Theorem 3.10.8 (Tychonoff).label Let $\curl{X_i}_{i\in I}$ be non-empty topological spaces, then $X=\prods{}{i\in I}X_{i}$ is compact (resp. kompact) if and only if $X_{i}$ is compact (resp. kompact) for each $i\in I$.
Proof. By Proposition 3.6.2 it suffices to prove the assertion for compact spaces. Suppose that $X$ is compact then for each $i\in I$, $X_{i}=\pi_{i}(X)$ is compact by Proposition 3.10.3(3). Conversely suppose $X_{i}$ is compact for each $i\in I$. Let $\scr{U}\suf \cali{P}(X)$ be an ultrafilter. For each $i\in I$, $\pi_{i}(\scr{U})$ is an ultrafilter base. Since $X_{i}$ is compact, there exists $x_{i}\in X_{i}$ such that $\pi_{i}(\scr{U})\to x_{i}$. By Axiom of Choice, there exists $x\in X$ such that $\pi_{i}(\scr{U})\to \pi_{i}(x)$. Therefore $\scr{U}\to x$.$\square$
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