Tychonoff's Theorem for Arbitrary Products of Compact Sets
Recall from the Alexander's Subbasis Theorem page that if $(X, \tau)$ is a topological space and $\mathcal S$ is a subbasis of the topology $\tau$ on $X$ then if every open cover consisting only of subbasis elements has a finite subcover, then $X$ is a compact space.
 Theorem 1 (Tychonoff's Theorem): Let $\{ X_i \}_{i \in I}$ be an arbitrary collection of topological spaces. Then X_i $]] is compact for all$i \in I$if and only if the topological product$\displaystyle{\prod_{i \in I} X_i}$is compact. • Proof:$\Rightarrow$Suppose that$X_i$is compact for all$i \in I$and consider the product$\displaystyle{\prod_{i \in I} X_i}. Recall from the Arbitrary Topological Products of Topological Spaces page that a subbasis for this topological product is given by: (1) \begin{align} \quad \mathcal S = \left \{ p_i^{-1}(U) : U \: \mathrm{is \: open \: in \:} X_i \mathrm{\: for \: some \:} i \in I \right \} \end{align} • We claim that any cover ofX$consisting only of subbasis elements has a finite subcover. Let$\mathcal F$be a cover of$X$consisting only of subbasis elements. • For each$i \in I$let$\mathcal U_i$be the set of all$U$for which$p_i^{-1}(U) \in \mathcal F, i.e.,: (2) \begin{align} \quad \mathcal U_i = \{ U : p_i^{-1}(U) \in \mathcal F \} \end{align} • We claim that there exists ani \in I$such that$\mathcal U_i$covers$X_i$. Suppose not, i.e., suppose that for all$i \in I$we have that$\mathcal U_i$does not cover$X_i$, i.e.,$\displaystyle{X_i \not \subseteq \bigcup_{U \in \mathcal U_i} U}$. Then there exists an$x_i \in X_i$that is not covered by$\mathcal U_i$, so$\displaystyle{x_i \not \in \bigcup_{U \in \mathcal U_i} U}$. • Let$(x_i)_{i \in I}$be the point in$\displaystyle{\prod_{i \in I} X_i}$constructed from taking coordinates as described above. Then$(x_i)_{i \in I} \not \in p_i^{-1}(U) \in \mathcal F$. But this is a contradiction since$\mathcal F$is supposed to be a cover of$\displaystyle{\prod_{i \in I} X_i}$. So the assumption that there does not exist an$i \in I$such that$\mathcal U_i$covers$X_i$is false. • So there exists an$i \in I$such that$\mathcal U_{i}$covers$X_{i}$. Each space in the product is compact, so$X_{i}$is compact, so there exists a finite subcover,$U_1, U_2, ..., U_n$of$\mathcal U_i$that also covers$X_i$. But then the following set is a finite subcover of$\displaystyle{\prod_{i \in I} X_i}: (3) \begin{align} \quad \{ p_i^{-1}(U_1), p_i^{-1}(U_2), ..., p_i^{-1}(U_n) \} \end{align} • Moreover, sinceU_1, U_2, ..., U_n \in \mathcal U_i$, this means that$p_i^{-1}(U_1), p_i^{-1}(U_2), ..., p_i^{-1}(U_n) \in \mathcal F$. So every open cover consisting only of elements from the subbasis$\mathcal S$has a finite subcover. • By Alexander's subbasis theorem, this implies that$\displaystyle{\prod_{i \in I} X_i}$is a compact space.$\blacksquare\$