Tychonoff's Theorem for Hausdorff Spaces

Theorem
Let:
 * $I$ be an indexing set.

Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty Hausdorff spaces.

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

Then $X$ is compact each $X_i$ is compact.

Proof
First assume that $X$ is compact.

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

are continuous.

From Continuous Image of Compact Space is Compact, it follows that the $X_i$ are compact.

Assume now that each $X_i$ is compact.

By Equivalence of Definitions of Compact Topological Space it is enough to show that every ultrafilter on $X$ converges.

Thus let $\FF$ be an ultrafilter on $X$.

From Image of Ultrafilter is Ultrafilter, for each $i \in I$, the image filter $\map {\pr_i} \FF$ is an ultrafilter on $X_i$.

Each $X_i$ is compact by assumption.

So by definition of compact, each $\map {\pr_i} \FF$ converges.

From Filter on Product of Hausdorff Spaces Converges iff Projections Converge, $\FF$ converges.

So, as $\FF$ was arbitrary, $X$ is compact.

However, Equivalence of Definitions of Compact Topological Space only invokes the Ultrafilter Lemma, so this theorem is safe to use when proving the Boolean Prime Ideal Theorem from the Ultrafilter Lemma.

Also see

 * Tychonoff's Theorem