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 $$x := \left({x_i}\right)_{i \in I}$$.