Proof of Zorn's lemma

Proof

Zorn's lemma implies the ultrafilter lemma, which in turn implies Tychonoff's theorem (together with the axiom of choice). However, it is also possible to prove Tychonoff's theorem directly from Zorn's lemma as follows (by using an ultrafilter implicitly).

Let $X_i$ be a family of compact spaces, not necessarily Hausdorff. We shall show that a family $F$ of closed subsets of $X := \prod_i X_i$ with the finite intersection property has nonempty intersection. Let $P$ be the collection of all the families of subsets of $X$ with the finite intersection property (not necessarily closed subsets). We let $P$ be ordered by set inclusion. If $C \subset P$ is a chain, then clearly the union of $C$ has the finite intersection property; so, the union is in $P$. Thus, by Zorn's lemma, there is a maximal element $M$ in $P$ containing $F$. We shall show that the intersection of all the closed sets in $M$ has nonempty intersection; a fortiori, $F$ has nonempty intersection.

For each finite subset $M' \subset M$, since $\cap M'$ is nonempty, $\cap p_i(M')$ is nonempty for the projection $p_i : X \to X_i.$ Thus, for each $i$, the set

$$N_i := \left \{ \overline{p_i(A)} \mid A \in M \right \}$$

has nonempty intersection since it has the finite intersection property and $X_i$ is compact. We note

1. If $M' \subset M$ is a finite subset, $\cap M'$ is in $M$.
2. For an open set $U_i$ intersecting $\cap N_i$, we have $p_i^{-1}(U_i)$ is in $M$.

Indeed, (1) holds since $M = M \cup \{ \cap M' \}$ by maximality, as the set on the right has the finite intersection property. Similarly, for a finite subset $M' \subset M$, since $A = \cap M'$ is in $M$, $U_i \cap p_i(A) \ne \emptyset$ or $p_i^{-1}(U_i) \cap A \ne \emptyset$. Thus, the set $M \cup \{ p_i^{-1}(U_i) \}$ has the finite intersection property and (2) follows by the maximality of $M$.

Now, by the axiom of choice, we can find an element $m$ in $\prod_i (\cap N_i)$. We claim this $m$ is in each closed set $A$ in $M$. Since $A$ is closed, it is enough to show $m$ is in the closure of $A$; i.e., each basic neighborhood of $m$ intersects $A$. A basic neighborhood of $m$ has the form $\cap_{j=1}^r p_{i_j}^{-1}(U_{i_j})$. But by the property (2) above, we have $\left \{ A, p_{i_j}^{-1}(U_{i_j}) \mid j = 1, \dots, r \right \} \subset M$, which gives the claim by the finite intersection property. $\square$