> [!theoremb] Theorem > > Let $\seqi{X}$ be a family of [[Compactness|compact]] [[Topological Space|topological spaces]]. Then their [[Product Topology|product topology]] $X = \prod_{i \in I}X_i$ is compact. > > *Proof*. Let $\net{x} \subset X$ be a [[Net|net]]. Let > $ > \mathcal{P} = \bigcup_{J \subset I}\bracs{p \in \prod_{j \in J}X_j: p \text{ is a cluster point of } \angles{\pi_B(x_\alpha)}_{\alpha \in A}} > $ > As each $X_i$ is [[Compactness|compact]], $\mathcal{P}$ is non-empty. Moreover, $\mathcal{P}$ is [[Partial Order|partially ordered]] by extensions. > > Let $\bracs{p_l: l \in L} \subset \mathcal{P}$ be a [[Total Order|chain]] where $B_l \subset I$ and $p_l \in \prod_{j \in B_l}X_j$. Let $B^* = \bigcup_{l \in L}B_l$, and let $p^* \in \prod_{j \in B^*}X_j$ be the unique element that extends every $p_l$. Let $U$ be a [[Neighbourhood|neighbourhood]] of $p^*$, then $U$ contains an [[Open Set|open set]] of the form $U' = \prod_{j \in B}\pi_j^{-1}(U_j)$ with $U_j \subset X_j$ open for all $j \in B$, and $B$ is finite. So there exists $l \in L$ such that $B_l \supset B$, and since $\angles{\pi_{B_l}(x_\alpha)}_{\alpha \in A}$ is frequently in $U' \subset U$, and $\angles{x_B(x_\alpha)}_{\alpha \in A}$ is frequently in $U$. Therefore $p^* \in \mathcal{P}$. > > By [[Zorn's Lemma]], there exists a maximal element $\ol{p} \in \prod_{i \in \ol{B}}X_i$. Suppose that $\ol{B} \ne I$, then let $i \in I \setminus \ol{B}$. Let $\netb{x}$ be a subnet such that $\angles{\pi_{\ol B}(x_\beta)}_{\beta \in B}$ converges to $\ol{p}$. Since $X_i$ is compact, there exists a subnet $\angles{x_\gamma}_{\gamma \in C}$ such that $\angles{\pi_i(x_\gamma)}_{\gamma \in C}$ converges to some $x_i \in X_i$. Let $p' \in \prod_{j \in \ol{\beta} \cup \bracs{i}}X_j$ be the unique extension of $\ol{p}$ and $x_i$, then $p'$ is an accumulation point of $\net{x}$, which contradicts the maximality of $\ol{p}$. Therefore $\ol{B} = I$, and $\ol{p} \in X$.