Compact products, ultrafilters and choice

Try

Finite products of compact spaces are compact, but an arbitrary product requires a proof that handles infinitely many coordinate choices. The product topology helps: each basic neighborhood still imposes only finitely many conditions.

Ultrafilters package compatible choices of large subsets. They also reveal the set-theoretic dependencies hidden by a casual instruction to “choose a point in every factor.” This lesson works in the usual set theory with the axiom of choice and states where maximal principles enter.

Predict

Predict why coordinatewise compactness alone is not a completed proof. One must obtain compatible coordinate limits for a single object on the product and show that every basic neighborhood of the resulting tuple is eventually or filter-large.

Predict whether the box topology supports the same argument. It does not: a box can impose infinitely many conditions, and a filter is only required to combine finitely many of its members by intersection.

Experiment

Track the logical tools.

Tool	Role
Proper filter	Upward-closed family, finite intersections, no empty set
Ultrafilter	Maximal proper filter
Zorn’s lemma	Extends a proper filter to an ultrafilter
Compactness	Makes every ultrafilter converge
Product basis	Combines finitely many coordinate neighborhoods

The axiom of choice implies Zorn’s lemma. The ultrafilter extension principle is a stated choice dependency, not an elementary consequence of the topology axioms.

Understand

A proper filter on X contains X, excludes the empty set, is upward closed, and is closed under finite intersections. To extend one to an ultrafilter, order its proper extensions by inclusion. The union of a chain is still proper; any two chosen members lie in one common chain element, so their intersection remains present. Zorn’s lemma gives a maximal extension.

An ultrafilter contains either A or its complement. If A is absent and met every filter member, adjoining A would generate a larger proper filter. Thus some member misses A, forcing the complement into the ultrafilter. Both cannot occur because their intersection is empty.

Compactness is equivalent to convergence of every ultrafilter. In a compact space the closures of filter members have the finite intersection property. Choose x in their total intersection. If an open neighborhood V of x were not in the ultrafilter, its closed complement would be, contradicting x belonging to that member’s closure. Conversely, extend the proper filter generated by any closed finite-intersection family to an ultrafilter. A limit belongs to each closed member, proving the compactness criterion.

For an ultrafilter on a product, push it through each coordinate projection; preimage identities show the result is an ultrafilter on that factor. Choose a limit x_i in each compact factor, using choice where needed. Each basic neighborhood of the tuple x imposes finitely many coordinate neighborhoods. Their preimages belong to the original ultrafilter, so their finite intersection does too. Hence it converges to x, and the product is compact. Empty cases are immediate.

Apply

Explain why a principal ultrafilter at a point always converges to that point, and why this observation alone does not prove compactness: nonprincipal ultrafilters must be handled too.

Repair “The proof uses only finite intersections, so no choice is involved.” Finite intersections explain the final product step. Extending filters by a maximal principle and selecting coordinate limits are separate set-theoretic ingredients. For compact Hausdorff factors, each coordinate limit is unique, but the ultrafilter extension dependency still remains.

Continue

The next construction uses a countable cube to turn continuous separating functions into a metric. Later, Stone–Čech uses a much larger cube and therefore the full product compactness mechanism. Optional depth: compare the finite tube-lemma proof with this ultrafilter proof and identify which steps genuinely require infinitely many factors.