# 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 if and only if each $X_i$ is compact.

## Proof

First assume that $X$ is compact.

$\pr_i : X \to X_i$

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.

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

$\blacksquare$

## Boolean Prime Ideal Theorem

This theorem depends on the Boolean Prime Ideal Theorem (BPI), by way of Equivalence of Definitions of Compact Topological Space.

Although not as strong as the Axiom of Choice, the BPI is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.

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.