Proofs of\Three Hard Theorems"

Fall 2004

Chapter x7 of Spivak's Calculus focuses on three of the most important

theorems in Calculus.In this note I will give alternative proofs of these

results.1 Preliminaries

We review a few important facts we have seen about sequences.

Lemma 1 If x

n

2 [a;b] for all n 2 N and (x

n

)

1n=1

converges to x.Then

x 2 [a;b].

Proof We ¯rst show a · x.Suppose x < a.Choose ² > 0 with ² < a ¡x.

Then no element of the sequence is in the interval (x¡²;x+²),a contradicton.

A similar argument shows b ¸ x.

Lemma 2 If f:[c;d]!R is continuous,a

n

2 [c;d] for n 2 N and (a

n

)

1n=1

converges to a 2 [c;d].Then (f(a

n

))

1n=1

converges to f(a).

Proof Let ² > 0.Since f is continuous,there is ± > 0 such that if jx¡aj < ±,

then jf(x) ¡ f(a)j < ².Since (a

n

)

1n=1

!a,there is N 2 N such that

ja

n

¡ aj < ± for all n ¸ N.Thus jf(a

n

) ¡ f(a)j < ² for all n ¸ N and

(f(a

n

))

1n=1

!f(a):

Theorem 3 (Nested Interval Theorem) Suppose I

n

= [a

n

;b

n

] where

a

n

< b

n

for n 2 N and I

1

¶ I

2

¶ I

3

¶:::.Then

1

\

n=1

I

n

6=;:

1

Proof Note that we have

a

1

· a

2

· a

3

:::· a

n

·::::::· b

n

·:::b

2

· b

1

:

Then each b

i

is an upper bound for the set A = fa

1

;a

2

;:::g.By the Com-

pleteness Axiom,we can ¯nd ® a least upper bound for A.

We claim that ® 2 I

n

for all n 2 N.Fix n 2 N.Since ® is an upper

bound for A,a

n

· ®.But b

n

is an upper bound for A and ® is the least

upper bound.Thus ® · b

n

.Hence ® 2 I

n

for all n 2 N and ® 2

1

\

n=1

I

n

.

Theorem 4 (Bolzano{Weierstrass Theorem) Every bounded sequence has

a convergent subsequence.

Proof Let (x

i

)

1i=1

be bounded.There is M 2 R such that jx

i

j · M for all

i 2 N.We inductively construct a sequence of intervals

I

0

¾ I

1

¾ I

2

¾:::

such that:

i) I

n

is a closed interval [a

n

;b

n

] where b

n

¡a

n

=

2M

2

n

;

ii) fi:x

i

2 I

n

g is in¯nite.

We let I

0

= [¡M;M].This closed interval has length 2M and x

i

2 I

0

for

all i 2 N.

Suppose we have I

n

= [a

n

;b

n

] satisfying i) and ii).Let c

n

be the midpoint

a

n

+b

n

2

.Each of the intervals [a

n

;c

n

] and [c

n

;b

n

] is half the length of I

n

.Thus

they both have length

1

2

2M

2

n

=

2M

2

n+1

If x

i

2 I

n

,then x

i

2 [a

n

;c

n

] or x

i

2 [c

n

;b

n

],

possibly both.Thus at least one of the sets

fi:x

i

2 [a

n

;c

n

]g and fi:x

i

2 [c

n

;b

n

]g

is in¯nite.If the ¯rst is in¯nite,we let a

n+1

= a

n

and b

n+1

= c

n

.If the

second is in¯nite,we let a

n+1

= c

n

and b

n+1

= b

n

.Let I

n+1

= [a

n+1

;b

n+1

]

Then i) and ii) are satis¯ed.

By the Nested Interval Theorem,there is ® 2

1

\

n=1

I

n

.We next ¯nd a

subsequence converging to ®.

2

Choose i

1

2 N such that x

i

1

2 I

1

.Suppose we have i

n

.We know

that fi:x

i

2 I

n+1

g is in¯nite.Thus we can choose i

n+1

> i

n

such that

x

i

n+1

2 I

n+1

.This allows us to construct a sequence of natural numbers

i

1

< i

2

< i

3

<:::

where i

n

2 I

n

for all n 2 N.

We ¯nish the proof by showing that the subsequence (x

in

)

1n=1

!®.Let

² > 0.Choose N such that ² >

2M

2

N

.Suppose n ¸ N.Then x

i

n

2 I

n

and

® 2 I

n

.Thus

jx

i

n

¡®j ·

2M

2

n

·

2M

2

N

< ²

for all n ¸ N and (x

i

n

)

1n=1

!®.

2 Bounding and the Extreme Value Theorem

Theorem 5 (Bounding Theorem) If f:[a;b]!R is continous,then

there is M 2 R such that jf(x)j · M for all x 2 [a;b].

Proof For purposes of contradiction,suppose not.Then for any n 2 N we

can ¯nd x

n

2 [a;b] such that jf(x

n

)j > n.By the Bolzano{Weierstrass The-

orem,we can ¯nd a convergent subsequence x

i

1

;x

i

2

;:::.Note that jf(x

i

n

j >

i

n

¸ n.Thus,replacing (x

n

)

1n=1

by (x

i

n

)

1n=1

,we may,without loss of gener-

ality,assume that (x

n

)

1n=1

is convergent.Suppose (x

n

)

1n=1

!x.By Lemma

1,x 2 [a;b].By Lemma 2,

(f(x

n

))

1n=1

!f(x):

But the sequence (f(x

n

))

1n=1

is unbounded,and hence divergent,a contradi-

cation.Theorem 6 (Extreme Value Theorem) Suppose a < b.If

f:[a;b]!R,then there are c;d 2 [a;b] such that f(c) · f(x) · f(d)

for all x 2 [a;b].

Proof Let A = ff(x):a · x · bg.Then A 6=;and,by the Bounding

Theorem,A is bounded above and below.Let ® = supA.We claim that

there is d 2 [a;b] with f(d) = ®.

3

Since ® = supA,for each n 2 N,there is x

n

2 [a;b] with

® ¡

1

n

< f(x

n

) · ®.Note that (f(x

n

))

1n=1

converges to ®.By the Bolzano{

Weierstrass Theorem,we can ¯nd a convergent subseqence.Replacing (x

n

)

1n=1

by a subsequence if necessary,we may assume (x

n

)

1n=1

!d for some d 2 [a;b].

Then (f(x

n

))

1n=1

!f(d).Thus f(d) = ®.Note that f(x) · ® = f(d) for all

x 2 [a;b].

Similarly,we can ¯nd c 2 [a;b] with f(c) = ¯ = inf A and f(c) · f(x)

for all x 2 [a;b].

3 Intermediate Value Theorem

Theorem 7 (Intermediate Value Theorem) If f:[a;b]!R is contin-

uous and f(a) < 0 < f(b),then there is a < c < b with f(c) = 0.

Proof We start to build a sequence of intervals

I

0

¶ I

1

¶ I

2

¶:::

such that I

n

= [a

n

;b

n

],f(a

n

) < 0 < f(b

n

) and b

n

¡ a

n

=

b¡a

=2

n

.Let a

0

=

a;b

0

= b and I

0

= [a

0

;b

0

].Then f(a

0

) < 0 < b

0

and b ¡a = (b ¡a)=2

0

.

Suppose we are given I

n

= [a

n

;b

n

] with f(a

n

) < 0 < f(b

n

) and b

n

¡a

n

=

b ¡a=over2

n

.Let d =

bn¡an

2

.If f(d) = 0,then we have found a < d < b with

f(d) = 0 and are done.If f(d) > 0,let a

n+1

= a

n

,b

n+1

= d.If f(d) < 0,let

a

n+1

= d and

Let I

n+1

= [b

n+1

;a

n+1

].Then I

n+1

½ I

n

,f(a

n+1

) < 0 < f(b

n+1

) and

b

n+1

¡a

n+1

=

b¡a

2

n

.

By the nested interval theorem,there is c 2

T

1n=0

I

n

.We claim that

f(c) = 0.

Since a

n

;c 2 I

n

,ja

n

¡ cj ·

b¡a

2

n

for all n 2 N.If ² > 0,choose N such

that

b¡a

2

N

< ².Then ja

n

¡cj < ² for all n ¸ N.Hence (a

n

)

1n=1

converges to c.

Thus,by Lemma 2,(f(a

n

))

1n=1

converges to f(c).Since f(a

n

) · 0 for all n,

we must have f(c) · 0.

Similarly,(b

n

)

1n=1

!c and (f(b

n

))

1n=1

!f(c).But each f(b

n

) > 0,thus

f(c) ¸ 0.Hence f(c) = 0.

Thus there is a < c < b with f(c) = 0.

4

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