[:extreme value theorem:]
• JRM:Top.2 §27.4
• Take continuous $f : X \to Y$ with $Y$ an order topology and $X$ compact. Then:
• Exists $a, b \in X$ such that for each $x \in X$ we have $f(a) \leq f(x) \leq f(b)$
• We can think of this as a generalization of the same fact for $\mathbb R$, which says that the image of a closed interval under a continuous function is bounded.
• Proof sketch. Since $f$ is continuous and $X$ compact, then $f(X)$ is compact. Since $f(X)$ is compact and in an order topology, we can show that it is bounded*. Call (elements from) the bounds' preimages $a$ and $b$.
• * Assume that $f(X)$ has no upper bound. Then $\{ (-\infty, a_i) \}$ forms an open covering $f(X)$; compactness guarantees a finite subcovering $\{ a_k \}$ which has upper bound $\max_k a_k$; contradiction