Tychonoff's Theorem

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 $X = \prod_{i \in I} X_i$ 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 the equivalent definitions of compact sets it is enough to show hat 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, each $\operatorname{pr}_i \left({\mathcal F}\right)$ therefore converges to a $x_i \in X_i$.

This implies that $\mathcal F$ converges to $x := \left({x_i}\right)_{i \in I}$.