Tychonoff's Theorem

Theorem
Let $$(X_i)_{i \in I}$$ be a family of non-empty topological spaces, where $$I$$ is an arbitrary index set.

Denote with $$X := \prod_{i \in I} X_i$$ 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 $$\pi_i : X \rightarrow X_i$$ 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 $$\pi_i(\mathcal{F})$$ then is an ultrafilter on $$X_i$$.

Since each $$X_i$$ is compact by assumption, each $$\pi_i(\mathcal{F})$$ therefore converges to a $$x_i \in X_i$$.

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