math 360: the real numbersdeturck/m360/week1-reals-short.pdf · math 360: the real numbers d....

Math 360: The real numbers

D. DeTurck

University of Pennsylvania

August 31, 2017

D. DeTurck Math 360 001 2017C: Real numbers 1 / 26

Some cautionary tales: R vs Q

“Analysis consists of the difficult proofs of obvious theorems”

Are the answers intuitive:

• What is the (general) solution of g ′(t) = t2? Are you sure?

• Are the “constant value theorem” and the “intermediate valuetheorem” obvious?

• If f ′(t) > 0 for all t does a > b imply that f (a) > f (b)?

Over Q, all of these are neither intuitive nor true! So what makesthe real numbers “better” for analysis?


Fields

Algebraic axioms first: A field F, or (F,+,×) is a set F with twobinary operations such that:

• (A1) a + (b + c) = (a + b) + c and a + b = b + a.

• (A2) There is a unique element 0 ∈ F with a + 0 = a for alla ∈ F.

• (A3) For each a ∈ F there is −a ∈ F with a + (−a) = 0.

So (F,+) is a commutative group; write a− b for a + (−b). Alsowrite ab for a× b. Then

• (M1) a(bc) = (ab)c and ab = ba

• (M2) There is a unique 1 ∈ F with a1 = a for all a ∈ F.

• (M3) For each a ∈ F with a 6= 0 there is a−1 with aa−1 = 1.

So (F− {0},×) is a commutative group, also write a/b for ab−1.

• (D1) a(b + c) = ab + ac (distributive law)


Ordered fields

Exercise: Show that a(−b) = (−a)b = −(ab) and(−a)(−b) = ab.

So far, F is a field. Now for the order axioms.

There is a nonempty subset P ⊂ F and write a > 0 iff a ∈ P:

• (P1) If a ∈ P and b ∈ P then a + b ∈ P and ab ∈ P.

• (P2) For all a ∈ F either a ∈ P or −a ∈ P or a = 0.(Trichotomy)

Write a > b iff a− b ∈ P and write a ≥ b iff a > b or a = b.

Now F is an ordered field. R and Q are examples of ordered fields(there are others).

Exercise: Show that a2 = (−a)2, so that a2 > 0 for all a 6= 0, andso 1 > 0. And by induction n > 0 for every positive integern = 1 + 1 + · · ·+ 1.


Absolute value

Definition of absolute value

|a| =

{a if a ≥ 0−a otherwise

Lemma 1

1 |a| ≥ 0.

2 |(|a|)| = |a|.3 | − a| = |a|4 |ab| = |a||b|5 |a|+ |b| ≥ |a + b|

Exercise: All of these except the last one are obvious. Prove thelast one (well, maybe also the next-to-last one).


Limit of a sequence

We work in the ordered field F (think R or Q, but there are others).

Definition

The sequence a1, a2, . . . tends to the limit L as n tends to infinity(or an → L as n→∞, or lim

n→∞an = L) if, given any ε > 0, we can

find an integer N(ε) such that

|an − L| < ε for all n > N(ε).


Properties of limits of sequences

Lemma 2

Suppose an → L as n→∞. Then:

1 The limit is unique (i.e., if also an → M then L = M).

2 (Subsequences) If n(1) < n(2) < n(3) < · · · then an(j) → L asj →∞.

3 If an = c for all n, then L = c .

4 If bn → M as n→∞ then an + bn → L+M and anbn → LM.

5 If an 6= 0 for each n and if L 6= 0, then a−1n → L−1

6 If an ≤ A for each n, then L ≤ A. Likewise if an ≥ A for eachn then L ≥ A.


Epsilontics; proof of lemma

We’ll give careful proofs of the properties on the previous slide toillustrate “epsilontics”.

1 Suppose an → L as n→∞. Then the limit is unique (i.e., ifalso an → M then L = M).

From the definition, we know that given ε > 0 we can find N1(ε)so that |an − L| < ε for all n > N1(ε), and we can find N2(ε) sothat |an −M| < ε for all n > N2(ε). If it were the case thatL 6= M, then picking ε = 1

2 |L−M| we have ε > 0, and thenchoosing any n > max(N1(ε),N2(ε)) we have:

|L−M| = |(L−an)+(an−M)| ≤ |an−L|−|an−M|<ε+ε = |L−M|.

This is impossible, so (by contradiction, or reductio ad absurdam)we must have L = M.


Proof of lemma

2 (Subsequences) Suppose an → L as n→∞. Ifn(1) < n(2) < n(3) < · · · then an(j) → L as j →∞.

Given ε > 0 we can find an N(ε) such that |an − L| < ε for alln > N(ε). Since n(j) ≥ j (proof by induction), we also have|an(j) − L| < ε for all j > N(ε).

3 If an = c for all n, then an → c as n→∞.

Given ε > 0 we can take N(ε) to be equal to 1, since

|an − c | = 0 < ε

for all n.


Limits of sums

4 If limn→∞

an = L and limn→∞

bn = M then limn→∞

an + bn = L + M.

This gives us a chance to use theε

2trick.

By hypothesis, given ε > 0 we can find N1(ε) such that|an − L| < ε for all n > N1(ε) and we can find N2(ε) such that|bn −M| < ε for all n > N2(ε). Now letN3(ε) = max(N1(ε/2),N2(ε/2)). Then for n > N3(ε) we have

|(an + bn)− (L + M)| = |(an − L) + (bn −M)|≤ |an − L|+ |bn −M|

<ε

2+ε

2= ε

which gives the result.


Limits of products

4 If limn→∞

an = L and limn→∞

bn = M then limn→∞

anbn = LM.

This gives us a chance to use theε

whatevertrick.

By hypothesis, given ε > 0 we can find N1(ε) such that|an − L| < ε for all n > N1(ε) and we can find N2(ε) such that|bn −M| < ε for all n > N2(ε). Now letN3(ε) = max(N1(1),N1(ε/(2(|M|+ 1))),N2(ε/(2(|L|+ 1))))(because it is possible that M = 0). Then for n > N3(ε) we have(noting that |an| < |L|+ 1)

|anbn − LM| = |anbn − anM + anM − LM|≤ |an||bn −M|+ |M||an − L|

<ε

2+ε

2= ε

which gives the result.D. DeTurck Math 360 001 2017C: Real numbers 11 / 26

Limits of reciprocals

5 If limn→∞

an = L, and if an 6= 0 for each n and if L 6= 0, then

a−1n → L−1

Let ε > 0 be given. By hypothesis, given ε > 0 we can find N1(ε)such that |an − L| < ε for all n > N1(ε). Since L 6= 0 we also have12 |L| 6= 0, so set N(ε) = max(N1(12L),N1(12L

2ε)). If n > N(ε), then∣∣∣∣ 1

an− 1

L

∣∣∣∣ =|L− an||L||an|

≤ 2|L− an||L|2

< ε

which gives the result.


Limits with bounds

6 f limn→∞

an = L Iand if an ≤ A for each n, then L ≤ A.

Suppose L > A. Then set N = n(L− A). If n > N, then

an = (an − L) + L ≥ L− |an − L|> L− (L− A) = A

which is a contradiction.


Functions and continuity

We’re still working in an ordered field F. Let E ⊂ F and f : E → Fbe a function from E to F.

Definition

f is continuous at x ∈ E if given any ε > 0 we can find δ(ε, x) > 0such that

|f (x)− f (y)| < ε for all y ∈ E such that |x − y | < δ(ε, x).

If f is continuous at every point of E then say that f : E → F is acontinuous function.

Lemma 3 (continuity and sequences)

If f is continuous at x ∈ E and an is a sequence of points of Ewith an → x as n→∞, then f (an)→ f (x) as n→∞.


Continuity properties

Lemma 4

Let E be a subset of the ordered fiend F and f and g are functionsfrom E to F.

1 If f (x) = c and g(x) = x for all x ∈ E then f and g arecontinuous on E .

2 Define f + g : E → F via (f + g)(x) = f (x) + g(x) andf × g : E → F via (f × g)(x) = f (x)g(x) for all x ∈ E . If fand g are continuous at y ∈ E then so are f + g and f × g .

3 If f (x) 6= 0 for all x ∈ E . If f is continuous at y ∈ E , then sois 1/f (defined in the obvious way).

Corollary 5

Polynomials are continuous on F, and rational functions arecontinuous on subsets E ⊂ F on which the denominator nevervanishes.


A cautionary example

First, note that the equation x2 = 2 has no solutions in Q.

Let f : Q→ Q be defined via:

f (x) =

{−1 if x2 < 2

1 if x2 > 2

Then f is continuous on Q but none of the “intuitive” theoremsabout continuous functions (“constant value theorem”,intermediate value theorem) hold for f , and the “positive derivativeimplies increasing” theorem does not hold for g(x) = x + f (x).

What property to we have to add to distinguish the real numbersfrom the rationals?


The fundamental axiom

The fundamental axiom

If an ∈ R and a1 ≤ a2 ≤ a3 ≤ · · · and if there is some numberA ∈ R such that an < A for all n, then there is an L ∈ R such thatan → L as n→∞.

In other words, every increasing sequence that is bounded abovetends to a limit.

Pretty much all of classical real analysis can be obtained from thealgebraic properties of R together with this axiom. There areseveral other equivalent statements, equivalent to the axiom,which we will prove. But we could have used any of them as ourfundamental axiom (and the textbook does this).


The Archimedian axiom

It’s called an axiom but we’re going to prove it, obvious as it seems:

Theorem 6 (Axiom of Archimedes)

1

n→ 0 as n→∞

Proof : 1/n is a decreasing sequence bounded below by 0 so it hasa limit, L. Then multiplying the sequence 1/n by the constantsequence all of whose terms are 1/2 gives that the limit of thesequence (1/2)× (1/n) = 1/(2n) is L/2. But 1/(2n) is also asubsequence of 1/n, so its limit must also be L. Therefore L = L/2which implies L = 0.

Corollary 7

There is no M ∈ R such that M > n for all n ∈ Z.


Q is dense in R

(a) Using the fact that every non-empty set of integers boundedabove has a maximum (with proof!), show that if x ∈ R, there isan integer m such that m ≤ x < m + 1 and so in particular|x −m| < 1.(b) Use this to show that if x ∈ R and q is a positive integer, then

there is an integer p such that

∣∣∣∣x − p

q

∣∣∣∣ < 1

n.

(c) If x ∈ R, then given any ε > 0 there exists y ∈ Q such that|x − y | < ε. (In other words, Q is dense in R.


The intermediate value theorem

Theorem 8 (Intermediate value theorem)

We work in R. If f : [a, b]→ R is continuous, and f (a) ≤ 0 ≤ f (b)then there exists c ∈ [a, b] such that f (c) = 0.

Proof : (by the “bisection method” or “lion hunting”). Define twosequences an and bn inductively as follows: Let a0 = a and b0 = b.Then set c0 = 1

2(a0 + b0). Either f (c0) ≥ 0 in which case seta1 = a0 and b1 = c0, or else f (c0) < 0 in which case set a1 = c0and b1 = b0. Either way, we have

f (a1) ≤ 0 ≤ f (b1) a0 ≤ a1 ≤ b1 ≤ b0 b1− a1 =1

2(b0− a0)

Continue inductively, so that

f (an) ≤ 0 ≤ f (bn) an−1 ≤ an ≤ bn ≤ bn−1 bn−an =1

2(bn−1−an−1)

for all n ≥ 1.D. DeTurck Math 360 001 2017C: Real numbers 20 / 26

Proof of the intermediate value theorem (continued)

Theorem 8 (Intermediate value theorem)

We work in R. If f : [a, b]→ R is continuous, and f (a) ≤ 0 ≤ f (b)then there exists c ∈ [a, b] such that f (c) = 0.

By their definition, we have a0 ≤ a1 ≤ a2 · · · so the an are anincreasing sequence that is bounded above by b0 = b. Thereforethere is a number c so that an → c as n→∞. We also havea ≤ c ≤ b. We can also note that bn − an = 2−n(b0 − a0), so thatbn − an → 0 (by Archimedes) and so

bn = ab + (bn − an)→ c + 0 = c

as n→∞.

Since f is continuous, we have f (an)→ f (c) as n→∞. Sincef (an) ≤ 0 for all n, we have f (c) ≤ 0. Likewise, f (bn)→ f (c) andf (bn) ≥ 0, so we also have f (c) ≥ 0. Therefore f (c) = 0,completing the proof.


The derivative

Definition of the derivative

Let f : (a, b)→ R (where (a, b) is the open interval a < x < b). fis differentiable at x ∈ (a, b) with derivative f ′(x) if, given ε > 0,we can find δ(x , ε) > 0 such that a < x − δ(x , ε) < x + δ(x , ε) < band ∣∣∣∣ f (y)− f (x)

y − x− f ′(x)

∣∣∣∣ < ε

whenever 0 < |y − x | < δ(x , ε).

Exercise: If f and g are differentiable at x with derivatives f ′(x)and g ′(x), and if α and β are real constants, show that αf + βg isdifferentiable at x with derivative αf ′(x) + βg ′(x).


The mean value inequality

Theorem 9 (The mean value inequality)

Suppose A < a < b < B, and f is differentiable on (at every pointof) (A,B) and that there is a real number M such that f ′(x) ≤ Mfor all x ∈ (A,B). Then

f (b)− f (a) ≤ (b − a)M.

Proof : We assume the contrary, namely thatf (b)− f (a) > M(b − a) and derive a contradiction, usingbisection. If that were true, then we can find an ε > 0 smallenough so that in fact f (b)− f (a) > (M + ε)(b − a) (we had toget an ε in there somewhere!). We’ll show that this is impossible.

So set a0 = a, b0 = b and let c0 = 12(a0 + b0). Observe:(

f (c0)− f (a0)− (M + ε)(c0 − a0))

+(f (b0)− f (c0)− (M + ε)(b0 − c0)

)=(f (b0)− f (a0)− (M + ε)(b0 − a0)

)> 0


Proof of the mean value inequality (continued)

Therefore, one of the two parenthesized expressions on the leftmust be positive.

If(f (b0)− f (c0)− (M + ε)(b0 − c0)

)> 0 then set a1 = c0 and

b1 = b0; otherwise set a1 = a0 and b1 = c0. Either way we nowhave:

f (b1)−f (a1) > (M+ε)(b1−a1), a0 ≤ a1 ≤ b1 ≤ b0, b1−a1 =1

2(b0−a0)

You can see where this is going: Continue inductively so that

f (bn)− f (an) > (M + ε)(bn − an), an−1 ≤ an ≤ bn ≤ bn−1,

and bn − an = 12(bn−1 − an−1). So as before, there is a c so that

both an → c and bn → c as n→∞.


Proof of the mean value inequality (concluded)

Because f is differentiable at c , and because an → c and bn → cwe know we can find a δ(c , ε) so that∣∣∣∣ f (y)− f (c)

y − c− f ′(c)

∣∣∣∣ < ε

whenever 0 < |y − c | < δ(c , ε), and an N so that bN − c < δ(c , ε)and c − aN < δ(c , ε). Therefore

f (bN)− f (c)

bN − c− f ′(c) < ε and

f (c)− f (aN)

c − an− f ′(c) < ε.

Combine these to conclude that

f (bN)− f (aN) <(f ′(c) + ε

)(bN − aN) ≤ (K + ε) (bN − aN)

which contradicts our assumption about f (bN)− f (aN), andcompletes the proof.


The constant value theorem

Theorem 10 (constant value theorem)

If U is an open interval in R (or if U = R) and f : U → R isdifferentiable with f ′(x) = 0 for all x ∈ U, then f is a constantfunction.

Proof :Suppose a and b are in U, with b > a. Applying the meanvalue inequality with M = 0 get that f (b)− f (a) ≤ 0. Then applyit to −f with M = 0 to get f (a)− f (b) ≤ 0, so f (a) = f (b). Sincea and b were arbitrary, f is constant.

Corollary 11

If f and g are differentiable on U and f ′(x) = g ′(x) for all x ∈ U,then f (x) = g(x) + c .

(Apply the preceding to f − g .)


math 360: the real numbersdeturck/m360/week1-reals-short.pdf · math 360: the real numbers d....

Documents