Proof of Tychonoff's theorem by maximal families with the finite intersection property

Proof

We use the finite intersection property and a maximal family given by Zorn's lemma.

1. Let $X = \prod_{i \in I} X_i$ with the product topology, and let $\pi_i \colon X \to X_i$ be the projections. If $X$ is empty, each open cover of $X$ has the empty subcover, so $X$ is compact. From now on, $X \neq \varnothing$.
2. A family $\mathcal{F}$ of subsets of $X$ has the finite intersection property (FIP) when each intersection of finitely many members is nonempty.
3. By the finite intersection property criterion with $K = X$, it suffices to show this. If $\mathcal{A}$ is a family of closed subsets of $X$ with FIP, then $\bigcap \mathcal{A} \neq \varnothing$. Fix such an $\mathcal{A}$.
4. Let $P$ be the set of families $\mathcal{F}$ of subsets of $X$ with $\mathcal{A} \subseteq \mathcal{F}$ and with FIP, ordered by inclusion. It contains $\mathcal{A}$, so it is nonempty.
5. Let $\mathcal{K}$ be a chain in $P$. If $\mathcal{K}$ is empty, $\mathcal{A}$ is an upper bound. Otherwise let $\mathcal{U}$ be the union of the members of $\mathcal{K}$. It contains $\mathcal{A}$.
6. The family $\mathcal{U}$ has FIP. Finitely many members $D_1, \ldots, D_n$ of $\mathcal{U}$ lie in members $\mathcal{F}_1, \ldots, \mathcal{F}_n$ of the chain. The largest of these contains all $D_k$, and it has FIP, so $D_1 \cap \cdots \cap D_n \neq \varnothing$.
7. So each chain of $P$ has an upper bound in $P$. By Zorn's lemma, $P$ has a maximal element $\mathcal{D}$.
8. Claim: each intersection $E$ of finitely many members of $\mathcal{D}$ is in $\mathcal{D}$. Indeed, $\mathcal{D} \cup \{E\}$ contains $\mathcal{A}$ and has FIP, since a finite intersection of its members is a finite intersection of members of $\mathcal{D}$. By maximality, $E \in \mathcal{D}$.
9. Claim: if $S \subseteq X$ meets each member of $\mathcal{D}$, then $S \in \mathcal{D}$. Take $D_1, \ldots, D_n \in \mathcal{D}$. By step 8, $D_1 \cap \cdots \cap D_n \in \mathcal{D}$, so $S$ meets it. Thus $\mathcal{D} \cup \{S\}$ has FIP, and maximality gives $S \in \mathcal{D}$.
10. Fix $i \in I$. Let $\overline{T}$ denote the closure of a set $T$. The family $\{\overline{\pi_i(D)} : D \in \mathcal{D}\}$ consists of closed subsets of $X_i$, by this result.
11. This family has FIP. For $D_1, \ldots, D_n \in \mathcal{D}$, we have $\pi_i(D_1 \cap \cdots \cap D_n) \subseteq \overline{\pi_i(D_1)} \cap \cdots \cap \overline{\pi_i(D_n)}$, and the left side is nonempty by FIP of $\mathcal{D}$.
12. The space $X_i$ is compact, so the finite intersection property criterion gives that the set $Z_i = \bigcap_{D \in \mathcal{D}} \overline{\pi_i(D)}$ is nonempty.
13. By the axiom of choice, choose $x_i \in Z_i$ for each $i \in I$, and let $x = (x_i)_{i \in I} \in X$.
14. Let $i \in I$ and let $U$ be open in $X_i$ with $x_i \in U$. Claim: $\pi_i^{-1}(U) \in \mathcal{D}$.
15. Proof of step 14. Let $D \in \mathcal{D}$. If $U \cap \pi_i(D) = \varnothing$, then $\pi_i(D) \subseteq X_i \setminus U$, which is closed. Then $\overline{\pi_i(D)} \subseteq X_i \setminus U$ by minimality of the closure, which contradicts $x_i \in \overline{\pi_i(D)}$.
16. So $U$ meets $\pi_i(D)$, and hence $\pi_i^{-1}(U)$ meets $D$. This holds for each $D \in \mathcal{D}$, so step 9 gives $\pi_i^{-1}(U) \in \mathcal{D}$.
17. Let $\mathcal{S}$ be the subbasis of sets $\pi_i^{-1}(U)$ with $U$ open in $X_i$. Let $\mathcal{B}$ be the set of intersections of finitely many members of $\mathcal{S}$, together with $X$.
18. Let $\mathcal{T}$ be the set of all unions of members of $\mathcal{B}$. It contains $\varnothing$ and $X$, and it is closed under unions.
19. The intersection of two unions $\bigcup_j B_j$ and $\bigcup_k B'_k$ is $\bigcup_{j, k} (B_j \cap B'_k)$, and each $B_j \cap B'_k$ is in $\mathcal{B}$. So $\mathcal{T}$ is a topology that contains $\mathcal{S}$.
20. The product topology is the smallest topology that contains $\mathcal{S}$, so each open set of $X$ is in $\mathcal{T}$. Thus, if $O$ is open and $x \in O$, there is $B \in \mathcal{B}$ with $x \in B \subseteq O$.
21. Each $B \in \mathcal{B}$ with $x \in B$ lies in $\mathcal{D}$. For $B = X$, note that $X$ meets each member of $\mathcal{D}$ and use step 9. Otherwise $B$ is a finite intersection of sets as in step 14, and step 8 applies.
22. Let $D \in \mathcal{D}$. Claim: $x \in \overline{D}$. Suppose not. Then $O = X \setminus \overline{D}$ is open and contains $x$, so by step 20 there is $B \in \mathcal{B}$ with $x \in B \subseteq O$.
23. By step 21, $B \in \mathcal{D}$, so $B \cap D \neq \varnothing$ by FIP. But $B \subseteq O$ and $D \subseteq \overline{D}$, so $B \cap D = \varnothing$. This contradiction proves the claim.
24. Let $A \in \mathcal{A}$. Then $A \in \mathcal{D}$, so $x \in \overline{A}$ by step 22. As $A$ is closed, $\overline{A} = A$, by minimality of the closure. So $x \in A$.
25. Thus $x \in \bigcap \mathcal{A}$, which is nonempty. By step 3, $X$ is compact.