The Extreme Value Theorem

Theorem(The Extreme Value Theorem):

Let be a continuous function then $\QTR{Large}{f\ }$ must assume an absolute maximum value and an absolute minimum value.

Proof:(below)

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Some Notes:

That $\QTR{Large}{[c,d]}$ is compact is essential in the proof .

Note: , where , assumes neither a maximum nor a minimum but is bounded above and below.
We will focus on functions on the Real Line, or a closed interval on the Real Line, but we will use "Open Ball" notation to indicate that much of what we will discuss generalizes to Compact Metric Spaces.

We will write for or, in the case of a closed interval,

for
A few formalities about sequences:

A sequence in a set $\QTR{Large}{S}$ is a function . We will denote this in the usual way as $\QTR{Large}{S.}$

Given a function then If $\QTR{Large}{g}$ is strictly increasing, that is for all $\QTR{Large}{i}$ , then we will

call $\QTR{Large}{f}$ $\QTR{Large}{g}$ a subsequence. We will be a bit informal an just write something like

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Definition:

Let . We write if for any there exists an $\QTR{Large}{n}$ such that

With the same meaning we sometimes write " converges to $\QTR{Large}{a}$ ."

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Theorem:

If then either $\QTR{Large}{a}$ is the unique limit point of or there exists an $\QTR{Large}{n}$ such

that for all

Proof:

Suppose that the sequence does not become constant. That $\QTR{Large}{a}$ is a limit point follows from the fact that indeed all we have to show is that at least one member of is in . But, in fact, we are given To see that the limit point is unique, let be some other point and it follows from the definition of " $\QTR{Large}{\lim }$ " that

can only contain a finite number of points of and hence $\QTR{Large}{b}$ cannot be a limit point of that set. Why?

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Homework Due March 16

Show that if is a continuous function and $\QTR{Large}{[c,d]}$ is such that then

Proof:

$\QTR{Large}{f}$ is continuous at $\QTR{Large}{a}$ means that for any there is a such that

means that for any there is an $\QTR{Large}{n}$ such that

Combining these we have that for any there is an $\QTR{Large}{n}$ such that

Which means

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Theorem:

In a closed interval every infinite set contains a subset that can be indexed as a sequence that converges to some .

Proof:

We will define a nested sequence of intervals beginning with

Begin by choosing any Since $\QTR{Large}{S}$ is infinite either or contains an infinite subset . Letting be that interval, choose

By induction we can find such that contains an infinite subset

and Choose and repeat the induction step.

Let MATH It is straight forward to show since

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Proof of The Extreme Value Theorem:

We show assumes an absolute maximum value. To find an absolute minimum value ,one finds an absolute maximum of and takes its negative.

Let Since we know that $\QTR{Large}{A}$ is bounded, let

We show that there exists an such that . We know that either or $\QTR{Large}{l}$ is a limit point of $\QTR{Large}{A}$ or both. To show that choose a sequence such that and for each choose such that Let Applying the previous Theorem we can find a subsequence and some such that To complete the proof one shows that:

It is still the case that

As we defined it, a subsequence of a convergent sequence is also convergent, to the the same value.

and hence by continuity
and implies