Compactness Characterized by Ultrafilters

Compactness Through Ultrafilter Convergence

Compactness is defined by open covers: every open cover must have a finite subcover. Ultrafilters give a different test, phrased in terms of convergence. The equivalence is useful because it turns a cover with no finite subcover into a single object that records the obstruction to compactness. The converse direction uses the ultrafilter extension principle, whose use of the axiom of choice will be made explicit.

Let $(X,\mathcal{T})$ be a topological space. Recall that $X$ is compact if every family of open sets whose union is $X$ has a finite subfamily whose union is $X$. A filter on $X$ 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 of $\mathcal{F}$ belongs to $\mathcal{F}$. These requirements make $\mathcal{F}$ a proper filter.

Definition: An ultrafilter on $X$ is a proper filter maximal under inclusion among proper filters on $X$. A filter $\mathcal{F}$ converges to $x\in X$ if every open neighborhood of $x$ belongs to $\mathcal{F}$.

Two elementary properties of ultrafilters will be used. First, for every $A\subseteq X$, exactly one of $A$ and $X\setminus A$ belongs to an ultrafilter $\mathcal{U}$. They cannot both belong, since their intersection is empty. If $A\notin\mathcal{U}$, adjoining $A$ cannot generate a proper filter, by maximality. Thus some $F\in\mathcal{U}$ has $F\cap A=\varnothing$; otherwise all finite intersections used to generate the larger filter would be nonempty. Then $F\subseteq X\setminus A$, so upward closure gives $X\setminus A\in\mathcal{U}$. Second, if a finite union $A_1\cup\cdots\cup A_m$ belongs to $\mathcal{U}$, at least one $A_i$ belongs to $\mathcal{U}$. If none did, all $X\setminus A_i$ would belong, and their intersection would be disjoint from the union, contradicting properness.

The Compactness Characterization

Theorem: A topological space $X$ is compact if and only if every ultrafilter on $X$ converges to at least one point of $X$.

Proof. Suppose first that $X$ is compact, and let $\mathcal{U}$ be an ultrafilter on $X$. For each $x\in X$, suppose there were an open neighborhood $U_x$ of $x$ with $U_x\notin\mathcal{U}$. The ultrafilter property gives $X\setminus U_x\in\mathcal{U}$. The family $\{U_x:x\in X\}$ is an open cover of $X$, so compactness gives points $x_1,\ldots,x_m$ for which $U_{x_1},\ldots,U_{x_m}$ cover $X$. Then $$ \bigcap_{i=1}^{m}(X\setminus U_{x_i}) = X\setminus\bigcup_{i=1}^{m}U_{x_i} = \varnothing. $$ Every set $X\setminus U_{x_i}$ belongs to $\mathcal{U}$, so their finite intersection must belong too. That would put $\varnothing$ in a proper filter, a contradiction. Therefore some $x\in X$ has every open neighborhood in $\mathcal{U}$, and $\mathcal{U}$ converges to $x$.

Conversely, suppose every ultrafilter on $X$ converges, and let $\{U_\alpha:\alpha\in J\}$ be an open cover of $X$. Assume for contradiction that it has no finite subcover. Set $C_\alpha=X\setminus U_\alpha$. Every finite intersection of the $C_\alpha$ is nonempty: if a finite intersection were empty, the corresponding finite collection of $U_\alpha$'s would cover $X$.

The supersets of finite intersections of the $C_\alpha$'s form a proper filter. Indeed, $X$ contains the empty finite intersection, the assumed nonemptiness ensures that the generated family excludes $\varnothing$, and intersecting two finite intersections gives another finite intersection. By the ultrafilter extension principle, this proper filter is contained in an ultrafilter $\mathcal{U}$. This principle follows from Zorn's lemma and therefore uses the axiom of choice. In particular, $C_\alpha\in\mathcal{U}$ for every $\alpha$.

By assumption, $\mathcal{U}$ converges to some $x\in X$. Since the $U_\alpha$'s cover $X$, choose $\alpha$ with $x\in U_\alpha$. Then $U_\alpha$ is an open neighborhood of $x$, so convergence gives $U_\alpha\in\mathcal{U}$. But $C_\alpha=X\setminus U_\alpha$ also belongs to $\mathcal{U}$, which would force $\varnothing=U_\alpha\cap C_\alpha\in\mathcal{U}$. This contradicts properness. Thus every open cover has a finite subcover, and $X$ is compact.

If $X=\varnothing$, there are no proper filters on $X$: the required membership $X\in\mathcal{F}$ would put $\varnothing$ in $\mathcal{F}$. Hence “every ultrafilter converges” is vacuously true. The empty space is compact because every open cover of it has the empty finite subcover. This handles the degenerate case as well. $\square$

Worked Applications

Worked Example: Ultrafilters on a Finite Space

Let $X=\{a,b,c\}$ with the discrete topology, in which every subset is open. Show that every ultrafilter on $X$ converges.

Let $\mathcal{U}$ be an ultrafilter. The singletons form the finite union $X=\{a\}\cup\{b\}\cup\{c\}$. Since $X\in\mathcal{U}$, the finite-union property shows that at least one singleton belongs to $\mathcal{U}$. At most one can belong: if, for example, both $\{a\}$ and $\{b\}$ belonged, their empty intersection would belong. Thus exactly one singleton, say $\{b\}$, is in $\mathcal{U}$.

Every set containing $b$ belongs to $\mathcal{U}$, by upward closure from $\{b\}$. No set omitting $b$ belongs, since its intersection with $\{b\}$ is empty. Thus $\mathcal{U}$ consists exactly of the subsets containing $b$. Every open neighborhood of $b$ contains $b$, so it belongs to $\mathcal{U}$, and $\mathcal{U}$ converges to $b$. This agrees with compactness: an open cover of a three-point space can be reduced to at most three cover members, one for each point.

Worked Example: The Cofinite Topology

Let $X$ be an infinite set with the cofinite topology: the open sets are $\varnothing$ and all subsets whose complements in $X$ are finite. First check compactness. In any open cover, choose a nonempty member $U$. Its complement $X\setminus U$ is finite. For each point of this finite complement, choose one cover member containing it. The chosen $U$, together with those finitely many members, covers $X$. So $X$ is compact.

Now let $\mathcal{U}$ be an ultrafilter on $X$. If some singleton $\{p\}$ belongs to $\mathcal{U}$, the same upward-closure argument as in the finite example shows that every open neighborhood of $p$ belongs to $\mathcal{U}$. Hence it converges to $p$.

If no singleton belongs, the ultrafilter property puts $X\setminus\{p\}$ in $\mathcal{U}$ for every $p\in X$. Any cofinite set $V$ is the intersection of $X\setminus\{p\}$ over the finitely many points $p\in X\setminus V$. Thus every cofinite set belongs to $\mathcal{U}$. Every open neighborhood of every point is cofinite, so $\mathcal{U}$ converges to every point. The two cases show directly how the ultrafilter criterion works here.

Worked Example: A Nonconvergent Ultrafilter on the Real Line

Give a nonconvergent ultrafilter on $\mathbb{R}$ with its standard topology. For each $n\in\mathbb{N}$, set $C_n=\mathbb{R}\setminus(-n,n)$. A finite intersection of these sets is nonempty: if $N$ is the largest index in that finite collection, then $N+1$ belongs to each $C_n$ in it, because $N+1\ge n$ and hence $N+1\notin(-n,n)$.

The supersets of these finite intersections form a proper filter. Extend it, by the ultrafilter extension principle, to an ultrafilter $\mathcal{U}$; thus $C_n\in\mathcal{U}$ for every $n$. Suppose $\mathcal{U}$ converges to $x\in\mathbb{R}$. Choose an integer $N>|x|+1$. Then $x\in(-N,N)$, so $(-N,N)$ is an open neighborhood of $x$ and convergence requires $(-N,N)\in\mathcal{U}$. But $C_N=\mathbb{R}\setminus(-N,N)$ is also in $\mathcal{U}$. Their intersection is empty, a contradiction. Therefore $\mathcal{U}$ does not converge. Correspondingly, the cover $\{(-n,n):n\in\mathbb{N}\}$ has no finite subcover: the union of any finite subcollection is $(-N,N)$ for its largest index $N$, and it omits $N+1$.

A Common Pitfall: Filters Are Not Enough

The theorem concerns ultrafilters, not all proper filters. For a concrete contrast, take $X=[0,1]$ with the standard topology and define $\mathcal{F}$ to consist of all subsets of $X$ whose complements are finite. This is a proper filter: finite intersections still have finite complement, supersets preserve that property, $X$ has finite complement, and $\varnothing$ does not.

It does not converge to any $x\in[0,1]$. For each such $x$, choose a relative open neighborhood $V$ with infinite complement. For example, if $x\le 1/2$, take $V=[0,1]\cap(x-1/4,x+1/4)$; it omits the interval $(3/4,1]$, apart from no points that could make this complement finite. If $x>1/2$, take $V=[0,1]\cap(x-1/4,x+1/4)$; it omits $[0,1/4)$, again an infinite set. In both cases $V$ is a neighborhood of $x$ with infinite complement, so $V\notin\mathcal{F}$. Thus $\mathcal{F}$ fails the convergence condition at every point.

This does not contradict compactness of $[0,1]$: compactness guarantees convergence for ultrafilters. In the proof of the converse, the extension principle is what turns the filter generated by a cover with no finite subcover into an ultrafilter, to which the assumed convergence condition applies. Treating an arbitrary filter as though it were an ultrafilter skips that essential step.

Why the Characterization Is Useful

Open covers are global objects: they involve a family of sets covering the whole space. Ultrafilters package compatible finite-intersection information into one maximal filter. The theorem says that compactness is precisely the condition preventing any such ultrafilter from escaping all possible limits. In particular, the forward proof needs only compactness and the ultrafilter complement property; the reverse proof needs the ultrafilter extension principle. Keeping track of these hypotheses matters when applying the criterion or discussing its reliance on choice.

Check Your Understanding

Use the definitions and the two directions of the theorem to answer these questions.

Why can an ultrafilter not contain both a set $A$ and its complement $X\setminus A$?
In the compact-to-convergence proof, where exactly is compactness used?
Why do the complements of an open cover with no finite subcover generate a proper filter?
For the cofinite topology on an infinite set, explain why an ultrafilter containing no singleton converges to every point.
In the real-line example, identify both the ultrafilter members that contradict convergence to $x$.