# Tychonoff's Theorem

## General Theorem

Let $\left \langle {X_i}\right \rangle_{i \in I}$ be a family of non-empty topological spaces, where $I$ is an arbitrary index set.

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

Then $X$ is compact if and only if each $X_i$ is.

## Proof

First assume that $X$ is compact.

Since the projections $\operatorname{pr}_i : X \to X_i$ are continuous, it follows that the $X_i$ are compact.

Assume now that each $X_i$ is compact.

By Equivalent Definitions of Compactness it is enough to show that every ultrafilter on $X$ converges.

Thus let $\mathcal F$ be an ultrafilter on $X$.

For each $i \in I$, the image filter $\operatorname{pr}_i \left({\mathcal F}\right)$ then is an ultrafilter on $X_i$.

Since each $X_i$ is compact by assumption, by Equivalent Definitions of Compactness, each $\operatorname{pr}_i \left({\mathcal F}\right)$ converges.

By Filter on Product Space Converges iff Projections Converge, $\mathcal F$ converges.

$\blacksquare$

## Tychonoff's Theorem for Hausdorff Spaces

Let $\left \langle {X_i}\right \rangle_{i \in I}$ be a family of non-empty Hausdorff spaces, where $I$ is an arbitrary index set.

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

Then $X$ is compact if and only if each $X_i$ is.

## Proof

First assume that $X$ is compact.

Since the projections $\operatorname{pr}_i : X \to X_i$ are continuous, it follows that the $X_i$ are compact.

Assume now that each $X_i$ is compact.

By Equivalent Definitions of Compactness it is enough to show that every ultrafilter on $X$ converges.

Thus let $\mathcal F$ be an ultrafilter on $X$.

For each $i \in I$, the image filter $\operatorname{pr}_i \left({\mathcal F}\right)$ then is an ultrafilter on $X_i$.

Since each $X_i$ is compact by assumption, by Equivalent Definitions of Compactness, each $\operatorname{pr}_i \left({\mathcal F}\right)$ converges.

By Filter on Product of Hausdorff Spaces Converges iff Projections Converge, $\mathcal F$ converges.

$\blacksquare$

## Source of Name

This entry was named for Andrey Nikolayevich Tychonoff.

## Also See

Tychonoff's Theorem Without Choice, a version that holds under more restrictive conditions but does not require the Axiom of Choice.