Tychonoff's Theorem for Arbitrary Products of Compact Sets

# 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.

We will use this great result to prove the famous Tychonoff's theorem.

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:

\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 of $X$ 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.,:

\begin{align} \quad \mathcal U_i = \{ U : p_i^{-1}(U) \in \mathcal F \} \end{align}

- We claim that there exists an $i \in I$ 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 $\displaystyle{\prod_{i \in I} X_i}$:

\begin{align} \quad \{ p_i^{-1}(U_1), p_i^{-1}(U_2), ..., p_i^{-1}(U_n) \} \end{align}

- Moreover, since $U_1, U_2, ..., U_n \in \mathcal U_i$, 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. $\blacksquare$