthermore, '

r>0

Ar = A.

Specifically, y # M is an element of the closure A of A if and only if y is anaccumulation point of A, which means that for every r > 0 there is an x # Asuch that d(x, y) < r, and which is the same as saying that y # Ar for everyr > 0. We can just as well restrict our attention to r > 0 of the form 1/n, wheren is a positive integer, to get that

#'

n=1

A1/n = A.

27

In particular, every closed set in M can be expressed as the intersection ofa sequence of open sets. If X is a topological space which satisfies the firstseparation condition, p # X , and X also satisfies the local countability conditionat p, then {p} can be expressed as the intersection of a sequence of open subsetsof X .

22.1 The Baire category theorem

Let (M, d(x, y)) be a complete metric space, and let U1, U2, . . . be a sequenceof dense open subsets of M . The Baire category theorem states that the inter-section

##

i=1 Ui is dense in M under these conditions. To see this, let x0 # Mand r0 > 0 be given, and consider the closed ball B0 = B(x0, r0). BecauseU1 is a dense open set in M , there is an x1 # U1 and an r1 > 0 such thatB1 = B(x1, r1) " B0 % U1 and r1 & 1. Similarly, there is an x2 # U2

and an r2 > 0 such that B2 = B(x2, r2) " B1 % U2 and r2 & 1/2. In gen-eral, for every positive integer i there is an xi # Ui and an ri > 0 such thatBi = B(xi, ri) " Bi!1 % Ui and ri & 1/i. By construction, {xi}#i=1 is a Cauchysequence in M and therefore converges to some point x # M . Since xi # Bl forevery i ) l and Bl is a closed ball, we get that x # Bl for every l ) 0. Hencex # B0 and x # Ul for every l ) 1, as desired. It follows from the Baire categorytheorem that the set of rational numbers is not the intersection of a sequenceof open subsets of the real line.

22.2 Sequences of open sets

Let B be the set of all binary sequences, and remember that there is a one-to-one correspondence between B and the real line. There is also a one-to-onecorrespondence between B and the set of sequences of integers, or even the setof sequences of elements of B.

If X is a topological space which satisfies the second countability condition,then the collection of open subsets of X has cardinal number less than or equalto that of B. If X is any topological space for which the collection of open setshas cardinality less than or equal to that of B, then the collection of sequences ofopen subsets of X also has cardinality less than or equal to the cardinality of B.Hence the collection of subsets of X which can be expressed as the intersection ofa sequence of open subsets of X has cardinality less than or equal to that of B inthis case. However, the collection of all subsets of X may have larger cardinality.This happens when X is the real line, for instance, since the cardinality of thecollection of all subsets of a set is always strictly larger than the cardinality ofthe set itself.

23 Open sets in R

Let U be a nonempty open set in the real line, with respect to the standardtopology. For every x # U , there are a, b # R such that a < x < b and

28

(a, b) " U , since U is an open set. Moreover, there are extended real numbersa", b" with a" < x < b" which satisfy the following three conditions:

(a", b") " U ;(23.1)

a" = (+ or a" $# U ;(23.2)

b" = ++ or b" $# U.(23.3)

Specifically, a" is the infimum of the a < x such that (a, x) " U , and b" is thesupremum of the b > x such that (x, b) " U . Any two open intervals satisfying(23.1), (23.2), and (23.3) are necessarily the same interval or are disjoint subsetsof the real line. For any such open interval, one can pick a rational number in theinterval, and di"erent intervals are then associated to di"erent rational numbersbecause of the disjointness of the intervals. It follows that every nonempty openset in the real line can be expressed as the union of a finite or countably-infinitecollection of disjoint open intervals, where the intervals may be bounded orunbounded, depending on the situation.

One can consider similar matters for the real line equipped with the exotictopologies "+, "!. If U " R is an open set with respect to "+ and x # U , thenthere is a b > x such that [x, b) " U . By taking the supremum of these b’s,it follows that there is an extended real number b" > x such that [x, b") " Uand either b" = ++ or b" $# U . By taking the infimum of the a & x such that[a, x] " U , one gets an extended real number a" & x such that either a" = (+and ((+, x] " U , or (+ < a" < x, (a", x] " U , and a" $# U , or (+ < a" & x,[a", x] " U , and (c, x] $" U when c < a". There are analogous statements for theexotic topology "! based on intervals of the form (a, b].

23.1 Collections of open sets

If X is a topological space which satisfies the second countability condition, thenthe cardinality of the collection of open subsets of X is less than or equal to thecardinality of the real line, because every open set in X can be described by asequence of elements of a countable base for the topology of X . In particular,the cardinality of the collection of open subsets of the real line is less than orequal to the cardinality of R. Of course the cardinality of R is less than or equalto the cardinality of the set of all open subsets of R, since R\{x} is an open setin R for every x # R. One can also show that the cardinality of the collection ofall open subsets of R is less than or equal to the cardinality of R using the factthat every open set in R is the union of a sequence of open intervals. Rememberthat the cardinality of the real line is the same as that of the set B of binarysequences. Also, the cardinality of B - B is equal to the cardinality of B. Itfollows that the cardinality of the collection of all open intervals in R is lessthan or equal to the cardinality of B. Hence the cardinality of the collectionof all open subsets of R is less than or equal to the cardinality of the set ofsequences of elements of B, and therefore less than or equal to the cardinality ofB. A similar argument can be applied to the real line equipped with the exotictopologies "+ or "!, which do not satisfy the second countability condition.

29

24 Compactness

Let X be a topological space and let E " X be given. By an open covering ofE we mean a family {Ui}i$I of open subsets of X such that

E "&

i$I

Ui.

We say that E is compact in X if for every open covering {Ui}i$I of E in Xthere is an finite subcovering. This means that there are i1, . . . , in # I such that

E " Ui1 ' · · · ' Uin.

If E has only finitely many elements, then E is automatically compact.If Y " X and E " Y , then one can also consider compactness of E relative to

Y , using the topology induced from the one on X . A nice feature of compactnessis that E is compact relative to X if and only if E is compact relative to Y ,basically because every open covering of E relative to X can be converted to anopen covering relative to Y and vice-versa. Suppose that "1, "2 are topologieson X such that the open subsets of X with respect to "1 are also open withrespect to "2. If E " X is compact with respect to "2, then E is compact withrespect to "1 as well, because every open covering of E with respect to "1 is anopen covering with respect to "2. If X is equipped with the discrete topologyand E " X is compact, then E has only finitely many elements.

24.1 Compactness and convergent sequences

Let X be a topological space, and let {xl}#l=1 be a sequence of elements of Xthat converges to x # X . Let K be the set consisting of the xl’s, l ) 1, and x.It is easy to see that K is a compact set in X . For suppose that {Ui}i$I is anarbitrary covering of K by open subsets of X . Since x # K, there is an i0 # Isuch that x # Ui0 . Because {xl}#l=1 converges to x in X , there is an L ) 1 suchthat xl # Ui0 when l ) L. If we choose i1, . . . , iL!1 # I such that xl # Uil

forl = 1, . . . , L ( 1, then K " Ui0 ' Ui1 ' · · · ' UiL"1

, as desired. Suppose nowthat X satisfies the local countability condition at every point. In this case, aset E " X is closed in X if and only if for every sequence {xl}#l=1 of elementsof E which converges to a point x # X we have that x # E. It follows fromthe previous remarks that E " X is a closed set in X if and only if E % K isrelatively closed in K for every compact set K " X . Consequently, U " X isan open set in X if and only if U %K is relatively open in K for every compactset K " X .

25 Properties of compact sets

Let X be a topological space, let K1, K2 be compact subsets of X , and let usshow that their union K1 'K2 is a compact set in X . Let {Ui}i$I be any open

30

covering of K1 ' K2 in X . Hence {Ui}i$I is an open covering of K1 and K2

individually, and it follows that there are finite subsets I1, I2 of I such that

K1 "&

i$I1

Ui, K2 "&

i$I2

Ui.

Therefore I1 ' I2 " I has only finitely many elements and

K1 ' K2 "&

i$I1%I2

Ui.

Thus there is a finite subcovering of K1 ' K2 from the open covering {Ui}i$I ,and it follows that K1 ' K2 is compact.

If K " X is compact and E " X is closed, then E % K is compact. Forif {Ui}i$I is any open covering of E % K in X , then the Ui’s together withX\E form an open covering of K, which implies that K is contained in theunion of finitely many Ui’s and X\E. Clearly E % K can be covered by thesame collection of finitely many Ui’s, and therefore E % K is compact. If X isHausdor" and K " X is compact, then K is a closed set in X . To see this, letp # X\K be given, and for every q # K let Uq, Vq be disjoint open subsets ofX such that p # Uq and q # Vq. Since K is compact, there are finitely manyelements q1, . . . , qn of K such that K is contained in the union of Vq1

, . . . , Vqn. If

U is the intersection of Uq1, . . . , Uqn

, then U is an open set in X such that p # Uand U " X\K, and thus p is not an accumulation point of K, and K is closed.If X is an infinite set equipped with the topology in which W " X is an openset exactly when W is the empty set or X\W has only finitely many elements,then X satisfies the first separation axiom and every E " X is compact.

25.1 The limit point property

If X is a topological space, then E " X has the limit point property if everyinfinite set A " E has a limit point in E. Similarly, E " X has the strong limitpoint property if every infinite set A " E has a strong limit point in E. Thestrong limit point property implies the limit point property, and the converseholds when X satisfies the first separation condition, since every limit point isthen a strong limit point. These two conditions are quite similar to compactness.For instance, if E has only finitely many elements, then E automatically hasthe strong limit point property. If E " Y " X , then E satisfies one of theseconditions relative to Y if and only if it satisfies the same condition relativeto X . If E1, E2 " X satisfy the limit point property or the strong limit pointproperty, then E1 ' E2 has the same feature. For if A " E1 ' E2 has infinitelymany elements, then at least one of A%E1, A%E2 has infinitely many elements,and hence a limit point or strong limit point in E1 or E2. The intersection ofa set which satisfies either of these versions of the limit point property with aclosed set in X also satisfies the same version of the limit point property, sincea closed set contains all of its limit points and therefore all limit points of itssubsets. If K " X is compact, then K satisfies the strong limit point property.

31

For if A " K is an infinite set without a strong limit point in K, then for everyp # K there is an open set U(p) " X such that p # U(p) and A has onlyfinitely many elements in U(p). Compactness implies that there are finitelymany elements p1, . . . , pn of K such that K " U(p1) ' · · · ' U(pn). Hence Ahas only finitely many elements, a contradiction. As a partial converse to thisstatement, suppose that E " X has the strong limit point property and thatV1, V2, . . . is a sequence of open subsets of X such that E "

$#

i=1 Vi. If for everyn ) 1 there is an xn # E\

$ni=1 Vi, then the set A " E of xn’s has infinitely

many elements. A strong limit point p # E of A would be in Vl for some l,which would imply that xn # Vl for some n > l, a contradiction. It follows thatE "

$ni=1 Vi for some n.

26 Countable compactness

Let X be a topological space. We say that E " X is countably compact if forevery open covering {Ui}i$I of E in X there is a set I0 " I of only finitely orcountably many indices such that

E "&

i$I0

Ui.

Thus compact sets and countably infinite sets are automatically countably com-pact. If X is an uncountable set equipped with the topology where W " X is anopen set exactly when W is the empty set or X\W has only finitely or count-ably many elements, then every E " X is countably compact. If E1, E2, . . .is a sequence of countably compact subsets of X , then their union

$#

l=1 El iscountably compact too. As in the case of compact sets, if Y " X and E " Y ,then E is countably compact relative to X if and only if E is countably com-pact relative to Y . If "1, "2 are topologies on X such that every open set in Xwith respect to "1 is an open set with respect to "2, and if E " X is countablycompact with respect to "2, then E is automatically countably compact withrespect to "1.

Suppose that X is a topological space, and that B is a base for the topologyof X . Let {Ui}i$I be any collection of open subsets of X , and let B" be the setof V # B such that V " Ui for some i # I. For every V # B", choose i(V ) # Isuch that V " Ui(V ), and let I0 " I be the set of i(V ), V # B". Since B is abase for the topology of X , each Ui is the union of the V # B such that V " Ui,and it follows that &

i$I0

Ui =&

i$I

Ui.

If B has only finitely or countably many elements, then the same holds for B",and hence for I0, and therefore every E " X is countably compact.

32

26.1 Related results

If X is a topological space, E " X is countably compact, and A " E is un-countable, then there is a condensation point p of A in E. This is analogous tothe fact that compactness implies the strong limit point property. We have seenpreviously that if E " X has the strong limit point property, then any coveringof E by a sequence of open sets in X can be reduced to a finite subcovering.Therefore, if E has the strong limit point property and is countably compact,then E is compact.

Suppose now that (M, d(x, y)) is a metric space. If E " M is compact,then E is totally bounded. For every $ > 0, the open balls B(x, $), x # E,cover E, and compactness implies that E can be covered by B(x, $) for finitelymany x # E, as desired. If E is countably compact, then the same argumentimplies that for every $ > 0, E can be covered by finitely or countably manyballs B(x, $), x # E. In particular, it follows that a countably compact metricspace is separable, and hence has a base for its topology with only finitely orcountably many elements. If E " M has the limit point property, then E istotally bounded. Otherwise, there is an $ > 0 such that E cannot be coveredby finitely many balls with radius $, and this implies that there is an infinitesequence x1, x2, . . . of elements of E such that d(xi, xl) ) $ when i < l. Theset A " E of xi’s then has infinitely many elements and no limit points, acontradiction. If M has the limit point property, then M is totally bounded,hence separable, hence countably compact, and consequently M is compact. IfE " M has the limit point property, then E is compact, since compactness andthe limit point property are preserved when one switches between a topologicalspace and an intermediate set equipped with the induced topology.

26.2 Additional related results

In general, for any topological space X , if B is a base for the topology of X ,Y " X , and BY consists of the sets of the form U %Y , U # B, then BY is a basefor the topology on Y induced from the topology on X . If there is a base for thetopology of X with only finitely or countably many elements, then it follows thatthe same holds for Y " X equipped with the topology induced from the one onX . For a metric space M , the following are equivalent: (i) M is separable, i.e.,there is a dense set A " M with only finitely or countably many elements; (ii)for every $ > 0, M can be covered by finitely or countably many balls of radius$; (iii) there is a base B of M with only finitely or countably many elements;and (iv) M is countably compact. Clearly (ii) implies that every E " Msatisfies the analogous property, and the remarks at the beginning say exactlythat (iii) implies the analogous condition for arbitrary subsets of a topologicalspace. Therefore separability of a metric space M implies that for every E " Mthere is a set AE " E with only finitely or countably many elements which isrelatively dense in E, or, equivalently, whose closure in M contains E. Also,countable compactness of a metric space M implies countable compactness ofevery E " M . Consequently, if M is a metric space and E1, E2 are subsets

33

of M such that E1 " E2 and E2 is countably compact, then E1 is countablycompact.

27 Continuity and compactness

Suppose that X and Y are topological spaces, and that f is a continuous map-ping from X to Y . If E " X is compact, then

f(E) = {y # Y : y = f(x) for some x # E}

is compact in Y . To see this, let {Vi}i$I be any open covering of f(E) in Y .For every i # I, put

Ui = f!1(Vi) = {x # X : f(x) # Vi}.

Because f : X . Y is continuous, Ui is an open set in X for every i # I. Thecompactness of E in X implies that there are finitely many indices i1, . . . , in # Isuch that

E " Ui1 ' · · · ' Uin.

This implies in turn that

f(E) " Vi1 ' · · · ' Vin.

This uses the observation that

f(f!1(A)) " A

for every A " Y . Hence we get a finite subcovering of f(E) from the initialopen covering {Vi}i$I in Y , and it follows that f(E) is compact in Y . Similarly,if E " X is countably compact, then f(E) is countably compact in Y .

28 Characterizations of compactness

Let X be a topological space, and suppose that K " X is compact. If {Ei}i$I

is a family of closed subsets of X such that

K % Ei1 % Ei2 % · · · % Ein$= !(28.1)

for every collection i1, . . . , in of finitely many indices in I, then

K %'

i$I

Ei $= !.(28.2)

Otherwise, the complements X\Ei of the Ei’s in X form an open covering ofK with no finite subcovering. Conversely, if K has this property, then one cancheck that K is compact, by considering the complements of the sets in an opencovering of K.

34

Suppose that K " X and that {Ei}i$I is a family of closed subsets of Xfor which (28.1) holds for any i1, . . . , in # I. Let A be the collection of finitenonempty subsets of I, ordered by inclusion. For every A # A, let xA be anelement of K %

#i$A Ei, which exists, by hypothesis. Thus {xA}A$A is a net

of elements of K indexed by A as a directed system. If there is an x # K suchthat for every open set U " X with x # U and every A # A there is an A" # Asuch that A " A" and xA! # U , then x # Ei for every i # I, because x is anaccumulation point of each Ei, and (28.2) holds.

Conversely, let K be a compact set in X , let (A,,) be a directed system,and let {xa}a$A be a net of elements of K. For every a # A, let Ea be theclosure in X of the set of xb with b # A and a , b. If a1, . . . , an # A, thenthere is an a # A such that ai , a for i = 1, . . . , n, and thus Ea " Eai

for eachi. Hence K % Ea1

% · · · % Eancontains K % Ea and is nonempty in particular.

Because K is compact, there is an x # K such that x # Ea for every a # A, andthus for every open set U " X with x # U and every a # A there is a b # Asuch that a , b and xb # U , as desired.

28.1 Sequential compactness

Let X be a topological space, and let E " X be given. We say that E issequentially compact if for every sequence {xi}#i=1 of elements of E there is asubsequence {xil

}#l=1 of {xi}#i=1 and an x # E such that {xil}#l=1 converges to

x. If E is sequentially compact, then E has the strong limit point property.For if A " E has infinitely many elements, then one can apply sequential com-pactness to a sequence of distinct elements of A, and a limit of a subsequenceof this sequence is a strong limit point of A. Conversely, if X satisfies the localcountability condition at every point and E " X has the strong limit pointproperty, then E is sequentially compact. To see this, let {xi}#i=1 be a sequenceof elements of E. If there is an x # E such that x = xi for infinitely manypositive integers i, then {xi}#i=1 has a constant subsequence which convergesto x trivially. Otherwise, the set A " E consisting of the xi’s has infinitelymany elements, and there is a strong limit point x # E of A. Using the localcountability condition for X at x, one can show that there is a subsequence of{xi}#i=1 that converges to x, as desired.

If E " Y " X , then E is sequentially compact relative to Y if and only ifE is sequentially compact with respect to X . If E1, E2 " X are sequentiallycompact, then E1 ' E2 is sequentially compact, because any sequence of ele-ments of E1 'E2 has a subsequence whose terms are contained in E1 or in E2.The intersection of a sequentially compact set with a sequentially closed set issequentially compact. If X is Hausdor", then sequentially compact subsets ofX are sequentially closed, because of the uniqueness of limits of convergent se-quences. If X , Y are topological spaces, f : X . Y is sequentially continuous,and E " X is sequentially compact, then f(E) is sequentially compact in Y .

35

28.2 Some variants

Let (M, d(x, y)) be a metric space. If {xi}#i=1 is a Cauchy sequence in M witha convergent subsequence {xil

}#l=1, then {xi}#i=1 converges to the same limit.Hence every Cauchy sequence in a sequentially compact set in M converges.

Let us say that E " M has property (/) if for every set A " E with infinitelymany elements and every $ > 0 there is a bounded set B " A with infinitelymany elements such that diamB < $. Equivalently, E has property (/) if forevery sequence of elements of E and every $ > 0 there is a subsequence of thesequence whose terms are contained in a bounded set with diameter less than $.If E is totally bounded, then E has property (/), and the converse holds by thesame argument used to show that a set with the limit point property is totallybounded. If every sequence in E has a subsequence which is a Cauchy sequence,then E has property (/). Conversely, suppose that E " M has property (/),and that a sequence of elements of E is given. Using property (/), one canget a subsequence of the sequence whose terms are contained in a bounded setwith diameter less than 1. By repeating the process, one can get a subsequenceof the subsequence whose terms are contained in a bounded set with diameterless than 1/2. For each positive integer n, one can apply property (/) to get asubsequence of the initial sequence which is also a subsequence of the previoussubsequence when n ) 2 whose terms are contained in a bounded set withdiameter less than 1/n. The sequence whose nth term is the nth term of thenth subsequence obtained in this way is a subsequence of the initial sequenceand is a Cauchy sequence. Therefore, E " M is totally bounded if and only ifevery sequence of elements of E has a subsequence which is a Cauchy sequence.If M is complete, then it follows that E " M is sequentially compact if andonly if E is closed in M and totally bounded.

29 Products of compact sets

Let X be a topological space, and suppose that B is a collection of open subsetsof X which forms a base for the topology of X . If K " X is compact withrespect to B, in the sense that every covering of K by elements of B can bereduced to a finite subcovering, then K is compact in the usual sense. For let{Ui}i$I be an arbitrary open covering of K in X , and let B" be the set of V # Bfor which there is an i # I such that V " Ui. Because B is a basis for thetopology of X , each Ui is equal to the union of the V # B such that V " Ui.Therefore K "

$i$I Ui =

$V $B! V , which is to say that the V ’s in B" cover

K. By hypothesis, there are V1, . . . , Vn # B" which cover K. The definition ofB" implies that there are i1, . . . , in # I such that Vl " Uil

for l = 1, . . . , n, andhence that Ui1 , . . . , Uin

cover K. This shows that K is compact. There is ananalogous statement for countable compactness, and this argument is actuallythe same in essence as the one used to show that every E " X is countablycompact if there is a countable base for the topology of X .

Now let X1, X2 be topological spaces, and consider the Cartesian product

36

X1 - X2 with the product topology. If K1, K2 are compact subsets of X1, X2,respectively, then K1-K2 is compact in X1-X2. To show this, it is enough toshow that every covering {Ui - Vi}i$I of K1 - K2 by products of open sets inX1, X2 can be reduced to a finite subcovering, by the remarks in the precedingparagraph. For every x # K1, let I(x) be the set of i # I such that x # Ui.Since {x} - K2 " K1 - K2, {Vi}i$I(x) is an open covering of K2 in X2. Thecompactness of K2 in X2 implies that there is a finite set I0(x) " I(x) such thatK2 is covered by Vi, i # I0(x). If U(x) is the intersection of Ui, i # I0(x), thenU(x) is an open set in X1 which contains x. The compactness of K1 implies thatthere are x1, . . . , xn # K1 such that K1 is covered by the U(xi)’s, 1 & i & n.Because U(x)-K2 is covered by Ui - Vi, i # I0(x), for every x # K1, it followsthat K1-K2 is covered by the Ui-Vi’s with i # I0(x1)' · · ·'I0(xn), as desired.

29.1 Sequences of subsequences

Let X1, X2, . . . be a sequence of topological spaces, and let X =%#

i=1 Xi betheir Cartesian product. Suppose that Ei " Xi, i ) 1, are sequentially com-pact, and consider their Cartesian product E =

%#

i=1 Ei. We would like toshow that E is sequentially compact in X when X is equipped with the producttopology. It follows from the definition of the product topology that a sequenceof elements of X converges to an element of X if and only if the correspondingsequence of coordinates in Xi to converges to the ith coordinate of the limit forevery i ) 1. Suppose that a sequence of elements of E is given. Because E1

is sequentially compact, there is a subsequence of this sequence whose coordi-nates in X1 converge to an element of E1. Similarly, because E2 is sequentiallycompact, there is a subsequence of the subsequence for which the coordinatesin X2 converge to an element of E2. Proceeding in this way, we get a sequenceof subsequences of the initial sequence, where each new subsequence is a subse-quence of the preceding subsequence, and where the coordinates in Xn of thenth subsequence converge to an element of En for every n ) 1. If we form asequence in which the nth term is equal to the nth term of the nth subsequence,then this sequence is a subsequence of the initial sequence which converges inX to an element of E.

29.2 Compactness of closed intervals

Let B be the set of all binary sequences, i.e., the set of sequences x = {xi}#i=1

such that xi = 0 or 1 for each i. Equivalently, B is the Cartesian product%#

i=1 Bi, with Bi = {0, 1} for each i. Using the product topology, based on thediscrete topology on the Bi’s, B becomes a Hausdor" topological space which issequentially compact, since the Bi’s are sequentially compact trivially. It followsthat B is a compact space, using the fact that the topology on B is determinedby a metric, or by observing that B has a countable base for its topology andtherefore is countably compact.

Consider the real-valued function f on B defined by f(x) =(#

i=1 xi 2!i, asusual. One can check that f : B . R is a continuous mapping. Consequently,

37

the closed unit interval [0, 1] is a compact set in the real line.

30 Filters

Let X be a set. By a filter on X we mean a nonempty collection F of nonemptysubsets of X such that E1, E2 # F implies that E1 % E2 # F and E " !E " Xand E # F imply that !E # F . For example, if x # X , then the collection of allsubsets of X that contain x is a filter. If F1, F2 are filters on X and F1 " F2,then we say that F2 is a refinement of F1. If X is a topological space, F is afilter on X , and x # X , then we say that F converges to x if F contains everyopen set U " X with x # U . Thus the filter consisting of all subsets of Xthat contain x converges to x automatically. The filter consisting of all subsetsof X that contain an open set which contains x also converges to x. If X isHausdor", then every convergent filter in X has a unique limit. If E " X andF is a filter on X which contains E and converges to a point x # X , then x isan accumulation point of E.

If (A,,) is a directed system and {xa}a$A is a net of elements of X , thenthere is a filter associated to the net in a natural way, consisting of the subsetsE of X for which there is a c # A such that xa # E for every a # A with c , a.In the case where X is a topological space, the net converges to a point x # X ifand only if the corresponding filter converges to x. Conversely, if F is a filter onX , then we can think of F as a directed system with respect to reverse inclusion,i.e., E1 , E2 when E1, E2 # F and E2 " E1. By picking elements from thesubsets of X in F , we get nets indexed by the filter as a directed system. If Xis a topological space and x # X , then F converges to x as a filter if and onlyif all of the corresponding nets also converge to x.

Let X be a topological space, and suppose that F is a filter on X , x # X ,and x is an element of the closure E of every E # F . Let F1 be the collectionof subsets of X which contain a set of the form E % U , where E # F , U " Xis an open set, and x # U . Under these conditions, F1 is a filter on X whichis a refinement of F and which converges to x. Conversely, if a filter F has arefinement which converges to x # X , then x # E for every E # F . Suppose nowthat K " X is compact and that F is a filter on X such that K # F . Considerthe collection of E, E # F . If E1, . . . , En # F , then E1 % · · ·%En %K # F andhence is nonempty. In particular, E1% · · ·%En %K $= !. The compactness of Kimplies that there is an x # K such that x # E for every E # F . Equivalently,there is a refinement of F that converges to a point x # K. Conversely, supposethat K " X and that for every filter F on X with K # F there is a point x # Kand a refinement of F that converges to x. Let {Ei}i$I be a family of closedsubsets of K such that Ei1 % · · · % Ein

% K $= ! for every collection i1, . . . , inof finitely many elements of I. Let F be the collection of all subsets of X thatcontain a set of the form Ei1 % · · ·%Ein

%K for some i1, . . . , in # I. It is easy tosee that F is a filter on X and that K # F . By hypothesis, there is a refinementof F that converges to a point x # K. Because Ei # F and Ei is a closed setsfor every i # I, one can check that x # Ei for every i # I. It follows that K is

38

a compact set in X , as desired.

31 Mappings and filters

Let X , Y be sets. Suppose that f is a mapping from X to Y , and that F is afilter on X . If f&(F) is the collection of subsets of Y which contain a set of theform f(E), E # F , where f(E) is the set of y # Y such that y = f(x) for somex # E, as usual, then f&(F) is a filter on Y . If X , Y are topological spaces,f : X . Y is continuous, x # X , and F converges to x in X , then one cancheck that f&(F) converges to f(x) in Y . One can also check that continuity ischaracterized by this property.

If X , Y are topological spaces, f : X . Y is continuous, and K " X iscompact, then we have seen that f(K) is compact in Y , as a consequence of thedefinition of compactness in terms of open coverings. One can look at this interms of nets as well, i.e., let (A,,) be a directed system, and let {ya}a$A be anet of elements of f(K) indexed by A. For every a # A, let xa be an element ofK such that f(xa) = ya, and consider the net {xa}a$A. By compactness, thereis an x # K such that for every open set U " X with x # U and every a # Athere is a b # A which satisfies a , b and xb # U . The continuity of f impliesthat for every open set V " Y with f(x) # V and every a # A there is a b # Asuch that a , b and f(xb) # V . There is an analogous argument in terms offilters too. Specifically, let F be a filter on Y such that f(K) # F . If F(K)is the collection of subsets of X which contain a set of the form f!1(E) % K,E # F , where f!1(E) consists of the x # X with f(x) # E, then F(K) is afilter on X which satisfies K # F(K). Because K is compact, there is an x # Kand a filter F(K)" on X which is a refinement of F(K) and which convergesto x. It follows that f&(F(K)") is a filter on Y which is a refinement of F andwhich converges to f(x), as desired.

32 Ultrafilters

Let X be a set and let F be a filter on X . We say that F is an ultrafilter ifit is maximal in the sense that any filter on X which is a refinement of F isequal to F . For example, if p # X , then the collection Up of subsets of X thatcontain p as an element is an ultrafilter on X . Ultrafilters of this form are saidto be principal ultrafilters. If F is a filter on X , A " X , and A % E $= ! forevery E # F , then the collection FA of subsets of X which contain a set of theform A % E, E # F , is a filter on X which is a refinement of F . Observe thatA # FA automatically, and hence FA = F implies that A # F . Conversely, ifA # F , then FA = F . If F is an ultrafilter on X , then it follows that FA = Funder these conditions. That is to say, if F is an ultrafilter on X , A " X , andA % E $= ! for every E # F , then A # F . For any filter F on X , if A " X andthere is an E # F such that A%E = !, then E " X\A and therefore X\A # F .Consequently, if F is an ultrafilter on X and A " X , then either A # F or

39

X\A # F . Now suppose that F ,F " are filters on X such that F " is a refinementof F . If A # F " and E # F , then A % E # F ", and thus A % E $= !. Hence ifF has a refinement which is di"erent from itself, then there is an A " X suchthat A $# F and A%E $= ! for every E # F . It follows that if F is a filter on Xand for every A " X either A # F or X\A # F , then F is an ultrafilter on X .

Using the axiom of choice, through the Hausdor" maximality principle orZorn’s lemma, one can show that for every filter F on X there is an ultrafilterU on X which is a refinement of F . For suppose that C is a nonempty chain offilters on X , which is to say a nonempty collection of filters on X such that forevery F1,F2 # C, either F2 is a refinement of F1 or the other way around. Let)F be the union of the filters in C, which means that E " X is an element of )Fif and only if E is an element of one of the filters in C. One can check that )F isa filter on X , and this permits one to use the Hausdor" maximality principle orZorn’s lemma to show that every filter has a refinement which is an ultrafilter.If X is a topological space and K " X , then we have seen that K is compactif and only if every filter F on X such that K # F has a refinement whichconverges to a point x # K. If U is an ultrafilter on X , K " X is compact, andK # U , then U converges to a point in K. Conversely, suppose that K " X andevery ultrafilter U on X with K # U converges to a point in K. If F is a filteron X with K # F , then the axiom of choice implies that there is an ultrafilterU which is a refinement of F . By hypothesis, U converges to a point in K, andhence K is compact.

33 Continuous real-valued functions

If X , Y are topological spaces, then the space of continuous mappings from Xto Y is denoted C(X, Y ). In the special case where Y is the real line with thestandard topology, we may use the abbreviation C(X) instead of C(X,R).

Suppose that X is a topological space, p # X , and f1, f2 are real-valuedfunctions on X which are continuous at p, and let us check that the sum f1 + f2

is continuous at p too. Let $ > 0 be given, and let U1, U2 be open subsets of Xsuch that p # U1, U2,

|f1(x) ( f1(p)| <$

2

for every x # U1, and

|f2(x) ( f2(y)| <$

2

for every x # U2. By the triangle inequality for the absolute value function,

|(f1(x) + f2(x)) ( (f1(p) + f2(p))| & |f1(x) ( f1(p)| + |f2(x) ( f2(p)| < $

for every x # U1 %U2. It follows that f1 + f2 is continuous at p, since U1 %U2 isan open set in X that contains p. Similarly, one can show that the product f1 f2

is continuous at p. It is a bit simpler to start with the special cases where f1 orf2 is a constant function, and where f1(p) = 0 or f2(p) = 0. The general casecan then be derived from these two special cases using the previous assertion for

40

sums. As a consequence, C(X) is a commutative ring with respect to pointwiseaddition and multiplication, since the sum and product of two continuous real-valued functions on X is a continuous function on X .

34 Compositions and inverses

Suppose that X1, X2, and X3 are topological spaces, f1 is a mapping from X1

to X2, and f2 is a mapping from X2 to X3. As usual, the composition f2 0 f1

is the mapping from X1 to X3 given by (f2 0 f1)(x) = f2(f1(x)). If p # X1,f1 is continuous at p, and f2 is continuous at f1(p), then it is easy to see thatf2 0 f1 is continuous at p. Consequently, f1 # C(X1, X2) and f2 # C(X2, X3)imply that f2 0 f1 # C(X1, X3), which one can also check by showing that(f2 0 f1)!1(W ) = f!1

1 (f!12 (W )) is an open set in X1 whenever W is an open

set in X3.Let X , Y be sets, and let IX , Iy be the identity mappings on X and y,

which is to say that IX(x) = x for every x # X and IY (y) = y for every y # Y .A mapping f : X . Y is invertible if there is a mapping f!1 : Y . X suchthat f!1 0 f = IX and f 0 f!1 = IY . Because f!1 0 f = IX , f is a one-to-one mapping from X to Y , or equivalently f : X . Y is an injection, whichmeans that f(x) = f(x") implies x = x" for every x, x" # X . Similarly, becausef 0f!1 = IY , f maps X onto Y , or equivalently f : X . Y is a surjection, whichmeans that for every y # Y there is an x # X such that f(x) = y. Conversely,if f is a one-to-one mapping from X onto Y , which is the same as saying that fis a one-to-one correspondence from X onto Y or that f : X . Y is a bijection,then f is invertible, and f!1 : Y . X can be defined by f!1(y) = x whenf(x) = y for x # X , y # Y . Observe in particular that the inverse f!1 of aninvertible mapping f is unique. If X1, X2, and X3 are sets and f1 : X1 . X2,f2 . X3 are invertible mappings, then one can check that f2 0 f1 : X1 . X3 isinvertible, and that (f20f1)!1 = f!1

1 0f!12 . If X , Y are topological spaces, then

a homeomorphism from X onto Y is an invertible mapping f : X . Y whichis continuous and whose inverse f!1 : Y . X is continuous. If X1, X2, X3

are topological spaces and f1 : X1 . X2, f2 : X2 . X3 are homeomorphisms,then f2 0 f1 : X1 . X3 is a homeomorphism too. Of course the inverse of ahomeomorphism is a homeomorphism as well.

35 Disjoint compact sets

Let X be a Hausdor" topological space. If p # X , K " X is compact, andp # X\K, then, as in the proof of the statement that compact subsets ofHausdor" spaces are closed, for every q # K there are disjoint open subsetsUq, Vq of X such that p # Uq and q # Vq. Because K is compact, there areq1, . . . , qn # K such that K "

$ni=1 Vqi

. If we put U =#n

i=1 Uqi, then U is an

open set in X such that p # U and U " X\K, as in the previous argument.If we also put V =

$ni=1 Vqi

, then we get the stronger statement that V is an

41

open set in X , K " V , and U %V = !, i.e., X behaves like a regular topologicalspace with respect to compact subsets instead of closed subsets. Suppose nextthat K1, K2 are disjoint compact subsets of X . For every z # K1, there aredisjoint open subsets W1(z), W2(z) of X such that z # W1(z) and K2 " W2(z),as in the preceding remarks. The compactness of K1 implies that there arez1, . . . , zr # K1 such that K1 "

$rl=1 W1(zl). Hence

$rl=1 W1(zl),

#rl=1 W2(zl)

are disjoint open subsets of X that contain K1, K2, respectively, which meansthat X behaves like a normal topological with respect to compact sets. Inparticular, a compact Hausdor" topological space is normal. Similarly, if X isregular and E, K are disjoint subsets of X with E closed and K compact, thenE, K are contained in disjoint open subsets of X .

36 Local compactness

A topological space X is said to be locally compact if for every p # X thereis an open set U " X such that p # U and U is contained in a compact setin X . Suppose that X is locally compact and that K " X is compact. Forevery p # K, there is an open set U(p) " X such that p # U(p) and U(p)is contained in a compact set. Because K is compact, there are finitely manyelements p1, . . . , pn of K such that K is contained in U =

$ni=1 U(pi). Of course

U is an open set in X , and U is contained in a compact set since the union offinitely many compact sets is a compact set.

If X is Hausdor", then X is locally compact if and only if for every p # Xthere is an open set U " X such that p # U and U is compact. Suppose that Xis a locally compact Hausdor" topological space, p # X , and U " X is an openset such that p # U and U is compact. Observe that 'U = U\U is a compactset in X , since it is the intersection of the compact set U with the closed setX\U . Because p # U " X\'U and X is Hausdor", there are disjoint opensubsets V , W of X such that p # V and 'U " W . Hence U1 = U %V is an openset in X , p # U , and U1 " U , which implies that U1 is compact. We also havethat U1 " X\W and hence U1 " X\W , since W is an open set. It follows thatU1 " U , since U1 " U and 'U " W , by construction. If U0 is any open set inX such that p # U0, there is an open set U " X such that p # U , U " U0, andU is compact, i.e., the intersection of U0 with any open set that contains p andhas compact closure. By the previous argument, there is an open set U1 " Xsuch that p # U1, U1 " U " U0, and U1 is compact. In particular, a locallycompact Hausdor" topological space is regular, since one can take U0 = X\Ewhen E " X is closed and p $# E, and then U1, X\U1 are disjoint open subsetsof X that contain p, E, respectively.

37 Localized separation conditions

Let X be a topological space. The Hausdor" property is equivalent to askingthat for every p, q # X with p $= q there is an open set U " X such that p # U

42

and q is not an element of the closure of U . Similarly, X is regular if and onlyif X satisfies the first separation condition and for every p # X and open setV " X with p # V there is an open set W " X such that p # W and the closureof W is contained in V . Also, normality is the same as saying that X satisfiesthe first separation condition and for every A, V " X with A closed, V open,and A " V there is an open set W " X such that A " W and W " V . If Xis regular and K, V are subsets of X with K compact, V open, and K " V ,then there is an open set W " X such that K " W and W " V , by the usualcovering arguments.

Suppose now that X is a locally compact Hausdor" topological space. Inthis case, X is regular, and every compact set in X is contained in an openset with compact closure. Using the observation at the end of the precedingparagraph, it follows that if K " X is compact and V " X is an open set withK " V , then there is an open set W " X such that K " W , W is compact, andW " V . If K1, K2 are disjoint compact subsets of X , then there are disjointopen sets V1, V2 " X such that Ki " Vi for i = 1, 2, as in any Hausdor" space.Hence there are open sets W1, W2 " X such that Ki " Wi for i = 1, 2 and W1,W2 are disjoint compact subsets of X .

38 !-Compactness

Let X be a topological space. A set E " X is said to be !-compact if thereis a sequence of compact subsets K1, K2, . . . of X such that E =

$#

i=1 Ki. Acompact set K is automatically !-compact, using Ki = K for every i ) 1. A!-compact set is countably compact, and the union of a sequence of !-compactsets is !-compact. If X is !-compact, X =

$#

i=1 Ki for some compact sets Ki,and E " X is closed, then E is !-compact, because E %Ki is compact for eachi and E =

$#

i=1(E %Ki). More generally, if X is !-compact and E is the unionof a sequence of closed sets, then E is !-compact. Consequently, if the topologyon X is determined by a metric, X is !-compact, and U " X is an open set,then U is !-compact, since every open set in a metric space can be expressedas the union of a sequence of closed sets.

Suppose now that X is a locally compact topological space which is !-compact, with X =

$#

i=1 Ki and Ki compact for every i ) 1. By local com-

pactness, for every i ) 1 there is an open set Ui " X and a compact set !Ki " Xsuch that Ki " Ui " !Ki. For convenience, put Vl =

$li=1 Ui, )Kl =

$li=1

!Ki for

every l ) 1. Hence the Vl’s are open subsets of X , the )Kl’s are compact subsetsof X , Vl " )Kl, Vl " Vl+1, and )Kl " )Kl+1 for every l ) 1, and

X =#&

l=1

Vl =#&

l=1

)Kl.

The Vl’s form an open covering of X , and therefore every compact set in X iscontained in Vn for some positive integer n, depending on the compact set. Ifone also assumes that X is Hausdor", then one can take !Ki = Ui for each i.

43

In this event one can make additional adjustments to the construction to get asequence of open subsets W1, W2, . . . of X such that X =

$#

j=1 Wj and Wj is acompact set contained in Wj+1 for every j ) 1. Note that a topological spaceis !-compact when it is locally compact and countably compact.

38.1 Subsets of R

Consider the real line with the standard topology. Of course R is !-compact,since closed and bounded intervals are compact. It follows that R is countablycompact, which one could also get from the fact that the open intervals (a, b)with a < b being rational numbers is a countable base for the topology of R.If the set of irrational numbers were !-compact, then it would be the unionof a sequence of closed subsets of R in particular. This would imply that theset of rational numbers is the intersection of a sequence of open subsets ofR, contradicting the Baire category theorem. However, the set of irrationalnumbers is countably compact. More precisely, every E " R is countablycompact, since there is a countable base for the topology of the real line.

39 Topological manifolds

Fix a positive integer n, and let Rn be the space of n-tuples of real numbers,as usual. Hence the elements of Rn are of the form x = (x1, x2, . . . , xn), whereeach coordinate xi of x, 1 & i & n, is a real number. Equivalently, Rn is theCartesian product of n copies of the real line, which leads to a natural topologyon Rn, based on the standard topology on the real line. This topology is alsothe one determined by the standard Euclidean metric on Rn.

A topological space X is said to be an n-dimensional topological manifold ifit satisfies the following three conditions. First, we ask that X be a Hausdor"topological space. Second, X is supposed to be locally Euclidean in the sensethat for every p # X there is an open set U " X such that p # U and U ishomeomorphic to an open set in Rn, with respect to the standard topology onRn, as in the previous paragraph. Third, we ask that X be !-compact, whichbasically means that X is not too big. The first two conditions imply thatX is locally compact, because closed and bounded subsets of Rn are compact.Therefore !-compactness is equivalent to countable compactness in this context.Clearly Rn is an n-dimensional topological manifold. Similarly, every open setin Rn is an n-dimensional topological manifold with the topology induced fromthe one on Rn. The unit sphere in Rn+1 is a compact n-dimensional topologicalmanifold. In general, topological manifolds have a lot of interesting features.

40 !-Compactness and normality

Let X be a locally compact Hausdor" topological space which is !-compact, andlet W1, W2, . . . be a sequence of open subsets of X such that Wj is a compactset contained in Wj+1 for every j ) 1 and X =

$#

j=1 Wj . Suppose that A, B

44

are disjoint closed subsets of X , and put Aj = A % Wj , Bj = B % Wj for everypositive integer j. Hence Aj , Bj are disjoint compact subsets of Wj+1 with

Aj " A " X\B and Bj " B " X\A

for each j. Because A1, B1 are disjoint compact subsets of X and X is Hausdor",there are disjoint open subsets U1, V1 of X such that A1 " U1, B1 " V1. Wemay as well suppose that U1 " W2%(X\B) and V1 " W2%(X\A), by replacingU1, V1 with their intersections with W2 % (X\B), W2 % (X\A), respectively.By shrinking U1, V1 further, using the hypothesis that X is a locally compactHausdor" topological space, we may suppose that U1, V1 are disjoint compactsubsets of X which satisfy U1 " W2 % (X\B), V1 " W2 % (X\B). In general,we would like to choose disjoint open subsets Uj , Vj of X recursively in such away that Aj " Uj , Bj " Vj , Uj!1 " Uj and Vj!1 " Vj when j ) 2,

Uj " Wj+1 % (X\B), Vj " Wj+1 % (X\A),

and Uj , Vj are disjoint compact sets for every positive integer j. If j ) 2 andwe have chosen Uj!1, Vj!1 in this manner, then

!Aj = Aj ' Uj!1, !Bj = Bj ' Vj!1

are disjoint compact subsets of X such that

!Aj " Wj+1 % (X\B), !Bj " Wj+1 % (X\A),

which permits us to repeat the process. It follows that U =$#

j=1 Uj, V =$#

j=1 Vj are disjoint open subsets of X which contain A, B, respectively. There-fore X is normal.

41 Unions of bases

Let X be a topological space, and suppose that {Ui}i$I is a collection of opensubsets of X such that X =

$i$I Ui. If for every i # I there is a collection Bi of

open subsets of Ui which form a base for the topology of Ui, then B =$

i$I Bi

is a base for the topology of X . For let p # X and an open set V " X withp # V be given. By hypothesis, p # Ui for some i # I, and hence there is anopen set W # Bi such that p # W and W " Ui % V , since Bi is a base for thetopology of Ui. In particular, W # B, p # W , and W " V , as desired.

Suppose now that X is an n-dimensional topological manifold. For everyopen set U " X which is homeomorphic to an open set in Rn, there is acollection BU of countably many open subsets of U which forms a base for thetopology of U . This follows from the fact that there is a countable base for thetopology of Rn, given by Cartesian products of open intervals in the real linewith rational endpoints, for instance. Countable compactness of X implies thatX can be covered by a family of finitely or countably many such open subsetsU , and thus there is a countable base for the topology of X , by the remarks ofthe previous paragraph.

45

42 Separating points

Let X be a topological space, and let C(X) be the space of continuous real-valued functions on X , as usual. We say that C(X) separates points in X iffor every p, q # X with p $= q there is an f # C(X) such that f(p) $= f(q). IfC(X) separates points in X , then X has to be a Hausdor" topological space.For suppose that p, q # X and f(p) $= f(q). If f(p) < f(q), then there is anr # R such that f(p) < r < f(q). In this case, U = {x # X : f(x) < r}and V = {x # X : f(x) > r} are disjoint open subsets of X containing p andq, respectively. The case where f(p) > f(q) can be handled similarly. On thereal line, the continuous function f(x) = x is su!cient to separate points. Forany positive integer n, the n standard coordinate functions on Rn, which sendx = (x1, . . . , xn) to xi, 1 & i & n, separate points. If (M, d(x, y)) is a metricspace, then one can use the triangle inequality to check that fp(x) = d(x, p) isa continuous function on M , and it is easy to see that these functions separatepoints on M . For any set X , the indicator function 1A associated to a setA " X is defined on X by putting 1A(x) equal to 1 when x # X and to 0 whenx # X\A. An indicator function 1A on a topological space X is a continuousfunction if and only if A is both open and closed in X , and in some situationsthese functions separate points in X . Suppose now that X is an n-dimensionaltopological manifold for some positive integer n, and let W " X be an openset which is homeomorphic to an open set W " " Rn. Let (" be a continuousreal-valued function on Rn for which the set of x # Rn with ("(x) $= 0 iscontained in a compact set K " with K " " W ". We can restrict (" to W " anduse the homeomorphism between W and W " to get a continuous real-valuedfunction ( on W corresponding to (" and a compact set K " W correspondingto K " such that K contains every x # W with ((x) $= 0. If we put ((x) = 0when x # X\W , then ( becomes a continuous real-valued function on X . Inparticular, one can use this construction to show that continuous real-valuedfunctions separate points on a topological manifold.

43 Urysohn’s lemma

Let X be a topological space. If f is a continuous real-valued function on Xand a, c are real numbers such that a < c, then the sets of x # X such thatf(x) & a, f(x) ) c are disjoint closed subsets of X . If moreover b # R anda < b < c, then the sets of x # X such that f(x) < b, f(x) > b are disjoint opensubsets of X which contain the previous closed sets, respectively.

Conversely, suppose that X is normal, and that E0, E1 are disjoint closedsubsets of X . Let {ri}#i=1 be an enumeration of the rational numbers r such that0 < r < 1, i.e., ri # Q% (0, 1) for every i ) 1, and every element of Q% (0, 1) isequal to ri for exactly one i. Because X is normal, there is an open set Ur1

" Xsuch that E0 " Ur1

and Ur1" X\E1. Similarly, there is an open set Ur2

" Xsuch that E0 " Ur2

, Ur2" X\E1, and Ur1

" Ur2if r1 < r2, Ur2

" Ur1if r1 > r2.

Continuing in this way, we get an open set Ur " X for every r # Q% (0, 1) such

46

that E0 " Ur, Ur " X\E1, and Ur " Ut when r < t. Consider the real-valuedfunction defined on X by putting f(x) equal to the infimum of r # Q % (0, 1)such that x # Ur when there is such an r, and f(x) = 1 otherwise. Equivalently,f(x) is equal to the supremum of r # Q% (0, 1) such that x # X\Ur when thereis such an r, and f(x) = 0 otherwise. By construction, f(x) = 0 when x # E0,f(x) = 1 when x # E1, and 0 & f(x) & 1 for every x # X . One can checkthat f is a continuous function on X , using the fact that Ur, X\Ur are opensubsets of X for every r # Q % (0, 1). If X is a locally compact Hausdor"topological space, K " X is compact, W " X is an open set, and K " W , thenone can use analogous arguments to show that there is a continuous real-valuedfunction h on X such that h(x) = 1 when x # K, h(x) = 0 when x # X\W ,and 0 & h(x) & 1 for every x # X .

43.1 R\{0}Consider the real line with the standard topology. The standard topology on Ris determined by the standard metric, and in particular R is a normal topologicalspace. The set R\{0} of nonzero real numbers is a topological space too, withthe topology induced by the standard topology on the real line. This topologyis also determined by the restriction of the standard metric to R\{0}, and thusR\{0} is normal. As subsets of the real line, the sets of positive and negativereal numbers are open sets. As subsets of the set of nonzero real numbers, thesesets are both open and closed. The function equal to 0 on the negative realnumbers and equal to 1 on the positive real numbers is a continuous functionon the set of nonzero real numbers. Of course this function is not the restrictionto the nonzero real numbers of a continuous function on the real line.

44 One-point compactification

If X is a locally compact Hausdor" topological space which is not compact,then the one-point compactification of X is the topological space X& defined asfollows. As a set, X& consists of the elements of X together with an additionalpoint p&, the point at infinity. The open subsets of X are also considered tobe open subsets of X&, and conversely they are the only open subsets of X&

contained in X , i.e., which do not contain p&. If V " X& and p& # V , then Vis considered to be an open set in X& if and only if V % X is an open set in Xand there is a compact set K " X such that X\K " V % X . It is easy to seethat this defines a topology on X&. Specifically, the intersection of finitely manyopen subsets of X& that contain p& is an open set in X& because the union offinitely many compact subsets of X is a compact set in X . By construction,X is a dense open set in X&. One can check that X& is Hausdor" using thehypothesis that X is locally compact. If {Vi}i$I is any open covering of X&,then there is an i& # I such that p& # Vi# , and thus there is a compact setK " X such that X\K " Vi# % X . Since {Vi % X}i$I is an open covering ofX , and hence of K, there is a collection i1, . . . , in of finitely many elements of I

47

such that K "$n

l=1(Vil%X). It follows that X& " Vi# '

$nl=1 Vil

, and thereforeX& is compact.

45 Supports of continuous functions

Let X be a topological space, and let C(X) be the space of continuous real-valued functions on X , as usual. The support of f # C(X) is denoted supp fand defined to be the closure of the set of x # X such that f(x) $= 0. Iff1, f2 # C(X), then

supp f1 + f2 " (supp f1) ' (supp f2)

andsupp f1 f2 " (supp f1) % (supp f2).

Let us specialize to the situation where X is a locally compact Hausdor" topo-logical space. The space of continuous real-valued functions on X with compactsupport is denoted Ccom(X). If f1, f1 # Ccom(X), then f1 + f2 # Ccom(X), bythe previous observation. Similarly, if f1, f2 # C(X) and at least one of f1, f2

has compact support in X , then the product f1 f2 also has compact support.Let us say that f # C(X) vanishes at infinity if for every $ > 0 there is a com-pact set K " X such that |f(x)| < $ when x # X\K. A continuous functionwith compact support automatically vanishes at infinity. The space of contin-uous real-valued functions on X which vanish at infinity is denoted C0(X). Iff1, f2 # C0(X), then it is not too di!cult to show that f1 + f2 # C0(X). Iff1 # C0(X), and if f2 # C(X) is bounded, which is to say that there is anA ) 0 such that |f2(x)| & A for every x # X , then f1 f2 # C0(X). Of courseCcom(X) = C0(X) = C(X) when X is compact. Suppose that X is not compact,and let X& be the one-point compactification of X . If f # C0(X), then f extendsto a continuous function on X& which is equal to 0 at the point p& at infinity,and conversely the restriction of any such function on X& to X lies in C0(X).In the same way, continuous functions on X with compact support correspondto continuous functions on X& which are equal to 0 on a neighborhood of p&.

46 Connectedness

46.1 Connected topological spaces

A topological space X is not connected if there are nonempty disjoint opensubsets U1, U2 of X such that

U1 ' U2 = X.

Equivalently, X is not connected if there are nonempty disjoint closed subsetsE1, E2 of X such that

E1 ' E1 = X.

48

This is also the same as having a set E " X which is both open and closedsuch that E $= !, X . Otherwise, if X does not have any of these equivalentproperties, then X is connected.

For example, if X is equipped with the discrete topology and X has at leasttwo elements, then X is not connected. If X is equipped with the indiscretetopology, then X is connected. If X is an infinite set with the topology in whichU " X is an open set when U = ! or X\U has only finite many elements, thenX is connected, because

U1 % U2 $= !for any nonempty open subsets U1, U2 of X .

The real line with the standard topology is connected. For suppose that E1,E2 are nonempty disjoint closed subsets of R whose union is R, and let x1, x2

be elements of E1, E2, respectively. We may as well suppose that x1 < x2, sinceotherwise we can relabel the indices. One can check that

sup(E1 % [x1, x2]) # E1 % E2,

contradicting the disjointness of E1 and E2.

46.2 Connected sets

Let X and Y be topological spaces, and suppose that f : X . Y is a continuousmapping such that f(X) = Y . If Y is not connected, then there are disjointnonempty open subsets V1, V2 of Y such that V1'V2 = Y . This implies that X isnot connected, because f!1(V1), f!1(V2) are disjoint nonempty open subsets ofX whose union is equal to X . This is the same as saying that the connectednessof X implies the connectedness of Y under these conditions.

A set E in a topological space X is said to be connected if it is connectedas a topological space itself, using the topology induced from the one on X . Amore direct characterization can be given as follows. A pair of subsets A, B ofX are said to be separated if

A % B = A % B = !,

where A and B are the closures of A and B in X . One can check that E " Xis not connected if and only if E is the union of a pair of nonempty separatedsubsets of X . This uses the fact that the closure of A " E relative to E is equalto the intersection of E with the closure of A relative to X . In particular, apair of sets A, B " E is separated relative to E if and only if they are separatedrelative to X . It is easy to see that a pair of disjoint closed subsets of a topo-logical space are separated, as well as a pair of disjoint open sets. Conversely,if a topological space is the union of a pair of separated sets, then the two setsare both open and closed.

If f : X . Y is continuous and E " X is connected, then f(E) is connectedin Y .

A set E in the real line with the standard topology is connected if and onlyif for every x, y # E with x < y, the interval [x, y] is contained in E.

49

46.3 Other properties

Let X be a topological space. If E1, E2 are connected subsets of X and E1%E2 isnot empty, then E1'E2 is also connected. For suppose that there are nonemptyseparated sets A, B of X such that

A ' B = E1 ' E2.

It is easy to see that

A1 = A % E1, B1 = B % E1

andA2 = A % E2, B2 = B % E2

are pairs of separated sets in X . The remaining point is that A1, B1 $= ! orA2, B2 $= !.

Similarly, if E is a connected set in X , then one can show that the closureE of E in X is connected. Otherwise, there are nonempty separated sets A, Bin X such that A'B = E. If A0 = A%E and B0 = B %E, then A0 and B0 areseparated sets in X too. Again the remaining point is to check that A0, B0 $= !under these conditions.

If E " X has the property that for every p, q # E there is a connected setE(p, q) " E such that p, q # E(p, q), then E is connected. As usual, if E is notconnected, then E is the union of a pair of nonempty separated sets A, B in X .Let p, q be elements of A, B, respectively. It is easy to see that

A % E(p, q), B % E(p, q)

are nonempty separated sets whose union is E(p, q), contradicting the connect-edness of E(p, q).

46.4 Pathwise connectedness

A set E in a topological space X is said to be pathwise connected if for everyp, q # E there are real numbers a, b with a & b and a continuous mappingf : [a, b] . X such that f(a) = p, f(b) = q, and f([a, b]) " E. Here [a, b] is theclosed interval in the real line with endpoints a, b, equipped with the topologyinduced by the standard topology on R. In particular, [a, b] is connected, andso

E(p, q) = f([a, b])

is connected. It follows from a previous observation that E is connected whenE is pathwise connected.

There are famous examples of closed sets in the plane that are connectedbut not pathwise connected. These examples also show that the closure of apathwise connected set may not be pathwise connected. Specifically,

E = {(x, y) # R2 : x > 0, y = sin(1/x)}

50

is pathwise connected, while

E = E ' {(0, y) # R2 : (1 & y & 1}

is not, because there is no continuous path in E that connects any point in Eto any point of the form (0, y), (1 & y & 1. Instead of the sine function, similarexamples could be based on any nonconstant continuous periodic real-valuedfunction on the real line.

46.5 Additional properties and examples

Let a, b, and c be real numbers with a & b & c, and let (, ) be continuousmappings from the closed intervals [a, b], [b, c] into a topological space X , re-spectively. If ((b) = )(b), then it is easy to see that the function on [a, c] equalto ( on [a, b] and to ) on [b, c] is also continuous as a mapping from [a, c] intoX . In particular, the union of two pathwise connected subsets of a topologicalspace is pathwise connected when they have a point in common, just as forconnected sets.

Remember that a set E " Rn is said to be convex if for every x, y # E, theline segment in Rn connecting x and y is contained in E. Thus convex sets arealways pathwise connected. Of course, a ball in Rn with respect to the standardEuclidean metric is convex. Subsets of Rn which are “rectangular boxes” in thesense that the are Cartesian products of intervals are also convex.

Let U be an open set in Rn with the standard topology. For each x # U ,let U(x) be the set of y # U for which there is a continuous path in U thatconnects x to y, and let us check that U(x) is also an open set in Rn. For anyy # U , there is a convex open set V in Rn such that y # V and V " U . Ify # U(x) and z # V , then z # U(x), because a continuous path from x to y inU can be continued to z along the line segment from y to z. Hence V " U(x)when y # U(x), which shows that U(x) is an open set. Similarly, if z # U(x) forany z # V , then y # U(x), and in fact w # U(x) for every w # V . This impliesthat U(x) is relatively closed in U . Equivalently, U\U(x) is an open set, sinceU is open. Note that U(x) is automatically pathwise connected, by the previousobservation about combinations of paths. If U is not pathwise connected, thenU $= U(x), and it follows from the preceding remarks that U is not connected.Hence U is pathwise connected when U is a connected open set in Rn.

46.6 Local connectedness

Let X be a connected topological space. If p, q # X are su!ciently close toeach other, then one might expect that there would be a connected set E(p, q)in X which contains p, q and stays close to p, q. Of course, one should be moreprecise about what exactly this might mean. For instance, one could ask thatfor each p # X and each open set U in X with p # U there be an open set V inX with p # V such that for every q # V there is a connected set E(p, q) " Uwith p, q # E(p, q). One can also ask that V be connected, so that one may

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simply take E(p, q) = V for every q # V . These local connectedness conditionsmake sense even if X is not connected, and hold for example when X is anyopen set in Rn with the induced topology. However, there are connected subsetsof the plane that are not locally connected in this way, including closed sets. Ifthe topology on X is determined by a metric, then one can consider uniformversions of local connectedness too. Instead of looking at neighborhoods of afixed point p, one can ask that for every $ > 0 there be a % > 0 such that everypair of elements of X at distance less than % from each other are contained ina connected set of diameter less than $. There are open sets in Rn that do nothave this property, including connected open sets. Convex sets automaticallysatisfy this property with % = $.

46.7 Local pathwise connectedness

A topological space X is said to be locally pathwise connected if for every p # Xand every open set U in X with p # U there is an open set V in X such thatp # V and every q # V can be connected to p by a continuous path in U . Forexample, Rn satisfies this property with respect to the standard topology. If Xis locally pathwise connected, then every connected open set in X is pathwiseconnected. This follows from essentially the same argument as when X = Rn.Note that if ( : [a, b] . X is a continuous path with values in U that connects pand q as before, then E(p, q) = (([a, b]) is a connected set in X which containsp, q and is contained in U . Thus local pathwise connectedness automaticallyimplies a version of local connectedness. A uniform version of local pathwiseconnectedness for metric spaces asks that for every $ > 0 there be a % > 0 suchthat any pair of points at distance less than % from each other can be connectedby a continuous path with values in a set of diameter less than $. This holdsfor Rn with % = $, as well as for convex subsets of R. However, there areopen subsets of Rn that do not satisfy this uniform version of local pathwiseconnectedness, even though they are automatically locally pathwise connected.This includes connected open subsets of Rn, which are also pathwise connected.

47 A little set theory

47.1 Mappings and their properties

Let A, B be sets, and let f be a function on A with values in B, also known as amapping from A into B. We say that f is one-to-one or injective if f(a) = f(a")implies that a = a" for every a, a" # A. We say that f maps A onto B, or thatf is a surjection, if for every b # B there is an a # A such that f(a) = b. If f isa one-to-one mapping of A onto B, then we say that f is a bijection, or that fis a one-to-one correspondence between A and B. In general, f(A) denotes theset of b # B for which there is an a # A such that f(a) = b.

Suppose that A, B, C are sets and f : A . B and g : B . C are mappingsbetween them. In this case, the composition g 0 f of f and g is the mapping

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from A to C defined by(g 0 f)(a) = g(f(a))

for every a # A. It is easy to see that the composition of two injections is aninjection, and that the composition of two surjections is a surjection. Hencethe composition of two bijections is a bijection as well. If f : A . B is abijection, then the inverse mapping f!1 : B . A is defined by f!1(b) = awhen f(a) = b. The inverse mapping is a bijection whose inverse is f . Byconstruction, f!1 0 f is the identity mapping on A, and f 0 f!1 is the identitymapping on B. Conversely, if there is a mapping f!1 : B . A such that f!1 0fis the identity mapping on A and f 0 f!1 is the identity mapping on B, thenf is a bijection, and f!1 is the same as the inverse mapping just described. Iff : A . B and g : B . C are both bijections, then it is easy to see that

(g 0 f)!1 = f!1 0 g!1.

47.2 One-to-one correspondences

Let A, B be sets. If there is a one-to-one correspondence from A onto B, thenwe may express this formally by

#A = #B.

Of course, #A = #A automatically, because the identity mapping on A is aone-to-one correspondence of A to itself. Also, #A = #B is equivalent to#B = #A, because the inverse of a bijection is a bijection. Similarly, #A = #Band #B = #C imply #A = #C for any sets A, B, and C, since the compositionof two bijections is a bijection. Note that A is a finite set with exactly n elementsfor some positive integer n if and only if #A = #{1, . . . , n}. A set A is countablyinfinite if and only if #A = #Z+, where Z+ denotes the set of positive integers.

If B is the set of infinite binary sequences and [0, 1] is the unit interval inthe real line, then #B = #[0, 1]. To see this, let f be the standard mappingfrom B onto [0, 1], which sends a binary sequence b = {bi}#i=1 to the real numberf(b) =

(#

i=1 bi 2!i. For each positive integer n, let An be the finite set of b # Bsuch that bi = 0 for every i ) n or bi = 1 for every i ) n. If A =

$#

n=1 An, thenA is countably infinite. Let E be the set of rational numbers r # [0, 1] such thatr is an integer multiple of 2!l for some nonnegative integer l. The restriction off to B\A is a one-to-one mapping onto [0, 1]\E. The restriction of f to A mapsonto E but is not one-to-one. However, A and E are both countably infinite,and so there is a one-to-one correspondence between them. Combining this withthe restriction of f to B\A leads to a one-to-one correspondence from B onto[0, 1], as desired.

47.3 One-to-one mappings

If there is a one-to-one mapping from a set A into a set B, then this may beexpressed formally by #A & #B. Thus #A = #B implies #A & #B and

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#B & #A. If A, B, C are sets such that #A & #B and #B & #C, then#A & #C, since the composition of two injections is also an injection.

Suppose that A, B are nonempty sets, and that f is a one-to-one mappingfrom A into B. Fix * # A, and let g : B . A be the mapping defined byg(b) = a when a # A and b = f(a) and g(b) = * when b # B\f(A). Clearly g isa surjection. Hence #A & #B implies that there is a mapping from B onto A.Conversely, suppose that there is a mapping g from B onto A. Assuming theaxiom of choice, there is a mapping f : A . B such that g 0 f is the identitymapping on A. Therefore f is one-to-one.

If A, B are sets such that #A & #B and #B & #A, then #A = #B.Without loss of generality, we may suppose that B " A, since #B & #A meansthat there is a one-to-one correspondence between B and a subset of A. In thiscase, #A & #B implies that there is an injective mapping f : A . A such thatf(A) " B. Let A1, A2, . . . be the sequence of subsets of A defined by A1 = f(A)and Al+1 = f(Al) for l ) 1. One can use induction to show that Al+1 " Al foreach l. If C =

##

l=1 Al, then it is easy to check also that f(C) = C. BecauseC " B, it su!ces to show that there is a one-to-one correspondence from A\Conto B\C. Put E1 = A\f(A1) and El = Al\Al+1 when l ) 1, and observe thatf(El) = El+1 for each l. By construction, the El’s are pairwise disjoint, and$#

l=1 El = A\C. For each x # E1, let O(x) be the orbit of x with respect tof , which is to say the set consisting of x, f(x), f(f(x)), etc. These orbits arepairwise disjoint, and their union is A\C. We would like to show that there isa one-to-one mapping from each orbit O(x) onto O(x) % B. All but possiblythe initial element x of O(x) is automatically contained in B, since f(A) " B.It follows that either the identity mapping or a simple shift is a one-to-onemapping of O(x) onto O(x) % B, as desired.

47.4 Some basic properties

Let A, B, C be sets, with B " A and A % C = !. If B is countably infiniteand C has only finitely or countably many elements, then B ' C is countablyinfinite. One can use this to show that #(A ' C) = #A. For example, one canapply this to check that

#(0, 1) = #[0, 1) = #(0, 1] = #[0, 1].

Also, f(x) = 1/2 + x/(1 + 2 |x|) is a one-to-one mapping from the real line Ronto the open interval (0, 1), and hence #R = #(0, 1). Remember that the setQ of rational numbers is countable. As another example, one can check that#(R\Q) = #R.

In general, if A, C are sets and #C & #A, then there is an argument basedon the axiom of choice which implies that #(A ' C) = #A. Another argumentbased on the axiom of choice implies that #A & #B or #B & #A for any pairof sets A, B.

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47.5 Bases of topological spaces

Let (X, ") be a topological space which has a countable base for its topology.Thus there is a sequence U1, U2, . . . of open subsets of X such that every openset in X can be expressed as a union of some collection of the Uj’s. Of course,the union of any collection of the Uj ’s is an open set in X , since each Uj is open.Let B be the set of all binary sequences, and consider the mapping ( : B . "which sends a binary sequence b = {bj}#j=1 to the open set in X that is theunion of the Uj’s such that bj = 1. If bj = 0 for each j, then we interpret ((b)as being the empty set. Because the Uj’s form a base for the topology of X , (maps B onto the collection " of all open subsets of X . Similarly, let ) : " . Bbe the mapping that sends an open set V in X to the binary sequence which isequal to 1 for each positive integer j such that Uj " V and equal to 0 otherwise.Because the Uj ’s form a base for the topology of X , each open set V in X isequal to the union of the Uj’s such that Uj " V . Equivalently, (()(V )) = Vfor each open set V in X , which implies that ) is a one-to-one mapping of "into B. Consequently, #" & #B.

Suppose now that (X, ") is the real line with the standard topology. Foreach real number t, the set ((+, t) of r # R with r < t is an open set inR. This defines a one-to-one mapping of R into " . Hence #R & #" , andtherefore #" = #R, since the open intervals in R with rational endpoints forma countable base for the topology.

47.6 Exponentials

Let A, B, be sets. The set of all functions on A with values in B is sometimesdenoted BA. This is the same as the Cartesian product of a family of copiesof B, with one copy of B for each element of A. This may also be denoted 2A

when B is the set consisting of exactly the two elements 0 and 1. The set ofall subsets of A may be identified with 2A, where a set E " A corresponds tothe function on A which is equal to 1 at elements of E and to 0 at elementsof A\E. If A and B are finite sets, then BA is a finite set. In this case, thenumber of elements of BA is the number of elements of B to the power whichis the number of elements of A.

47.7 Properties of exponentials

Let A, B, and C be sets. Remember that the Cartesian product A-B consistsof the ordered pairs (a, b) with a # A and b # B. There is a natural one-to-onecorrespondence between the set CA'B of mappings from A-B into C and theset (CB)A of mappings from A into the set of mappings from B into C. For iff(a, b) is a function on A-B with values in C, then for each a # A we can viewfa(b) = f(a, b) as a function on B with values in C. Equivalently, a 1. fa is amapping from A into CB . Thus we get a mapping from CA'B into (CB)A. Itis easy to see that this is a one-to-one correspondence, as desired. If A and B

55

are countably infinite, then A - B is countably infinite, and

#(CB)A = #CA'B = #CZ+ .

By definition, the set B of binary sequences is the same as 2Z+ . Considerthe set Z

Z+

+ of all sequences of positive integers. One can check that

#B = #2Z+ & #ZZ+

+ & #BZ+ ,

since 2 & #Z+ & #B. However, the remarks in the previous paragraph imply

that #BZ+ = #(2Z+)Z+ = #2Z+'Z+ = #2Z+ = #B. Therefore #ZZ+

+ = #B.

47.8 Countable dense sets

Suppose that X is a Hausdor" topological space which satisfies the local count-ability condition at each point and contains a countable dense set E. Underthese conditions, for each x # X there is a sequence {xl}#l=1 of elements of Ethat converges to x. Because X is Hausdor", the limit of any convergent se-quence of elements of X is unique. Let B be the set of all binary sequences, andlet C be the set of convergent sequences of elements of E. Thus C " EZ+ , andhence #C & #EZ+ = #B. There is a natural mapping from C into X , whichsends a convergent sequence to its limit in X . By hypothesis, this is a mappingfrom C onto X . Therefore #X & #C & #B.

If X is the real line with the standard topology, then X satisfies the condi-tions described in the previous paragraph, and #X = #B.

47.9 Additional properties of exponentials

Let A, B, and C be sets, with A % B = !. There is a natural one-to-onecorrespondence between the set CA%B of mappings from A ' B into C and theCartesian product CA - CB of the set CA of mappings from A into C withthe set CB of mappings from B into C. Basically, a mapping from A ' B intoC is the same as a combination of a mapping from A into C and a mappingfrom B into C. If A and B are countable sets, then A ' B is countable, and itfollows that #CA = #CB = #CA%B. Thus #CA - CB = #CZ+ , and hence#CZ+ - CZ+ = #CZ+ .

In particular, the set B of all binary sequences is the same as CZ+ withC = {0, 1}. Therefore #B - B = #B.

A Additional homework assignments

Note that the first homework assignment was mentioned in Section 1.

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A.1 Limit points

Let X be a topological space, and for E " X , put

E" = {p # X : p is a limit point of E}.

Discuss the question of whether E" is closed, with examples or general results,as appropriate, e.g., how might this be related to the separation conditions?

A.2 Dense open sets

Let (X, ") be a topological space. Suppose that " " is the collection of opensubsets U of X with respect to " such that either U = ! or U is dense in X withrespect to " . For the homework, please show the following statements. First, " "

also satisfies the requirements of a topology on X . Second, X is automaticallynot Hausdor" with respect to " " when X has at least two elements. Third,suppose that X satisfies the first separation condition with respect to " . Ifevery element of X is a limit point of X with respect to " , then X satisfies thefirst separation condition with respect to " " too.

A.3 Combining topologies

Let X be a set, and suppose that "1, "2 are topologies on X . Let " = "1 2 "2 bethe collection of subsets W of X such that for every x # W there are subsetsU , V of X with x # U , x # V , U # "1, V # "2, and U % V " W . Show thefollowing statements. First, " is a topology on X . Second, "1, "2 " " . Third,if !" is any other topology on X such that "1, "2 " !" , then " " !" . Fourth, ifd1(x, y), d2(x, y) are metrics on X for which the associated topologies are "1,"2, respectively, then d(x, y) = max(d1(x, y), d2(x, y)) defines a metric on X forwhich the associated topology is " .

A.4 Complements of countable sets

Let X be a set. Let us say that U " X is an open set if U = ! or X\U has onlyfinitely or countably many elements. Show that this defines a topology on X .Of course, this is analogous to the situation mentioned previously in which theopen sets are the empty set and the sets whose complements have only finitelymany elements. Discuss some of the properties of this topology. For instance,under what conditions is X a Hausdor" space? Which sets E " X are dense inX? When does a sequence of elements of X converge?

A.5 Combining topologies again

Let X be a set equipped with a topology "1, and let "2 be the topology on Xdescribed in Homework # 5 in which a set V " X is an open set if V = !or X\V has only finitely or countable many elements. Consider the topology" = "1 2 "2 on X as defined in Homework # 4. What happens when X has

57

only finitely or countably many elements? In general, when is a point p # X alimit point of a set E " X with respect to "? When is a set E " X dense inX with respect to "? Which sequences in X converge with respect to "? Does(X, ") satisfy the first separation condition? Is (X, ") Hausdor" when (X, "1) isHausdor"? If (X, "1) satisfies the local countability condition at a point p # X ,then under what conditions does (X, ") satisfy the local countability conditionat p?

A.6 Combining topologies on R

Let "1 be the standard topology on the real line, and let "2 be the topology onR defined in Homework # 5. Thus U " R is an open set with respect to "2 ifU = ! or R\U has only finitely or countably many elements. Let " = "1 2 "2 bethe topology on R obtained by combining these two topologies as in Homework# 4. Show that (R, ") is not a regular topological space, so that there is a pointp # R and a closed set E " R with respect to " such that p $# E and U %V $= !for every pair of open subsets U , V of R with respect to " which satisfy p # Uand E " V .

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[2] P. Halmos, Naive Set Theory, Springer-Verlag, 1974.

[3] J. Hocking and G. Young, Topology, 2nd edition, Dover, 1988.

[4] I. James, Introduction to Uniform Spaces, Cambridge University Press,1990.

[5] I. James, Topologies and Uniformities, Springer-Verlag, 1999.

[6] I. Kaplansky, Set Theory and Metric Spaces, 2nd edition, Chelsea, 1977.

[7] J. Kelley, General Topology, Springer-Verlag, 1975.

[8] J. Kelley, I. Namioka, et al., Linear Topological Spaces, Springer-Verlag,1976.

[9] S. Krantz, The Elements of Advanced Mathematics, 2nd edition, Chapman& Hall / CRC, 2002.

[10] S. Krantz, Real Analysis and Foundations, 2nd edition, Chapman & Hall/ CRC, 2005.

[11] S. Krantz, A Guide to Real Variables, Mathematical Association of Amer-ica, 2009.

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[12] S. Krantz, A Guide to Topology, Mathematical Association of America,2009.

[13] S. Krantz, Essentials of Topology with Applications, CRC Press, 2010.

[14] J. Lee, Introduction to Topological Manifolds, Springer-Verlag, 2000.

[15] G. McCarty, Topology: An Introduction with Application to TopologicalGroups, 2nd edition, Dover, 1988.

[16] B. Mendelson, Introduction to Topology, 3rd edition, Dover, 1990.

[17] W. Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill,1976.

[18] L. Steen and J. Seebach, Jr., Counterexamples in Topology, Dover, 1995.

[19] P. Warren, Topological Uniform Structures, Dover, 1988.

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[21] S. Willard, General Topology, Dover, 2004.

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