# 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 Equivalent Definitions of Compactness 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 Equivalent Definitions of Compactness, each $\map {\pr_i} \FF$ converges.

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

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$\blacksquare$

## 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.

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $5$: Compact spaces: $5.6$: Compactness and Constructions: Remark $5.6.3$ - 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties