Tychonoff's theorem is the central compactness theorem of general topology: arbitrary products of compact spaces are compact. Its proof explains why the product topology is the right topology.
Statement and finite intuition
Finite products can be handled by open covers; arbitrary products need subbases or ultrafilters.
Theorem. If \(\{X_i:i\in I\}\) is any family of compact spaces, then \(\prod_{i\in I}X_i\) is compact in the product topology.
Proof. The subbasic proof below proves the theorem directly. The product topology has the projection subbasis \(\pi_i^{-1}(U)\). Alexander's subbase theorem reduces compactness to subbasic covers, and compactness of each factor prevents such a cover from failing finite subcover coordinate by coordinate.
Remark. For finite products this is elementary. For arbitrary products, the theorem is equivalent over weak set theories to forms of the axiom of choice.
Subbasic proof
Alexander's subbase theorem reduces compactness to covers by inverse images under projections.
Proof. Let \(X=\prod_iX_i\). The product topology has subbasis \(\pi_i^{-1}(U)\). By Alexander's subbase theorem, it suffices to show that any subbasic cover has a finite subcover. Suppose \(\{\pi_{i_\alpha}^{-1}(U_\alpha)\}\) is a subbasic cover with no finite subcover. For each fixed \(i\), the sets \(U_\alpha\subseteq X_i\) with \(i_\alpha=i\) cannot cover \(X_i\), otherwise compactness of \(X_i\) would give finitely many and hence a finite subcover of \(X\). Choose \(x_i\in X_i\) outside all such \(U_\alpha\). Then \(x=(x_i)\in X\) lies in none of the subbasic cover sets, contradiction.
Remark. The finite-coordinate nature of the product topology is exactly what makes the subbasic argument possible.
Ultrafilter proof
The ultrafilter proof is often the cleanest conceptual proof.
Theorem. An ultrafilter \(\mathcal U\) on \(\prod_iX_i\) converges if and only if each projected ultrafilter \((\pi_i)_\ast\mathcal U\) converges in \(X_i\).
Proof. If \(\mathcal U\to x\), then every neighborhood of \(x_i\) pulls back to a neighborhood of \(x\), so the projected ultrafilter converges to \(x_i\). Conversely, if \((\pi_i)_\ast\mathcal U\to x_i\) for every \(i\), then every subbasic neighborhood \(\pi_i^{-1}(V_i)\) of \(x=(x_i)\) belongs to \(\mathcal U\). Finite intersections of such subbasic neighborhoods also belong to \(\mathcal U\), so every basic neighborhood of \(x\) belongs to \(\mathcal U\).
Corollary. Tychonoff's theorem follows because every ultrafilter on the product has compact coordinate projections and therefore converges coordinatewise.
Proof. Each \(X_i\) is compact, so \((\pi_i)_\ast\mathcal U\) converges to some \(x_i\). The theorem above gives convergence of \(\mathcal U\) to \((x_i)\). The ultrafilter compactness criterion then gives compactness of the product.
Consequences
Many compactness results in analysis are disguised applications of Tychonoff.
Corollary. The cube \([0,1]^I\) is compact for every index set \(I\).
Proof. Each factor \([0,1]\) is compact; apply Tychonoff.
Example. Banach-Alaoglu can be viewed as compactness of a weak-star closed subset of a product of compact scalar disks.
Remark. The theorem is powerful but topological compactness in huge products is often nonsequential; nets or filters are the natural language there.