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MAT425 Real Analysis L17

Prof. Thistleton

From your studies at home you know the following:

A set S is called compact if
Let S be a compact set. Then S is closed and bounded.

We will spend the next two lectures trying to understand the behavior of continuous functions. Specifically, recall the $\epsilon - \delta$ definition of continuity.

We might wonder, for what functions f may we find a $\delta$ which does the job uniformly for any given $\epsilon$ everywhere on the domain of the function? Also, given a function f, can we restrict the domain of f in such a way as to find this $\delta$ ? Besides just a technical interest, these ideas lead to two of the most important results in Calculus: the Extreme Value Theorem and the Mean Value Theorem.

We will prove the Heine-Borel theorem. This theorem will show us that a topological idea, open covers and finite subcovers, is connected to our concept of compactness. The chain of ideas will be

$\begin{displaymath} compact \Leftrightarrow closed\ and\ bounded \Leftrightarrow finite\ subcover\ property\end{displaymath}$

The first definition we need is that of an open cover.

Definition 191

Let S be a subset of $\Re$ . Define an open cover of S to be a collection of open sets, $\{ O_{\alpha} \}$ ,such that $S\ \subseteq\ \cup\ O_{\alpha}$ .

It is the case for some sets S and their open covers $\{ O_{\alpha} \}$ that we may find a finite number of these open sets which will also cover S.

Definition 201

Given a set S and an open cover of S, $\{ O_{\alpha} \}$ ,we say that there is a finite subcover of S if there exists a finite collection of open sets from $\{ O_{\alpha} \}$ which also covers S.

What do we mean when we say that a subset $S \subseteq \Re$ is bounded?

The proof which follows is classic and will remind you of the proof of Bolzano-Weierstrass. Recall the Nested Intervals Property of $\Re$ .

Theorem 207

Heine-Borel Let $S \subseteq \Re$ be closed and bounded. Then any open cover of S, $\{ U_{\alpha} \}$ , has a finite subcover.

We now present the proof.

Theorem 210

Let S be a set with the finite subcovering property. Then S is closed and bounded.

We have established that

$\begin{displaymath} compact \Rightarrow closed\ and\ bounded \Leftrightarrow finite\ subcover\ property\end{displaymath}$

Now show that continuous functions preserve the finite subcovering property.

Theorem 215

Let S be a set with the finite subcovering property and let $f : S \rightarrow \Re$ be a continuous function. Then f(S) has the finite subcovering property.

To complete this circle of ideas we show that

$\begin{displaymath} finite\ subcover\ property \Rightarrow compact\end{displaymath}$

Theorem 219

Let S be a set with the finite subcovering property. Then S is compact.

We present two more ideas:

Theorem 220

The intersection of any collection of compact sets is compact.

Theorem 224

Let S be a compact set. If $f : S \rightarrow \Re$ is continuous then f(S) is compact.

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William Thistleton
11/10/1998