Why Choice Appears in Product Compactness
Tychonoff’s theorem says that products of compact spaces are compact. Its ultrafilter proof involves two different tasks: obtaining ultrafilters and choosing limits in the factors. These tasks do not always require the same strength of choice. In particular, the compact-Hausdorff version and the unrestricted version of the theorem have different standard set-theoretic relationships to the axiom of choice.
We work in set theory without assuming the axiom of choice unless it is explicitly invoked. A family \((A_i)_{i\in J}\) of sets is nonempty-valued if \(A_i\ne\varnothing\) for each \(i\in J\). A choice function for this family is a function \(c\) with domain \(J\) such that \(c(i)\in A_i\) for every \(i\in J\).
The UEP is a choice principle: it promises an extension, but does not specify a unique extension. This matters when the principle is used for a whole family of sets or filters. The UEP is not the axiom of choice itself.
How the Axiom of Choice Gives Ultrafilter Extensions
Proof. Let \(\mathcal{F}\) be a proper filter on \(S\). Consider the set of all proper filters on \(S\) that contain \(\mathcal{F}\), ordered by inclusion. This collection is nonempty because it contains \(\mathcal{F}\). If \(\mathcal{C}\) is a chain in this ordered set, its union \(\mathcal{G}=\bigcup_{\mathcal{H}\in\mathcal{C}}\mathcal{H}\) is a proper filter containing \(\mathcal{F}\). Indeed, \(S\in\mathcal{G}\), while \(\varnothing\notin\mathcal{G}\). If \(A,B\in\mathcal{G}\), they belong to two members of the chain, one of which contains the other; hence both belong to one filter in the chain, so \(A\cap B\in\mathcal{G}\). The upward-closure condition follows in the same way from the filter containing the given member. Thus every chain has an upper bound. By Zorn’s lemma, the ordered set has a maximal member \(\mathcal{U}\). Zorn’s lemma is equivalent to AC, so its use here is legitimate under the stated assumption.
It remains to check that \(\mathcal{U}\) is an ultrafilter, rather than just a maximal filter in the restricted collection. Suppose \(A\subseteq S\) and \(A\notin\mathcal{U}\). If every \(B\in\mathcal{U}\) met \(A\), then the sets \(B\cap A\), for \(B\in\mathcal{U}\), would generate a proper filter containing \(\mathcal{U}\) and \(A\). It would be proper because every finite intersection of its generators contains a set \(B\cap A\ne\varnothing\), using closure of \(\mathcal{U}\) under finite intersections and the supposition that every member meets \(A\). This would contradict maximality. Therefore some \(B\in\mathcal{U}\) satisfies \(B\cap A=\varnothing\), so \(B\subseteq S\setminus A\), and upward closure gives \(S\setminus A\in\mathcal{U}\). Thus for every \(A\subseteq S\), either \(A\in\mathcal{U}\) or \(S\setminus A\in\mathcal{U}\), which is the ultrafilter property. \(\square\)
In the proof, AC enters through Zorn’s lemma. The UEP is weaker: the standard set-theoretic result is that it is equivalent to the Boolean prime ideal theorem (also called the ultrafilter theorem), and it does not imply full AC. The strictness is a theorem of set theory: in appropriate models of ZF, the Boolean prime ideal theorem holds while AC fails. That independence result is not proved here.
What the Ultrafilter Proof Needs
We use the following facts from the earlier material on ultrafilters and compactness. A space is compact if and only if every ultrafilter on its underlying set converges. The image of an ultrafilter under a coordinate projection is an ultrafilter on that coordinate space. In a compact space every ultrafilter converges, and in a Hausdorff space a convergent ultrafilter has at most one limit.
Proof. Let \(X=\prod_{i\in J}X_i\), where each \(X_i\) is compact and Hausdorff. The case \(J=\varnothing\) is a singleton and is compact. If some factor is empty, then \(X=\varnothing\), which is compact. Otherwise suppose \(X\ne\varnothing\), and let \(\mathcal{U}\) be any ultrafilter on \(X\). For each \(i\in J\), its coordinate image \((\pi_i)_*\mathcal{U}\) is an ultrafilter on \(X_i\), so compactness gives a limit \(x_i\in X_i\). Hausdorffness makes that limit unique. Therefore the assignment \(i\mapsto x_i\) defines a point \(x=(x_i)_{i\in J}\in X\), without making any further choices among multiple possible limits. For every basic product neighborhood of \(x\), only finitely many coordinates are restricted; convergence of the coordinate ultrafilters implies that the corresponding inverse images under those finitely many projections belong to \(\mathcal{U}\). Their finite intersection belongs to \(\mathcal{U}\), so the basic neighborhood belongs to \(\mathcal{U}\). Hence \(\mathcal{U}\) converges to \(x\). Since every ultrafilter on \(X\) converges, the ultrafilter characterization of compactness gives that \(X\) is compact. \(\square\)
The uniqueness step is the key point. For general compact spaces, an ultrafilter can have more than one limit. To define a point of the product, one would need to select a limit in each factor. The UEP alone does not provide such a family of selections. The standard set-theoretic relationship is that Tychonoff’s theorem for compact Hausdorff spaces is equivalent, over ZF, to the Boolean prime ideal theorem, hence to the UEP. The unrestricted product theorem has a different formulation and strength, as the next examples clarify.
Worked Example: Two Ultrafilter Extensions on a Finite Set
Let \(S=\{a,b,c\}\), and let \(\mathcal{F}=\{S,\{a,b\}\}\). This is a proper filter: it contains \(S\), excludes \(\varnothing\), and the intersection of any two of its members is one of its members.
The principal ultrafilter at \(a\) is \(\mathcal{U}_a=\{A\subseteq S:a\in A\}\). Its members are \(S,\{a,b\},\{a,c\},\{a\}\). It contains \(\mathcal{F}\), since both \(S\) and \(\{a,b\}\) contain \(a\). The principal ultrafilter at \(b\) is \(\mathcal{U}_b=\{A\subseteq S:b\in A\}\). Its members are \(S,\{a,b\},\{b,c\},\{b\}\), so it also contains \(\mathcal{F}\). Each is an ultrafilter because for every subset \(A\subseteq S\), exactly one of \(A\) and \(S\setminus A\) contains its designated point.
These extensions are distinct: \(\{a\}\in\mathcal{U}_a\), while \(\{a\}\notin\mathcal{U}_b\). Thus the extension principle asserts existence, not a uniquely specified extension. On this finite example the alternatives can be listed directly; for an arbitrary family of filters, making simultaneous selections is a separate issue.
Two Formulations of Tychonoff’s Theorem
In ordinary topology, the phrase “the product of compact spaces is compact” is commonly used with AC in the background. In set theory without AC, it is important to say whether the theorem asserts only compactness or also nonemptiness when every factor is nonempty. The distinction is not cosmetic: the empty space is compact under the open-cover definition.
Worked Example: The Empty Product and an Empty Factor
The product over an index set \(J\) is the set of functions \(f\) with domain \(J\) such that \(f(i)\in X_i\) for each \(i\in J\). If \(J=\varnothing\), there is exactly one such function: the empty function. Thus the empty product is a singleton, not the empty set. Its only topologies are necessarily \(\{\varnothing,\{\varnothing\}\}\), and it is compact because every open cover of a singleton has a one-set subcover.
If \(J\ne\varnothing\) and some factor \(X_k=\varnothing\), no function can satisfy \(f(k)\in X_k\), so the product is empty. The empty space is compact: every open cover has the empty finite subfamily as a subcover, since its union is the whole empty space. These two cases differ as sets, even though both products are compact.
Check: for \(J=\varnothing\), the product contains the empty function; for \(J\ne\varnothing\) with \(X_k=\varnothing\), it contains no function. In neither case does compactness itself imply that a product of nonempty factors is nonempty.
Worked Example: Compactness Does Not Supply a Choice Function
Let \((A_i)_{i\in J}\) be any nonempty-valued family of sets, and give each \(A_i\) the indiscrete topology \(\{\varnothing,A_i\}\). Each \(A_i\) is compact. In fact, any open cover of nonempty \(A_i\) must contain \(A_i\) itself: the only other open set is empty, and empty sets cannot cover a point. Thus one member of the cover suffices.
The product \(\prod_{i\in J}A_i\), as a set, consists exactly of the choice functions for the family. If no choice function exists, this product is empty and is still compact. Consequently, the assertion “the product is compact” alone does not imply that the family has a choice function, even though each factor is compact.
If instead the unrestricted product theorem is stated to assert that every product of nonempty compact spaces is both compact and nonempty, then applying it to these indiscrete spaces gives a choice function for every nonempty-valued family. That nonemptiness clause implies AC directly. Conversely, AC supplies nonemptiness and the usual unrestricted Tychonoff theorem. This is the sense in which the full unrestricted product theorem, with nonemptiness included, has the strength of AC.
For compact Hausdorff factors, the standard set-theoretic formulation concerns the compactness conclusion; it is equivalent to the UEP, not to full AC. This theorem-level equivalence is a standard result in set theory, and its proof is beyond this tutorial. It should not be confused with the unrestricted theorem that also asserts products of nonempty factors are nonempty.
Worked Example: A Product Whose Point Is Explicit
Let \(J=\{1,2\}\), \(X_1=\{0,1\}\), and \(X_2=\{u,v\}\), each with the discrete topology. Every finite topological space is compact: given an open cover, choose for each point one covering member; there are only finitely many points, so the chosen members form a finite subcover. The product set has the four points \((0,u),(0,v),(1,u),(1,v)\), and it is compact by the same finite-set argument.
For instance, \((0,u)\) is a point of the product because its two coordinate values satisfy \(0\in X_1\) and \(u\in X_2\). This verification involves only two explicit choices. For an arbitrary family, the corresponding assignment \(i\mapsto x_i\) is precisely a choice function, and it cannot be silently assumed merely because each factor is nonempty.
Common Pitfall: Conflating Extension and Selection
A frequent error is to say, “The UEP chooses a limit in every factor, so the product is compact.” The UEP extends a filter to an ultrafilter; it does not select one point from each of an arbitrary family of nonempty sets. For compact Hausdorff factors, each coordinate ultrafilter has a unique limit, so the family of limits is determined. For general compact factors, the limit sets may contain several points, and selecting one from each is an additional choice issue.
A second error is to infer that a compact product must be nonempty. Compactness means every open cover has a finite subcover, and the empty space satisfies this definition. When discussing choice, state separately whether nonemptiness of products of nonempty factors is part of the theorem.
| Statement | Set-theoretic relationship | Important qualification |
|---|---|---|
| AC implies the UEP | Proved using Zorn’s lemma | The UEP does not imply full AC |
| Tychonoff for compact Hausdorff spaces | Equivalent over ZF to the UEP, or Boolean prime ideal theorem | This is a standard set-theoretic equivalence |
| Unrestricted product theorem with nonemptiness included | Equivalent in strength to AC | Nonempty products are explicitly asserted to be nonempty |
| Compactness of an empty product | Always holds | The empty product is a singleton |
Check Your Understanding
For each question, keep separate the existence of an ultrafilter, the selection of limits, and the nonemptiness of a product.
- State the ultrafilter extension principle and explain why it does not assert a unique extension.
- In the finite example above, verify that \(\mathcal{U}_a\) and \(\mathcal{U}_b\) both contain \(\mathcal{F}\), but are unequal.
- Why does Hausdorffness remove the coordinatewise selection issue in the ultrafilter proof?
- Describe the empty product as a set, and distinguish it from a product with an empty factor.
- How does the nonemptiness clause for products of nonempty indiscrete compact spaces yield the axiom of choice?