Додаток: чи можна об'єднати ці підходи до теореми про екстремальне значення?

The Heine-Borel and Bolzano-Weierstrass approaches both rely on the same underlying principle - compactness.

Definition: An open covering of a set $X$ is a collection of open subsets $U_\alpha\subseteq X$ such that $\bigcup_{\alpha\in\mathcal{A}}U_\alpha=X$. The indexing set $\mathcal{A}$ may be uncountable.

Example: Let $X=(0,1)$. An example of an open cover of $X$ is given by $U_n=(\frac{1}{n},1)$ for $n\in\mathbb{N}$, since $\bigcup_{n\in\mathbb{N}}(\frac{1}{n},1)=(0,1)$.

Definition: A set $X$ is compact iff every open covering of $X$ has a finite subcover; that is, there is a finite subcollection that also covers $X$.

Non-example: The interval $(0,1)$ is not compact. Using the cover in the previous example, there is no finite subcover that covers all of $(0,1)$. Any finite subcover will only cover $(\frac{1}{N},1)$ for some (fixed) $N\in\mathbb{N}$.

The Heine-Borel theorem is pretty clearly related to compactness since it characterizes compact sets in $\mathbb{R}$. But what about the Bolzano-Weierstrass theorem? How does that relate to compactness?

It turns out that the Bolzano-Weierstrass property, which is the property that every sequence in $X$ has a convergent subsequence in $X$, is equivalent to compactness in $\mathbb{R}$. That is:

Theorem: Let $X\subseteq\mathbb{R}$. The following are equivalent:

1. $X$ is compact.

2. $X$ is closed and bounded. (Heine-Borel)

3. Every sequence in $X$ has a convergent subsequence in $X$. (Bolzano-Weierstrass property)

We won't provide the proof (see a textbook on real analysis or point-set topology for the proof), but one direction is easy to see: If $X$ is compact, then it has the Bolzano-Weierstrass property:

Take $X$ to be any closed and bounded set (hence compact by Heine-Borel), and consider any sequence $(x_n)\subset X$. By the Bolzano-Weierstrass theorem, this sequence has a convergent subsequence since it is bounded. Moreover, since $X$ is closed, the subsequence has its limit in $X$.

The proofs of the Extreme Value Theorem using the Heine-Borel and Bolzano-Weierstrass didn't use compactness directly - they used the Heine-Borel theorem and Bolzano-Weierstrass theorem, respectively.

However, we can formulate and prove a more general version of the Extreme Value Theorem just from using the definition of compactness.

Theorem (Extreme Value Theorem): If $X$ is compact, a continuous function $f:X\to\mathbb{R}$ attains its maximum and minimum.

Proof: Since $X$ is compact, so is $f(X)$. Suppose $f(X)$ has no maximum; then $\{(-\infty,f(x)):x\in X\}$ is an open cover of $f(X)$. (This is since for every $f(x_0)\in f(X)$, there is some other $f(x_1)>f(x_0)$ since $f$ is unbounded; then $f(x_0)\in(-\infty,f(x_1))$.)

By compactness, there is a finite subcover $\{(-\infty,f(x_1)),\ldots,(-\infty,f(x_n))\}$ of $f(X)$. Let $f(x_1)$ be the maximum over $\{f(x_1),\ldots,f(x_n)\}$. Then $f(x_1)$ is not in the subcover of $f(X)$, a contradiction! Thus $f(X)$ must have a maximum. (And $f(X)$ having a maximum means there is some $f(x_M)\ge f(x)$ for all $x\in X$, so that $f$ attains its maximum at $x_M$.)

For the minimum, apply the same proof or use $-f$.

For the Extreme Value Theorem we had before with $f:[a,b]\to\mathbb{R}$, we will need to use the Heine-Borel Theorem to know that $[a,b]$ is compact. But this new version of the Extreme Value Theorem is more general, and applies to any compact set $X\subset\mathbb{R}$. (Actually, we don't even need $X\subset\mathbb{R}$, just that $X$ is a compact topological space - a term defined in point-set topology.) Nevertheless, we can get rather weird non-interval compact sets $X\subset\mathbb{R}$, and this more general Extreme Value Theorem applies even to those sets!