more topological compactness and the limit inferior and limit...

MATH 4530: Analysis One

More Topological Compactness and the LimitInferior and Limit Superior of a Function

James K. Peterson

Department of Biological Sciences and Department of Mathematical SciencesClemson University

November 2, 2018


Outline

1 More Topological Compactness

2 Limit inferior and Limit superior of functions

3 A Function with limx→p(f ) = limx→p(f ) Only Once

4 A Function with limx→p(f ) = limx→p(f ) Never

5 Homework


More Topological Compactness

Let’s review:

Definition

We say a collection of open sets U = Oα : α ∈ Ω indexed by anarbitrary set Ω ( so it can be finite, countably infinite or uncountable)is a open cover of a set A if A ⊆ ∪α∈Ω Oα.

We say the collection Oα1 , . . . ,OαN from U , for some natural

number N, is a finite subcover or fsc of A if A ⊆ ∪Nn=1 Oαn .

Definition

We say a set A is topologically compact if every open cover of A hasa finite subcover.

We need to show this idea in our setting < is equivalent to sequential

compactness.



We have just proven this:

Theorem

The closed and bounded finite interval [a, b], a < b, is topologicallycompact.

Now we are ready for to show:

Theorem

A closed subset B of [a, b], a < b, is topologically compact.

Proof

Let U be an open cover of B. Then the collection of open setsO′ = O ∪ BC is an open cover of [a, b] and hence has a fscV ′ = O1, . . . ,ON ,B

C. for some N with each On ∈ U . This showsB is topologically compact.



Theorem

A closed and bounded set A is topologically compact.

Proof

Since A is bounded, there is a positive number so that A ⊆ [−B,B].Since A is a closed subset of [−B,B], by the previous theorem, A istopologically compact.

We have now done what we set out to do. We already knew a set A issequentially compact IFF it is closed and bounded and now we knowA is topologically compact IFF it is closed and bounded. Hence A issequentially compact IFF topologically compact IFF closed andbounded. Hence, from now on we will just say a set is compact andunderstand we can use these different ways of looking at it asconvenient to our arguments.


Limit inferior and Limit superior of functions

Recall if a nonempty set S is unbounded below, we set inf S = −∞ andif it is unbounded above, we set sup S =∞. We all know what functionsare: they are a way to assign each point of one set D to a point inanother set R.

Definition

Let f be a mapping that takes each point in the domain set D intothe range set R. We say f is a function if to each x ∈ D, thecorresponding value y ∈ R is unique. We use the notation y = f (x) todenote this value in the range. Hence, R = f (x) : x ∈ D.

Comment

It is ok for two x values to go to one y value: for example f (x) = x2

sends ±1 to 1. But if you look at the relationship defined by x = y2

which you know is a parabola going sideways rather than up, for x = 1,there are two possibilities: y = 1 and y = −1. This relationship does notdefine a function. There are two functions here: y =

√x and y = −

√x .


Limit inferior and Limit superior of functions

Comment

We have seen a lot of functions already: every sequence is a function ffrom a RII domain. But we now want to look at functions whose domainis an interesting subset of <. These are functions are at least locallydefined.

Definition

We say a function f is locally defined at the point p, if there is a circleBr (p) for some radius r > 0 so that f (x) is defined as a range valuefor all x in Br (p). Of course, f may have a larger domain than this,but at p, f is defined at least on a circle about p.

Let a be a real number. If f is locally defined at p except possibly pitself, we say a is a cluster point of f at p if there is a sequence(xn) ⊂ Br (p) so that xn → p and f (xn)→ a. We letS(p) = a : a is a cluster point of f at p. If S(p) is nonempty, wedefine the limit inferior of f at p to be limx→p(f ) = inf S(p). We

also define limit superior of f at p to be limx→p(f ) = sup S(p).


A Function with limx→p(f ) = limx→p(f ) Only Once

Now let’s begin a long extended example.

Example

f (x) =

x , x ∈ Q ∩ [0, 1]−x , x ∈ IR ∩ [0, 1] = QC ∩ [0, 1]

where IR denotes the set of all irrational numbers. Look at the point 0.This point is locally defined because it is defined on any Br (0) for apositive r . Pick any such r . Let (xn) be any sequence such that xn → 0.We suspect 0 is a cluster point of f at 0. Consider

|f (xn)− 0| =

|xn|, xn ∈ Q ∩ [0, 1]| − xn|, xn ∈ IR ∩ [0, 1]

= |xn|, ∀n.

Pick ε > 0. then since xn → 0, there is a N so thatn > N ⇒ |xn − 0| < ε. Thus n > N ⇒ |f (xn)− 0| < ε also. We seef (xn)→ 0 and so 0 is a cluster point of f at 0.



Example

(Continued)Can any other point a 6= 0 be a cluster point of f at 0? The set up weused above still works, except now we consider

|f (xn)− a| =

|xn − a|, xn ∈ Q ∩ [0, 1]| − xn − a|, xn ∈ IR ∩ [0, 1]

=

|xn − a|, xn ∈ Q ∩ [0, 1]|xn + a|, xn ∈ IR ∩ [0, 1]

Since xn → 0, for ε = |a|/2, there is an N so that n > N implies|xn − 0| < |a|/2. Let’s do the case a > 0 for convenience. The other caseis similar and is left to you. We have for n > N,

−a/2− a < xn − a < a/2− a and − a/2 + a < xn + a < a/2 + a

or

−3a/2 < xn − a < −a/2 and a/2 < xn + a < 3a/2



Example

(Continued)This says |f (xn)− a| can not be made small for n sufficiently large.Hence, a 6= 0 can not be a cluster point of f . Thus S(0) = 0 telling uslimx→0 f (x) = inf S(0) = 0 and limx→0 f (x) = sup S(0) = 0.

Here is another way to show this. Can any other point a 6= 0 be a clusterpoint of f at 0? The set up we used above still works, except now weconsider

f (xn)− a =

xn − a, xn ∈ Q ∩ [0, 1]−xn − a, xn ∈ IR ∩ [0, 1]

If (xn) ⊂ Q except for finitely many values in IR, then f (xn)− a→ 0− a.This is nonzero unless a = 0. So these sequences will converge to 0.

If (xn) ⊂ IR except for finitely many values in Q, then f (xn)− a→ 0− aand this is not zero unless a = 0 also. Thus these sequences converge to0.



Example

(Continued):If (xn) has infinitely many values in both Q and IR, then

f (xn)− a →−a, xn ∈ Q ∩ [0, 1]−a, xn ∈ IR ∩ [0, 1]

and these two limits are not zero unless a = 0. Thus again, the onlycluster point for f at 0 is a = 0. So S(0) = 0 telling uslimx→0 f (x) = 0 and limx→0 f (x) = 0.



Example

(Continued)What about a point p 6= 0? What are the cluster points there? Theargument is very similar. We now have xn → p. We have

f (xn)− a =

xn − a, xn ∈ Q ∩ [0, 1]−xn − a, xn ∈ IR ∩ [0, 1]

If (xn) ⊂ Q, except for finitely many irrationals, then f (xn)− a→ p − a.This is nonzero unless a = p. So these sequences will converge to p.

If (xn) ⊂ IR, except for finitely many rationals, then f (xn)− a→ −p − aand this is not zero unless a = −p. Thus these sequences converge to−p.

If (xn) has infinitely many values in both Q and IR, then since xn → p,for any ε > 0, ∃N so that n > N =⇒ p − ε < xn < p + ε. Thus, forn > N,

f (xn)− a ∈

(p − a− ε, p − a + ε), xn ∈ Q ∩ [0, 1](−p − a− ε,−p − a + ε), xn ∈ IR ∩ [0, 1]



Example

(Continued)This says f (xn)− a has values in two sets with no common values as longas p − a 6= −p − a or p 6= 0. Hence if (xn) has infinitely many values inboth Q and IR, then the limit of f (xn) can not exist.

Thus the only sequence limits are those for sequences with infinitelymany rational values and finitely many irrational values which will havelimit p and the other case for sequences with infinitely many irrationalvalues and finitely many rational values which will have limit −p.

Let S(p) denote the cluster points of f at p. We have

S(p) =

−p, p, p 6= 00, p = 0

and

limx→p

f (x) =

−p, p 6= 00, p = 0

and limx→p

f (x) =

p, p 6= 00, p = 0



In this example, note there is only one point where the limit inferior andlimit superior values match. This occurs at 0 andlimx→0 f (x) = limx→0 f (x) = 0.

We might conjecture that the limit of f as x → p exists IFFlimx→p f (x) = limx→p f (x) and this common value is the value oflimx→p f (x).

We could also conjecture that what we call continuity for f at thepoint p should mean limx→p f (x) = limx→p f (x) = a and f (p) = a.That is, limx→p f (x) exists and equals a and this matches thefunction value f (p). In short, we would say, f is continuous at p iflimx→p f (x) = f (p).

So our function here is continuous at only one point.

So far no ε’s in how we define things. That comes later.



Here is another example of this type of function, but we do the workfaster.

Example

f (x) =

3x , x ∈ Q ∩ [0, 1]−2x , x ∈ IR ∩ [0, 1]

Look at the point p. Let (xn) be any sequence such that xn → p. Let abe any number and consider

f (xn)− a =

3xn − a, xn ∈ Q ∩ [0, 1]−2xn − a xn ∈ IR ∩ [0, 1]

If (xn) ∈ Q for infinitely many n and in IR for only finitely many n,f (xn)− a→ 3p − a which is nonzero unless a = 3p. Hence these type ofsequences f (xn) converges to 3p and so 3p is a cluster point of f at anyp.



Example

(Continued):If (xn) ∈ IR for infinitely many n and in Q for only finitely many n,f (xn)− a→ −2p− a which is nonzero unless a = −2p. Hence these typeof sequences f (xn) converges to −2p and so −2p is a cluster point of fat any p.

If (xn) has infinitely many values in both Q and IR, then

f (xn)− a →

3p − a, xn ∈ Q ∩ [0, 1]−2p − a xn ∈ IR ∩ [0, 1]

and these two values are distinct unless 3p − a = −2p − a or p = 0.Hence,

S(p) =

−2p, 3p, p 6= 00, p = 0



Example

(Continued):and

limx→p

f (x) =

−3p, p 6= 0

0, p = 0and lim

x→pf (x) =

3p, p 6= 00, p = 0

In this example, again note there is only one point where the limit inferiorand limit superior values match. This occurs at 0 andlimx→0 f (x) = limx→0 f (x) = 0.

So our function here is continuous at only one point.


A Function with limx→p(f ) = limx→p(f ) Never

Example

f (x) =

1, x ∈ Q−1, x ∈ IR

Let a be real number. Let (xn) be a sequence with xn → p. We don’tassume p is rational or irrational.

f (xn)− a =

1− a, xn ∈ Q−1− a, xn ∈ IR

Let (xn) ∈ Q for infinitely many n and in IR for only finitely many n, Forthis sequence, even though xn → p, it is still possible that p is irrational.For example, we can easily find a sequence of rational numbers thatconverge to

√2. f (xn)− a→ 1− a which is nonzero unless a = 1. Hence

for these type of sequences f (xn) converges to 1 and so 1 is a clusterpoint of f at any p.

If (xn) ∈ IR for infinitely many n and in Q for only finitely many n,f (xn)− a→ −1− a which is nonzero unless a = −1.



Example

(Continued):Hence these type of sequences f (xn) converges to −1 and so −1 is acluster point of f at any p.

If (xn) is a sequence with infinitely many rational and irrational values,then

f (xn)− a →

1− a, xn ∈ Q−1− a, xn ∈ IR

Since these two values are distinct for all values of a, we see for none ofthese sequences does f (xn) converge.



Example

(Continued)Thus the only sequence limits are those for sequences with infinitelymany rational values and finitely many irrational values which will havelimit 1 and the other case for sequences with infinitely many irrationalvalues and finitely many rational values which will have limit −1.

We have S(p) = −1, 1 and limx→p f (x) = −1 and limx→p f (x) = 1

Note the limit inferior and limit superior never match. So wesuspect this is a function which is continuous nowhere.

We suspect that limits of functions and continuity of functions arepointwise concepts, not interval based concepts. Of course,continuity for all points in an entire interval should have powerfulconsequences.

If the domain of f is compact, really powerful things happen!



Example

f (x) =

−14, x ∈ Q10, x ∈ IR

Let a be real number. Let (xn) be a sequence with xn → p.

f (xn)− a =

−14− a, xn ∈ Q10− a, xn ∈ IR

If (xn) ∈ Q for infinitely many n and in IR for only finitely many n,f (xn)− a→ −14− a which is nonzero unless a = −14. Hence these typeof sequences f (xn) converges to −14 and so −14 is a cluster point of fat any p.

If (xn) ∈ IR for infinitely many n and in Q for only finitely many n,f (xn)− a→ 10− a which is nonzero unless a = 10. Hence these type ofsequences f (xn) converges to 10 and so 10 is a cluster point of f at anyp.



Example

(Continued):If (xn) is a sequence with infinitely many rational and irrational values,then

f (xn)− a →−14− a, xn ∈ Q10− a, xn ∈ IR

Since these two values are distinct for all values of a, we see for none ofthese sequences does f (xn) converge. Thus, the only cluster points atany p are −14 and 10 and limx→p f (x) = −14 and limx→p f (x) = 10


Homework

Homework 13

Remember, we can just use the term compact now for a set.

13.1 If a nonempty subset A is compact, is it true that any subsetof A is also?If a nonempty subset A is sequentially compact, it is true thatany closed and bounded subset of A is topologically compact?If a nonempty subset A is closed, is it true that any subset ofA is also?If a nonempty subset A is open, is it true that any subset of Ais also?

13.2 Prove if A and B are topologically compact, then A ∪ B isalso. Do this proof using topologically compact ideas only.Prove if A and B are sequentially compact, then A ∪ B is also.Do this proof using sequentially compact ideas only.


Homework

Homework 13

(Continued):

13.3

f (x) =

2, x ∈ Q−3, x ∈ IR

Mimic what we have done to show the cluster points of f at p areS(p) = −3, 2 and limx→p f (x) = −3 and limx→p f (x) = 2.

13.4

f (x) =

3, x ∈ Q7, x ∈ IR

Mimic what we have done to show the cluster points of f at p areS(p) = 3, 7 and limx→p f (x) = 3 and limx→p f (x) = 7.

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