# Tychonoff's Theorem

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## Theorem

### General Theorem

Let $I$ be an indexing set.

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

### Tychonoff's Theorem for Hausdorff Spaces

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.

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.

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

$\blacksquare$

## Also known as

Tychonoff's Theorem is also seen presented as Tikhonov's theorem, based on an alternative transliteration of Tychonoff's name,

## Source of Name

This entry was named for Andrey Nikolayevich Tychonoff.