Continuous real function on a compact set attains its bounds

Statement

Let $X$ be a topological space, let $K \subseteq X$ be a nonempty compact set and let $f \colon X \to \mathbb{R}$ be a continuous map. Then there are points $a, b \in K$ with $f(a) \le f(x) \le f(b)$ for all $x \in K$.

Remarks

Proof: the set $f(K)$ is compact in $\mathbb{R}$, because the continuous image of a compact set is compact. By the Heine–Borel theorem, it is closed and bounded. Thus it contains its supremum and its infimum. For $K = [a, b]$, this is the extreme value theorem of calculus. Compactness is needed: $f(x) = x$ on $(0, 1)$ has no maximum, and $f(x) = x$ on $\mathbb{R}$ is not bounded.

Explanations

L1 human

A continuous function on a compact set has a largest value and a smallest value, and it reaches them at some points. On an open interval, the function can come closer and closer to a value without reaching it.