Tychonoff's Theorem

Theorem
Let $\left \langle {X_i}\right \rangle_{i \in I}$ be a family of non-empty topological spaces, where $I$ is an arbitrary index set.

Let $\displaystyle X = \prod_{i \in I} X_i$ be the corresponding product space.

Then $X$ is compact if and only if each $X_i$ is.

Proof

 * First assume that $X$ is compact.

Since the projections $\operatorname{pr}_i : X \to X_i$ are continuous, it follows that the $X_i$ are compact.


 * Assume now that each $X_i$ is compact.

By the equivalent definitions of compact sets it is enough to show that every ultrafilter on $X$ converges.

Thus let $\mathcal F$ be an ultrafilter on $X$.

For each $i \in I$, the image filter $\operatorname{pr}_i \left({\mathcal F}\right)$ then is an ultrafilter on $X_i$.

Since each $X_i$ is compact by assumption, each $\operatorname{pr}_i \left({\mathcal F}\right)$ therefore converges to a $x_i \in X_i$.

This implies that $\mathcal F$ converges to $x := \left({x_i}\right)_{i \in I}$.

Preliminaries
A basic open set in a Cartesian product $\prod_{i\in I}X_i$ of topological spaces is a set of the form $\prod_{i\in I}U_i$, where each $U_i$ is a nonempty open subset of $X_i$ and $U_i = X_i$ for all but finitely many $i\in I$.

The word '"collection'" will be used to mean set of sets.

Statement
Let $(I,<)$ be a well-ordered set and $(X_i)_{i\in I}$ be a family of compact topological spaces. Denote $X = \prod_{i\in I}X_i$. Let $F$ be the set-theoretic tree $\bigl(\bigcup_{i\in I}\prod_{j<i}X_j\bigr)\cup X$ of functions defined on initial intervals of $I$ ordered by inclusion. Consider the set of all subtrees $T\subset F$ with the following property: for every $i\in I$ and every $f\in T\cap\prod_{j<i}X_j$, the set $\{\,g(i)\mid g\in T,\ f\subsetneqq g\,\}$ is closed in $X_i$; suppose that every such tree $T$ has a branch. Then $\prod_{i\in I}X_i$ is compact.

Proof
To prove that every open cover of $X$ has a finite subcover, it is enough to prove that every open cover by basic open sets has a finite subcover.

Let $\mathcal O$ be a collection of basic open subsets of $X$ such that no finite subcollection of $\mathcal O$ covers $X$. It is enough to prove that $\mathcal O$ does not cover $X$.

Let $T_{\mathcal O}$ be the set of all elements $f\in F$ such that the set $\{\,x\in X\mid f\subset x\,\}$ is not covered by any finite subcollection of $\mathcal O$. Then $T_{\mathcal O}$ is a subtree of $F$. For every $i\in I$ and $f\in T_{\mathcal O}\cap\prod_{j<i}X_j$, denote \[ C_{\mathcal O}(f) = \{\,g(i)\mid f\subsetneqq g\in T_{\mathcal O}\,\}. \]

Step 1. For every $i\in I$ and every $f\in T_{\mathcal O}\cap\prod_{j<i}X_j$, $C_{\mathcal O}(f)$ is closed in $X_i$ and nonempty. To prove this, let $\mathcal W$ be the collection of all open subsets $U$ of $X_i$ such that there exist a finite subcollection $\mathcal P\subset\mathcal O$ such that \[ \{\,x\in X\mid f\subset x,\ x(i)\in U\ \,\}\subset\bigcup\mathcal P. \] Then $X_i\setminus C_{\mathcal O}(f) = \bigcup\mathcal W$ and hence $C_{\mathcal O}(f)$ is closed. It also follow that $C_{\mathcal O}(f)$ is nonempty, because otherwise, by compactness of $X_i$, $\mathcal W$ would have a finite subcover for $X_i$, which would yield a finite collection of $\mathcal O$ that covers $\{\,x\in X\mid f\subset x\ \,\}$ in contradiction with the fact that $f\in T_{\mathcal O}$. (Here the compactness of $X_i$ is used similarly to the usual proof that the product of two compact spaces is compact.) It is left to show that indeed $X_i\setminus C_{\mathcal O}(f) = \bigcup\mathcal W$. Consider an arbitrary $a\in X_i\setminus C_{\mathcal O}(f)$ and define $g\in\prod_{j\le i}X_j$ by: $f\subset g$ and $g(i)=a$. Then $g\notin T_{\mathcal O}$, and therefore there is a finite collection $\mathcal P\subset\mathcal O$ such that \begin{gather*} \{\,x\in X\mid f\subset x,\ x(i)=a\,\} =\{\,x\in X\mid g\subset x\,\} \subset\bigcup\mathcal P\\ \text{and}\\ \{\,x\in X\mid f\subset x,\ x(i)=a\,\}\cap V\ne\varnothing \quad\text{for every}\quad V\in\mathcal P. \end{gather*} Let $U=\bigcap\{\,\operatorname{pr}_i(V) \mid V\in\mathcal P\,\}$. Then $U$ is an open subset of $X_i$, $a\in U$, and \[ \{\,x\in X\mid f\subset x,\ x(i)\in U\ \,\}\subset\bigcup\mathcal P. \] Therefore $U\cap C_{\mathcal O}(f)=\varnothing$ and $a\in U\in\mathcal W$.

Step 2. Every branch of $T_{\mathcal O}$ has the greatest element. To prove this, suppose that $B$ is a branch of $T_{\mathcal O}$ without the greatest element. Let $f=\bigcup B$. Let $i$ be the least element of $I$ that is not in the domain of any element of $B$; then $f\in\prod_{j<i}X_j$. Since $B$ has no greatest element, $f\notin B$, and since $B$ is a maximal chain in $T_{\mathcal O}$, $f\notin T_{\mathcal O}$. Let $\mathcal P\subset\mathcal O$ be a finite collection such that \[ \{\,x\in X\mid f\subset x\,\}\subset\bigcup\mathcal P. \] Let $m$ be the greatest element of the finite set $\{\, j\in I\mid j<i\text{ and }\exists V\in\mathcal P\ \operatorname{pr}_j(V)\ne X_j\,\}$. Let $g$ be any element of $B$ that is defined on $m$. Consider an arbitrary $x\in X$ such that $g\subset x$. Let $y\in X$ be defined by $f\subset y$ and $y(j)=x(j)$ for every $j\ge i$, and choose $V\in\mathcal P$ such that $y\in V$. Then $x\in V$ (because $V$ does not "take into account" the values of $x(j)$ for $m<j<i$). Thus \[ \{\,x\in X\mid g\subset x\,\}\subset\bigcup\mathcal P, \] in contradicts with the fact that $g\in T_{\mathcal O}$.

Step 3. Every maximal element of $T_{\mathcal O}$ is an element of $X$: an element $f\in T_{\mathcal O}\setminus X$ cannot be maximal in $T_{\mathcal O}$ because $C_{\mathcal O}(f)\ne\varnothing$.

Now it can be shown that there is $f\in X$ such that $f\notin\bigcup\mathcal O$. Indeed, according to the hypotheses, $T_{\mathcal O}$ has a branch $B$. Let $f$ be the greatest element of $B$. Then $f$ is a maximal element of $T_{\mathcal O}$. Therefore $f\in X$. Therefore the set $\{f\}=\{\,x\in X\mid f\subset x\,\}$ is not covered by any finite subcollection of $\mathcal O$, and hence $f\notin\bigcup\mathcal O$.

Corollary 1
The Cartesian product of a finite family of compact topological spaces is compact.

Corollary 2
Let $I$ be a well-orderable set and $(X_i)_{i\in I}$ a family of compact topological spaces. Suppose that the Cartesian product of all nonempty closed subsets of all $X_i$ is nonempty: \[ \prod\{\,C\mid\text{$C$ is nonempty and closed in $X_i$ for some $i\in I$}\,\}\ne\varnothing. \] Then $\prod_{i\in I}X_i$ is compact.

Proof
Let $<$ be a well-order relation on $I$. Denote $X = \prod_{i\in I}X_i$. Let $F$ be the tree $\bigl(\bigcup_{i\in I}\prod_{j<i}X_j\bigr)\cup X$ ordered by inclusion. To apply the theorem, it is enough to verify that if $T$ is a subtree of $F$ such that for every $i\in I$ and every $f\in T\cap\prod_{j<i}X_j$, the set $\{\,g(i)\mid g\in T,\ f\subsetneqq g\,\}$ is closed in $X_i$, then $T$ has a branch.

Let \[ e\in\prod\{\,C\mid \text{$C$ is nonempty and closed in $X_i$ for some $i\in I$}\,\} \] be a "choice function". Let $B_e$ be the minimal tree among all subtrees $S$ of $T$ with the property that for every $i\in I$ and every $f\in S\cap\prod_{j<i}X_j$, \begin{gather*} e(\{\,g(i)\mid g\in T,\ f\subsetneqq g\,\}) \in\{\,g(i)\mid g\in S,\ f\subsetneqq g\,\}\\ \text{unless}\\ \{\,g(i)\mid g\in T,\ f\subsetneqq g\,\}=\varnothing. \end{gather*} Such subtrees of $T$ exist because $T$ itself is such, and the minimal such subtree is the intersection of all such subtrees. It can be shown that $B_e$ is a branch by assuming that it is not, considering the minimal $i\in I$ where it "branches", and arriving at a contradiction with its minimality.

Applications
The proof that $[0,1]^{\mathbb Z}$ is compact does not require the Axiom of Choice, because the product of all nonempty closed subsets of $[0,1]$ contains, for example, the greatest lower bound function $\inf$ (restricted to the collection of closed subsets of $[0,1]$).