Proof of the Extreme Value Theorem in Topological Spaces 📂Topology

Proof of the Extreme Value Theorem in Topological Spaces

Theorem ¹

For a compact space $X$, if a function $f : X \to \mathbb{R}$ is continuous, then there exist $c,d \in X$ satisfying $f(c) \le f(x) \le f(d)$ for all $x \in X$.

Explanation

In $\mathbb{R}$, being compact is equivalent to being a closed interval $[a,b]$, so this ends up being a generalization of the theorem we learned in high school and in analysis. Precisely because it uses the difficult theories of topology, the proof is instead simple and easy.

Proof

Lemma on compact spaces: For $f : X \to Y$, suppose $X$ is compact and $f$ is continuous.
[1]: If $f$ is surjective, then $Y$ is compact. Even if $f$ is not surjective, $f(X)$ is compact.
[2]: If $Y$ is Hausdorff, then $f$ is a closed map. For a closed set $C \subset X$, $f(C) \subset Y$ is a closed set.

Since $f(X)$ is compact, for the open cover $\mathcal{O} := \left\{ (-n,n) \ | \ n \in \mathbb{N} \right\}$ there exists $m$ satisfying $$ f(X) \subset \bigcup_{n=1}^{m} (-n , n) \subset \mathcal{O} $$ By Lemma [1], $f(X)$ is bounded, and since $\mathbb{R}$ is a Hausdorff space, by Lemma [2] the image $f(X)$ of the whole space $X \subset X$ is a closed set in $\mathbb{R}$. Since $f(X)$ is bounded, by the completeness axiom, $$ u := \sup f(X) \\ l : = \inf f(X) $$ exist. Since $f(X)$ is a closed set, $$ u, l \in f(X) \\ f(c) = l \\ f(d) = u $$ there must exist some $c,d \in X$ satisfying these. These $c,d$ satisfy $f(c) \le f(x) \le f(d)$ for all $x \in X$.

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