Compactness and Tychonoff's Theorem

We use the characterization: a space is compact iff every ultrafilter on it converges. (Recall a filter is a nonempty family $\mathcal{F}$ of nonempty sets closed under supersets and finite intersections; an ultrafilter is a maximal filter, equivalently one with $A \in \mathcal{U}$ or $X \setminus A \in \mathcal{U}$ for every $A$.)

Step 1 — Compactness ⟺ ultrafilter convergence. Suppose $X$ is compact and $\mathcal{U}$ is an ultrafilter. The closures $\{\overline{F} : F \in \mathcal{U}\}$ form a family of closed sets with the FIP, so by [Compactness] there is $x \in \bigcap_{F} \overline{F}$. Then $\mathcal{U} \to x$: any open $V \ni x$ meets every $F \in \mathcal{U}$, so $X \setminus V \notin \mathcal{U}$, hence $V \in \mathcal{U}$ by maximality. Conversely, if every ultrafilter converges, take a family $\mathcal{C}$ of closed sets with FIP, extend it (by Zorn's Lemma, the choice ingredient) to an ultrafilter $\mathcal{U} \supseteq \mathcal{C}$; its limit point lies in every $\overline{C} = C$.

Step 2 — Push forward the ultrafilter. Let $\mathcal{U}$ be an ultrafilter on $X = \prod_i X_i$. For each $i$, the pushforward

Derivation hint: A continuous image of an ultrafilter under any map is again an ultrafilter — check the dichotomy $A$ or $X_i\setminus A$ transports through preimages. is an ultrafilter on $X_i$.

Step 3 — Choose limits coordinatewise. Each $X_i$ is compact, so $(\pi_i)_*\mathcal{U}$ converges to some $x_i \in X_i$. Using the Axiom of Choice, select one such $x_i$ for every $i \in I$ and define $x = (x_i)_{i\in I} \in X$.

Step 4 — The chosen point is a limit. A basic neighborhood of $x$ has the form $W = \bigcap_{k=1}^n \pi_{i_k}^{-1}(U_{i_k})$ with $x_{i_k} \in U_{i_k}$. Since $(\pi_{i_k})_*\mathcal{U} \to x_{i_k}$, we have $U_{i_k} \in (\pi_{i_k})_*\mathcal{U}$, i.e. $\pi_{i_k}^{-1}(U_{i_k}) \in \mathcal{U}$. A filter is closed under finite intersection, so

Derivation hint: Finitely many filter members intersect to a filter member; this is exactly where "finitely many coordinates" in the product topology is essential. Every basic (hence every) neighborhood of $x$ belongs to $\mathcal{U}$, so $\mathcal{U} \to x$. By Step 1, $X$ is compact. $\qquad\blacksquare$

Given nonempty sets $\{A_i\}_{i \in I}$, adjoin a point $\infty_i \notin A_i$ and topologize $Y_i = A_i \cup \{\infty_i\}$ so that the closed sets are $Y_i$, $\varnothing$, and all sets containing $\infty_i$... more precisely give $Y_i$ the topology whose open sets are $\varnothing, Y_i$, and $A_i$ together with cofinite-type structure making each $Y_i$ compact.

Step 1. Each $Y_i$ is compact (one verifies finitely many opens suffice because every nonempty open contains points forcing finite reduction).

Step 2. By Tychonoff, $Y = \prod_i Y_i$ is compact.

Step 3. The sets $C_i = \pi_i^{-1}(A_i)$ are closed and have the FIP: a finite intersection $\bigcap_{k} C_{i_k}$ is nonempty since we only need to choose elements of finitely many $A_{i_k}$ (no choice needed for finite collections), padding the rest with $\infty_j$.

Step 4. Compactness gives $\bigcap_{i \in I} C_i \neq \varnothing$. Any element $f$ of this intersection satisfies $f(i) \in A_i$ for all $i$, i.e. $f$ is a choice function:

Derivation hint: A point in the full intersection has all coordinates in the original sets, which is precisely the statement of the Axiom of Choice. Hence Tychonoff ⟹ AC. Combined with the forward proof (which used AC), the two are ZF-equivalent. $\qquad\blacksquare$