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.

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


Also see


Source of Name

This entry was named for Andrey Nikolayevich Tychonoff.


Sources

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