Theorem 4.16.6 (Tychonoff). Let $\seqi{X}$ be compact topological spaces, then $X = \prod_{i \in I}X_{i}$ is compact.
Proof. Let $\fU \subset 2^{X}$ be an ultrafilter. By Proposition 4.2.4, for each $i \in I$, $\pi_{i}(\fU)$ is an ultrafilter base. Since $X_{i}$ is compact, there exists $x_{i} \in X_{i}$ such that $\pi_{i}(\fU) \to x_{i}$. By the axiom of choice, there exists $x \in X$ such that $\pi_{i}(\fU) \to \pi_{i}(x)$ for all $i \in I$. Therefore $\fU \to x$ by Proposition 4.7.2.$\square$