Stone-Cech Compactifications Of Discrete Spaces

Why Ultrafilters Describe the Compactification

A discrete space has every subset open, so any function from it to a topological space is continuous. The difficulty in extending such a function to a compact space is not continuity on the original domain; it is deciding what value to assign to each new point. Ultrafilters provide those points and specify, for every map into a compact Hausdorff space, the value forced at each one.

We will construct a compact Hausdorff space \(\beta X\) containing a discrete space \(X\) as a dense subspace. Its key property is that every map from \(X\) into a compact Hausdorff space extends uniquely to a continuous map on \(\beta X\). Throughout, \(X\) has the discrete topology unless otherwise stated.

A useful consequence of maximality is that for every \(A\subseteq X\), exactly one of \(A\) and \(X\setminus A\) belongs to an ultrafilter \(\mathcal{U}\). They cannot both belong, because their intersection is empty. To see that at least one belongs, suppose \(A\notin\mathcal{U}\). If every \(F\in\mathcal{U}\) met \(A\), adjoining \(A\) would generate a proper filter strictly containing \(\mathcal{U}\), contrary to maximality. Thus some \(F\in\mathcal{U}\) has \(F\cap A=\varnothing\); then \(F\subseteq X\setminus A\), so \(X\setminus A\in\mathcal{U}\).

Proof. The sets containing a finite intersection of members of \(\mathcal{F}\) form a proper filter: the finite intersection property prevents the empty set from entering it, and the filter closure conditions follow from taking unions of finite lists of members of \(\mathcal{F}\). Consider the partially ordered set of proper filters containing this filter, ordered by inclusion. The union of any chain is a proper filter: any two of its members lie together in one filter in the chain, and no member of the chain contains the empty set. Thus every chain has an upper bound. Zorn’s lemma gives a maximal proper filter containing \(\mathcal{F}\), which is an ultrafilter. \(\square\)

The Ultrafilter Space \(\beta X\)

For now assume \(X\ne\varnothing\). Define \(\beta X\) to be the set of all ultrafilters on \(X\). Give it the topology with basis consisting of the sets \([A]\), for all \(A\subseteq X\). This is a basis because \([X]=\beta X\), and \([A]\cap[B]=[A\cap B]\). In particular, \([A]\) is both open and closed: its complement in \(\beta X\) is \([X\setminus A]\), by the ultrafilter dichotomy.

Proof. First, \(\delta_x\) is an ultrafilter: it is a proper filter, and for every \(A\subseteq X\), it contains \(A\) or \(X\setminus A\) according as \(x\in A\) or \(x\notin A\). Distinct points give distinct ultrafilters, since \(\{x\}\in\delta_x\) but \(\{x\}\notin\delta_y\) when \(y\ne x\). Also, \([A]\cap\{\delta_x:x\in X\}=\{\delta_x:x\in A\}\). In particular \([\{x\}]\) meets the embedded copy of \(X\) in just \(\delta_x\), and it contains \(\delta_x\). Thus that copy has the discrete subspace topology.

To prove density, take a nonempty open subset of \(\beta X\). It contains a nonempty basis set \([A]\), so some ultrafilter contains \(A\), which implies \(A\ne\varnothing\). Choose \(x\in A\). Then \(\delta_x\in[A]\), so every nonempty open subset meets the embedded copy of \(X\).

If \(\mathcal{U}\ne\mathcal{V}\), choose \(A\in\mathcal{U}\) with \(A\notin\mathcal{V}\), interchanging the two ultrafilters if necessary. The ultrafilter dichotomy gives \(X\setminus A\in\mathcal{V}\). The disjoint open neighborhoods \([A]\) of \(\mathcal{U}\) and \([X\setminus A]\) of \(\mathcal{V}\) show that \(\beta X\) is Hausdorff.

For compactness, first suppose a cover consists of basis sets \([A_i]\), indexed by \(i\in J\). If it has no finite subcover, then no finite collection of the \(A_i\) covers \(X\). Consequently the family \(\{X\setminus A_i:i\in J\}\) has the finite intersection property. By the lemma, an ultrafilter \(\mathcal{U}\) contains all the \(X\setminus A_i\). It therefore belongs to none of the \([A_i]\), contradicting that they cover \(\beta X\). So every basis-set cover has a finite subcover. For any open cover, take all basis sets contained in its members; these form a basis-set cover, and a finite subcover of those sets yields a finite subcover from the original cover. Hence \(\beta X\) is compact. \(\square\)

When \(X=\varnothing\), there are no ultrafilters, and we define \(\beta X=\varnothing\). It is compact and Hausdorff, and the empty space embeds densely in it. The construction and extension property below are then immediate for the unique map with empty domain.

Ultrafilter Limits in Compact Hausdorff Spaces

To extend a map, we need a precise meaning for the limit of an ultrafilter. A filter \(\mathcal{V}\) on a topological space \(Y\) converges to \(y\in Y\) if every open neighborhood of \(y\) belongs to \(\mathcal{V}\). For a map \(f:X\to Y\) and an ultrafilter \(\mathcal{U}\) on \(X\), define its image ultrafilter by \(f_*\mathcal{U}=\{B\subseteq Y:f^{-1}(B)\in\mathcal{U}\}\). Preimages preserve inclusions, intersections, the whole set and the empty set, so \(f_*\mathcal{U}\) is a proper filter. The ultrafilter dichotomy for \(\mathcal{U}\), applied to \(f^{-1}(B)\), shows that \(f_*\mathcal{U}\) is an ultrafilter.

Proof. Let \(\mathcal{V}\) be an ultrafilter on compact \(Y\). The closed sets \(\overline{B}\), for \(B\in\mathcal{V}\), have the finite intersection property. Indeed, for any finite list \(B_1,\ldots,B_n\in\mathcal{V}\), the intersection \(B_1\cap\cdots\cap B_n\) belongs to the proper filter and is nonempty, and it lies in \(\overline{B_1}\cap\cdots\cap\overline{B_n}\). Compactness implies that these closed sets have a common point \(y\); otherwise their open complements would cover \(Y\), and a finite subcover would contradict the finite intersection property. If an open neighborhood \(V\) of \(y\) did not belong to \(\mathcal{V}\), then \(Y\setminus V\in\mathcal{V}\). Since \(Y\setminus V\) is closed, \(y\in\overline{Y\setminus V}=Y\setminus V\), contradicting \(y\in V\). Thus \(\mathcal{V}\) converges to \(y\). If it converged to two distinct points, the Hausdorff property would give disjoint open neighborhoods of them. Both would belong to \(\mathcal{V}\), so their empty intersection would belong to the proper filter, a contradiction. \(\square\)

The Extension Property

Proof. For each ultrafilter \(\mathcal{U}\in\beta X\), the image ultrafilter \(f_*\mathcal{U}\) converges to exactly one point of \(Y\), by the proposition. Define \(\widetilde f(\mathcal{U})\) to be that point. For a principal ultrafilter \(\delta_x\), the image ultrafilter is \(\delta_{f(x)}\), which converges to \(f(x)\); hence \(\widetilde f(\delta_x)=f(x)\).

We prove continuity. We use the following consequence of compactness and Hausdorffness: if \(y\in U\) and \(U\) is open in \(Y\), there is an open \(V\) with \(y\in V\) and \(\overline V\subseteq U\). To verify it, the closed set \(K=Y\setminus U\) is compact. For each \(z\in K\), choose disjoint open neighborhoods \(O_z\) of \(z\) and \(W_z\) of \(y\), using Hausdorffness. Finitely many \(O_z\) cover \(K\), by compactness. Intersect the corresponding finitely many \(W_z\) to obtain an open neighborhood \(V\) of \(y\). The union of those finitely many \(O_z\) is open, contains \(K\), and is disjoint from \(V\). Its closed complement contains \(\overline V\) and lies in \(U\), as required. If \(K\) is empty, take \(V=Y\).

Now let \(\mathcal{U}\in\widetilde f^{-1}(U)\), where \(U\) is open, and put \(y=\widetilde f(\mathcal{U})\). Choose \(V\) as above, and set \(A=f^{-1}(V)\). Since \(f_*\mathcal{U}\) converges to \(y\), it contains \(V\), so \(A\in\mathcal{U}\). If \(\mathcal{W}\in[A]\), then \(V\in f_*\mathcal{W}\), and \(f_*\mathcal{W}\) converges to \(z=\widetilde f(\mathcal{W})\). This implies \(z\in\overline V\): if not, the open neighborhood \(Y\setminus\overline V\) of \(z\) would also belong to \(f_*\mathcal{W}\), contradicting that a proper filter cannot contain two sets with empty intersection. Therefore \(z\in\overline V\subseteq U\). We have found an open neighborhood \([A]\) of \(\mathcal{U}\) contained in \(\widetilde f^{-1}(U)\). This proves continuity.

Finally, suppose \(g,h:\beta X\to Y\) are continuous and agree on the embedded \(X\). The set \(\{p\in\beta X:g(p)=h(p)\}\) is closed: the diagonal \(\{(y,y):y\in Y\}\) is closed in \(Y\times Y\) because \(Y\) is Hausdorff, and the set is its preimage under the continuous map \(p\mapsto(g(p),h(p))\). This closed set contains the dense embedded copy of \(X\), so it is all of \(\beta X\). Thus \(g=h\), proving uniqueness. \(\square\)

Common Pitfall: A Compactification Is Not Just the Original Set

It is wrong to identify every point of \(\beta X\) with a point of \(X\). For infinite \(X\), ultrafilters need not be principal. For example, the cofinite filter on \(\mathbb{N}\) extends to an ultrafilter containing all cofinite sets, and the argument above shows that this ultrafilter is not \(\delta_n\) for any \(n\). The extra points are essential: they are where the extension takes limits.

Another error is to assume that compactness alone makes an ultrafilter limit unique. Compactness gave existence; Hausdorffness gave uniqueness, because it supplies disjoint neighborhoods of distinct points. The extension theorem uses both hypotheses on \(Y\). For discrete \(X\), the construction above supplies a compact Hausdorff space in which those limits are available and proves the extension property directly.