Tychonoff's Theorem/General Case/Proof 2

Proof
First assume that $X$ is compact.

From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous.

It follows from Continuous Image of Compact Space is Compact that the $X_i$ are compact.

Assume now that each $X_i$ is compact.

By Compact Space satisfies Finite Intersection Axiom, it suffices to show that:


 * for each family $\CC$ of closed subsets of $X$ with:


 * $\ds \bigcap_{C \in \CC} C = \O$


 * there exists a finite subset $\SS \subseteq \CC$ such that:


 * $\ds \bigcap_{S \in \SS} S = \O$

We show the contrapositive.

That is, let $\CC$ be a family $\CC$ of closed subsets of $X$ such that:


 * $\ds \bigcap_{S \in \SS} S \ne \O$

for all finite subsets $\SS \subseteq \CC$.

We say that $\CC$ has the "finite intersection property".

We aim to show that:


 * $\ds \bigcap_{C \in \CC} C \ne \O$

Let:


 * $\ds S = \set {\CC \subseteq \AA : \AA \text { has the finite intersection property} }$

We aim to show that $S$ has a $\subseteq$-maximal element.

We will use Zorn's Lemma for this.

Clearly $S \ne \O$, since $\CC \in S$.

Now let $\sequence {U_\alpha}_{\alpha \mathop \in A}$ be a chain in $S$.

Let:


 * $\ds U = \bigcup_{\alpha \mathop \in A} U_\alpha$

We have $U_\alpha \subseteq U$ for each $\alpha \in A$.

It remains to show that $U$ has the finite intersection property.