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