Ultrafilter Limits in Compact Spaces

Ultrafilter Convergence

Sequences are useful for describing limits in many familiar spaces, but they do not capture all the ways that points can be approached in general topological spaces. An ultrafilter gives a more flexible notion of convergence: instead of recording one point at each natural-number stage, it records a coherent collection of subsets. Compactness then has a concise consequence: every ultrafilter on a compact space converges.

We first recall the set-theoretic definitions needed here. A proper filter on a nonempty set \(X\) is a family \(\mathcal{F}\subseteq\mathcal{P}(X)\) such that \(X\in\mathcal{F}\), \(\varnothing\notin\mathcal{F}\), the intersection of any two members belongs to \(\mathcal{F}\), and every superset in \(X\) of a member belongs to \(\mathcal{F}\). An ultrafilter is a proper filter maximal under inclusion among proper filters.

The definition requires every neighborhood, not merely some neighborhood, to lie in the ultrafilter. Maximality gives a useful test for this condition. For every subset \(A\subseteq X\), an ultrafilter \(\mathcal{U}\) contains exactly one of \(A\) and \(X\setminus A\). Indeed, they cannot both belong to a proper filter, because their intersection is empty. If \(A\notin\mathcal{U}\), maximality implies that adjoining \(A\) cannot generate a proper filter. Thus some \(B\in\mathcal{U}\) satisfies \(B\cap A=\varnothing\); otherwise the supersets of sets \(B\cap A\), for \(B\in\mathcal{U}\), would form a proper filter containing both \(\mathcal{U}\) and \(A\). Since \(B\subseteq X\setminus A\), upward closure gives \(X\setminus A\in\mathcal{U}\).

A Closed-Set Test for Limits

Proof. Suppose first that \(\mathcal{U}\) converges to \(x\). If a closed set \(C\in\mathcal{U}\) did not contain \(x\), then \(X\setminus C\) would be an open neighborhood of \(x\), so convergence would give \(X\setminus C\in\mathcal{U}\). The filter would then contain \(C\cap(X\setminus C)=\varnothing\), contradicting properness. Hence every closed member contains \(x\).

Conversely, suppose every closed member of \(\mathcal{U}\) contains \(x\). Let \(V\) be any open neighborhood of \(x\). If \(V\notin\mathcal{U}\), the ultrafilter property gives \(X\setminus V\in\mathcal{U}\). This complement is closed, but it does not contain \(x\), contradicting the supposition. Therefore \(V\in\mathcal{U}\) for every open neighborhood \(V\) of \(x\), which is precisely convergence to \(x\). \(\square\)

This test converts convergence into a statement about closed sets. Compactness is naturally suited to it because compactness controls families of closed sets whose finite intersections are nonempty. A family has the finite intersection property if the intersection of every finite subfamily is nonempty. The empty finite intersection is understood to be \(X\).

Proof. Let \(\{C_\alpha:\alpha\in J\}\) be a family of closed sets with the finite intersection property. Suppose, to obtain a contradiction, that \(\bigcap_{\alpha\in J}C_\alpha=\varnothing\). Then the open sets \(X\setminus C_\alpha\) cover \(X\): each point fails to belong to at least one \(C_\alpha\). By compactness, finitely many of these open sets cover \(X\), say \(X\setminus C_{\alpha_1},\ldots,X\setminus C_{\alpha_n}\). Taking complements, this says \(C_{\alpha_1}\cap\cdots\cap C_{\alpha_n}=\varnothing\), contradicting the finite intersection property. Thus the total intersection is nonempty. \(\square\)

Compactness Forces Convergence

Proof. Consider the family \(\mathcal{C}\) of all closed sets that belong to \(\mathcal{U}\). This family has the finite intersection property. In fact, if \(C_1,\ldots,C_n\in\mathcal{C}\), then their intersection belongs to \(\mathcal{U}\), because a filter is closed under finite intersections. It cannot be empty, because \(\mathcal{U}\) is proper. The closed-set lemma therefore gives a point \(x\in\bigcap_{C\in\mathcal{C}}C\). Every closed set in \(\mathcal{U}\) contains \(x\), so the closed-set test shows that \(\mathcal{U}\) converges to \(x\). \(\square\)

The proof uses compactness exactly once: it turns the finite intersection property of the closed members into a common point. The point obtained need not be unique in every compact space. If \(X\) is Hausdorff, however, a convergent ultrafilter has at most one limit. To verify this, suppose it converges to distinct points \(x\) and \(y\). The Hausdorff condition supplies disjoint open neighborhoods \(V\) of \(x\) and \(W\) of \(y\). Convergence would put both \(V\) and \(W\) in the ultrafilter, so it would also put \(V\cap W=\varnothing\) in the ultrafilter, which is impossible for a proper filter. Thus in a compact Hausdorff space every ultrafilter has exactly one limit.

A Common Pitfall: Dropping “Ultrafilter”

It is incorrect to replace “ultrafilter” by “filter” in the compactness theorem. For a concrete counterexample, let \(X=\{a,b\}\) again have the discrete topology and consider \(\mathcal{F}=\{X\}\). This is a proper filter: its only member is \(X\), and the filter axioms hold. But it does not converge to \(a\), because the open neighborhood \(\{a\}\) is not in \(\mathcal{F}\). It does not converge to \(b\), because the open neighborhood \(\{b\}\) is not in \(\mathcal{F}\). The space is compact, since every open cover of a finite space has a finite subcover. The missing hypothesis is maximality: \(\mathcal{F}\) is not an ultrafilter, as it can be properly enlarged to \(\mathcal{U}_a\) or to the principal ultrafilter at \(b\).

The closed-set perspective makes the distinction visible. The closed members of \(\mathcal{F}\) consist only of \(X\), whose intersection is \(X\); that does not select a point whose every neighborhood belongs to \(\mathcal{F}\). For an ultrafilter, the dichotomy between each set and its complement makes the closed-set test equivalent to convergence. In a compact space, finite intersections then provide the common point required by that test.