Filters V: Wallman compactifications

The Wallman compactification.

In Filters IV, we have shown that we could realize the Stone-Čech compactification ßX of a discrete space X as its space of ultrafilters of subsets UX.  The topology on the latter had the sets U as a basis, where U = {F in UX | U in F}.

In 1938, Henry Wallman gave a more general construction [1], which applies to any T1 space X, and produces the Stone-Čech compactification ßX of X (up to isomorphism) whenever X is T4.  I will describe it in detail.  Many extensions were given since thence, notably by Orrin Frink in 1964 [2], but I will not say anything about them.

Last time, I’ve also warned you that this post would be technical.  I guess you’ll agree it holds to its promises.  I’ve tried to postpone the hard, Stone-duality based material, as much as I could.  Still, this post is going to be packed.

Instead of considering ultrafilters of subsets, that is, maximal non-trivial filters of subsets, Wallman considers maximal non-trivial filters in the complete lattice HX of closed subsets of X.  (The H is for Hoare: HX is the so-called Hoare powerdomain over X, up to a small detail.  I will talk about it another time.)  Let us write ωX for the subset of maximal non-trivial filters of closed subsets of X.  I’ll say Wallman filter (over X) instead of “maximal non-trivial filter of closed subsets of X”, which starts to be lengthy.

We do as in Filters IV, replacing subsets by closed subsets, ultrafilters of subsets by Wallman filters, and UX by ωX.

In UX, we had defined A as the set of ultrafilters of subsets that contained A, and these were the opens (and also the closed sets) of UX.  We would like to adapt this to ωX.  Reasoning with opens is awkward, here, however, since Wallman filters are filters of closed subsets.  So we shall define C only when C is a closed subset of X, and declare these sets C to be closed.  In other words, the complements of these sets form a subbasis for a topology on ωX.  We shall call it the Wallman topology.

Theorem. ωX is a compact T1 space.

Proof.  As for UX, we do this with filters.

  • [T1] If F and F’ are two distinct Wallman filters, then there is a closed subset C in F and outside F’.  Otherwise, F would be included in F’, hence equal to it by maximality.  Then F’ is in the open complement of C, and F is not (i.e., F is in C), by definition.  By a symmetric argument, from an element of F’ that is not in F, we build an open neighborhood of F that does not contain F’.
  • [Compact] As for UX, we shall use the Kowalsky sum, or a slight variant.  We now define F as the set {C in HX | C is in F}.  Using the fact that # commutes with intersections, we see that F is a non-trivial filter.  It may fail to be maximal, but, by Zorn’s Lemma, F is contained in some Wallman filter F♭’.
    The rest of the argument for compactness is as in the discrete case: F♭’ is a limit of F. It suffices to show that any basic open neighborhood (complement U of a basic closed subset C) of F♭’ is in F… and this is similar to what we did in Filters IV: since F♭’ is in U, it is not in C, so C is not in F♭’ [definition of #], in particular C is not in the smaller filter F; it follows that C is not in F [definition of F], and because F is an ultrafilter, the complement U of C is in F.
    In particular, every ultrafilter F of subsets of ωX has a limit. So ωX is compact.  ◻︎

 

Proposition.  If X is T1, then X embeds into ωX through the map η that sends every point x to the filter of all the closed subsets that contain x.

The assumption that X is T1 is necessary: if X embeds into ωX, since ωX is T1, X must be, too.  Note that I am reusing the notation η for the embedding.  This not quite the same as the one of Filters IV, but close.

Proof.  We use the fact that X is T1 to show that the non-trivial filter η(x) is maximal: if it was not maximal, then there would be a strictly larger filter F, and a closed set C in F that does not contain x; since cl(x)={x} is also in F, and F is non-trivial, the intersection C ∩ {x} cannot be empty, contradiction.

The inverse image η-1(C) is C, so the inverse image of every closed set is closed, hence η is continuous.  The map η is injective: for xy in X, {x} is a closed set in η(x) that is not in η(y).  Finally, the direct image of C is Im η ∩ C, so the inverse η-1 : Im η → X is continuous (the inverse images of closed subsets by η-1 are closed).  Therefore η is an embedding.  ◻︎

 

To go further, we need the following observation: on any distributive lattice, every maximal non-trivial filter F is prime, that is, if F contains the sup of finitely many elements ai, then it contains some ai. In particular, every Wallman filter is prime: a Wallman filter that contains a finite union of closed sets ∪i=1n Ci must contain one of them.

(Here is a proof for two closed sets C and C‘.  I’ll let you generalize.  We first show: (*) C intersects every element of F, or C’ does. Otherwise, for every C” in F, CC” and C‘ ∩ C” are both empty; so (CC’) ∩ C” is, too [this is where distributivity is needed]; but this is impossible since CC’ and C” are both in F, hence also their intersection, but F is non-trivial. Using (*), we show the claim that F is prime. By symmetry, we may assume that C intersects every element of F. Add C to F, and complete so as to obtain a filter: that is, consider the filter F’ of all supersets of sets of the form AC with A in F. F’ is a strictly larger filter than F because C was not in F, and is non-trivial because C intersects every A in F. This is impossible since F is maximal.)

It follows that (∪i=1n Ci) = ∪i=1n Ci, so the sets C form a basis (not just a subbasis) of closed sets: every closed set in the Wallman topology is an intersection of such sets.  (In fact, since # also commutes with finite intersections, every closed set is a filtered intersection of such sets.)

Using this, we can show that every non-empty open subset U of ωX intersects Im η.  U must contain a non-empty basic set, obtained as the complement of some C.  Since this complement is non-empty, there is a Wallman filter F that is not in C.  This means that C is not in F.  In particular, C cannot be the whole of X.  Let x be a point outside C.  Then C is not in η(x), so η(x) is not in C, hence in U.  This implies the following:

Lemma. If X is T1, and modulo the embedding η : XωX, X is a dense subset of ωX.  ◻︎

 

This does not make ωX a compactification of X yet, as we have defined compactifications as compact T2 spaces.  And we know that the spaces that have compactifications are exactly the T3 1/2 spaces.  So we should at least assume that.  We shall in fact require X to be T1 and normal, that is, T4.

This will require us to give an equivalent characterization of the Wallman topology.  For an arbitrary subset A of X, let us define A* as the collection of Wallman filters that contain a closed subset of A.

This allows us to define another topology on ωX: the Wallman* topology has the sets U* as basic open sets, when U ranges over the open subsets of X.  Let us write ωX* for the space of all Wallman filters on X with the Wallman* topology.

Lemma.  For every normal space X, ωX* is T2.

Proof.  Let F, F’ be two distinct Wallman ultrafilters.  There is a closed subset C in F and outside F’, say.

There must be a closed subset C’ in F’ that does not intersect C.  Indeed, otherwise, the collection F” of all the closed subsets that contain the intersection of C with an (arbitrary) element of F’ is again a non-trivial filter: since all the closed subsets C’ in F’ intersect C, all the elements of F” are non-empty; that F” is a filter is by construction. By Zorn’s Lemma, F” is included in a maximal non-trivial filter, and since F” is strictly larger than F’ (as it contains C), this contradicts the maximality of F’.  This proves the claim: there is a closed subset C’ in F’ that does not intersect C.

Since X is normal, we can find two disjoint open neighborhoods, U of C, and U’ of C’.  By construction, F is in U*, and F‘ is in U’*.  It remains to show that the intersection of U* and U’* is empty: any Wallman filter F” that is in both should contain a closed subset of U and a closed subset of U’, hence also their intersection, which is empty; this is impossible since F” is non-trivial.  ◻︎

 

Proposition.  If X is normal, then the subsets of the form U*, when U ranges over the open subsets of X, form a basis of the Wallman topology on ωX — in other words, ωX=ωX*.  Moreover, ωX is compact T2.

Proof.  We first check that U* is open in the Wallman topology.  Let F be a Wallman filter in U*.  By definition, it contains an element C that is included in U.  By normality, we can find an open set U’ and a closed set C’ such that CU’C’U.  The complement U of (complement of U’) is an open neighborhood of F: that means that the complement of U’ is not in F, and indeed it cannot be in F because F is non-trivial and C ⋂ (complement of U’) is empty.  Also, U is included in U*: for every Wallman filter F’ in U, F’ does not contain the complement of U’, but contains (complement of U’) ⋃ C’ = X; since F’ is a prime filter (see above: every maximal non-trivial filter is prime), it must contain C’, and therefore F’ is in U*.

We have shown that U* is a neighborhood of any of its points F.  Therefore U* is open (in the Wallman topology).  This shows that the Wallman* topology is coarser than the Wallman topology.  By the preceding Lemma, the Wallman* topology is T2, and we know already that the Wallman topology is compact.  It follows immediately that the two topologies are identical!  Indeed, any compact topology finer than a T2 topology must coincide with it: see Theorem 4.4.27 in the book.  ◻︎

 

We are slowly working our way toward proving that if X is T4, then ωX is (isomorphic to) the Stone-Čech compactification ßX.  For now, we know that ωX is compact T2, and that X is dense in ωX.  This is a good omen!  But this is not enough.

This is where we shall need to explore some facets of Stone duality — some new, some old.

The Stone duality view.

Let us first rephrase what we have done in terms of more familiar constructions — more familiar, at least, if you have read Chapter 8 and, possibly, Section 9.5 of the book.

Any filter F of closed subsets of X defines a set I, of all opens whose complements are in F: I = {U | complement of U in F}.  One checks easily that I is an ideal in the lattice OX of open subsets of X, namely a directed, downward closed set of opens; and that this construction defines an order isomorphism between the lattice of filters of closed subsets of X and the ideal completion I(OX) of the lattice of open sets of X.

Let us rephrase ωX through this isomorphism: ωX is the space of all maximal non-trivial ideals of opens subsets of X.  By non-trivial we mean those ideals different from the top ideal, OX itself.  And the topology of ωX has as basic open sets the complements of U={I in ωX | U is in I}, where U ranges over the open subsets of X.

We have seen a similar construction in proving Johnstone’s Theorem (Theorem 9.5.14).  There we considered the topological space pt I(OX), and we shall look at it with a different view.  The elements of pt I(OX) can be described as the prime elements of I(OX).  Remember that, in a distributive lattice, the maximal non-trivial filters are all prime.  Since ωX is the set of maximal non-trivial elements of I(OX), ωX occurs as a subset of pt I(OX).

Oh, for this to hold, we should check that I(OX) is indeed a distributive lattice.  And I said that ωX occurs as a subset of pt I(OX), not as a subspace, and we should check that as well.

For the first item, OX is a distributive pointed lattice, so we can rely on Exercise 9.5.11, and conclude that I(OX) is an algebraic fully arithmetic lattice.  In particular, it is distributive, but this says much more!  See Exercise 9.5.6: this means that I(OX) is an algebraic, distributive lattice in which the top element (OX) is finite and where the greatest lower bound of any two finite elements is finite.

By Exercise 9.5.5, pt I(OX) is then a spectral space — whatever X is (no need for normality or for any separation axiom).  This was one of our first steps in proving Johnstone’s Theorem, which states that X occurs as a retract of pt I(OX) as soon as X is stably compact.  (We shall make use of this later.)  Of course, in the cases we are interested in here, X will be at best T4, not stably compact.

Let us proceed to the second item.

Lemma.  ωX is a subspace of pt I(OX) [not just a subset].

Proof.  Let us look at the topology on pt I(OX).  For each ideal J in I(OX), there is an open subset OJ in pt I(OX), defined as the set of prime elements I of I(OX) such that J is not included in I.  (In the book, we defined pt I(OX) as a space of completely prime filters of elements of I(OX).  Using this presentation, a point y, namely a completely prime filter, would be in OJ  if and only if J is in y.  We are profiting from the fact that completely prime filters are exactly the complements of the downward closures of prime elements I [see Corollary 8.1.21], yielding the above formula for OJ.)

This topology on pt I(OX) is generated by the open subsets O↓U, where U is in OX.  This is because, for every ideal J, OJ is the union of the opens O↓U with U in J.  Writing C for the (closed) complement of U, O↓U then appears to be the complement of the set of prime elements I of I(OX) that contain U.  By a legitimate abuse of language, write U for the latter set, {I in pt I(OX) | U is in I}.  The latter therefore form a base of closed sets for the topology of pt I(OX).

Does this ring a bell?  We had written U, until now, for the basic closed sets {I in ωX | U is in I} of the topology of ωX.  So our former sets U are the intersection of ωX with our new sets U.  It follows that ωX is a subspace, not just a subset, of the spectral space pt I(OX).  ◻︎

We therefore have a sequence of maps:

  • η : XωX, which (in our reading of ωX as a space of maximal non-trivial ideals) maps x in X to {U open in X | x is not in U}
  • the inclusion ωXpt I(OX).

Both are embeddings as soon as X is T1.  In fact, their composition from X to pt I(OX) is always an embedding (provided X is T0), and factors through ωX if and only if X is T1.  We shall write the composite map, X to pt I(OX), again, as η.  This should not cause any confusion.

 

Let us come back to our original problem.  We would like to show that ωX is a Stone-Čech compactification of X.  For this, it is enough to show that it satisfies the universal property of Stone-Čech compactifications: every continuous map g : XY, where Y is compact T2, has exactly one continuous extension g’ from ωX to Y, where by extension we mean that g’(η(x))=g(x) for every x in X.

Uniqueness is easy, because Im η is dense in ωX.  Equating x with η(x) in ωX, and therefore seeing X as a dense subspace of ωX, we have that if g’ exists, then for every F in ωX, F will be a limit of some net (xi)i in I, ⊑ of elements of X, so g’(F) will be the limit of (g(xi))i in I, ⊑; and this limit is unique because Y is T2.

The real problem is showing that g’ exists.  Oh well, you might say: the previous paragraph gives us a formula for g’!  However, it is almost unusable.  We don’t even know whether g’(F) thus defined is independent of the chosen net (xi)i in I, ⊑.  Replacing nets by filters does not help much here: try it for yourself if you are not convinced.

However, there is a very principled way of showing that g’ exists, through Stone duality.  We show the following more general result first.  This is actually a consequence of the already cited Johnstone Theorem.

Theorem.  Every continuous map g : XY, where Y is stably compact, has at least one continuous extension g’ from pt I(OX) to Y, where by extension we mean that g’(η(x))=g(x) for every x in X.

Proof.  Apply Stone duality, that is, look at the counterparts of g and η, in the world of frames.  The counterpart of g is the frame homomorphism Og : OYOX, which maps every open subset V of Y to g-1(V).

Similarly, there is a Stone dual counterpart to η.  Let us compute what it does.  The open subsets of pt I(OX) are the subsets OJ, for each ideal J in I(OX), defined as the set of prime elements I of I(OX) such that J is not included in I.  Then η-1(OJ) is the set of points x in X such that J is not included in η(x) = {U open in X | x is not in U}, i.e., such that there is an U in J such that x is in U: this is just the union of all the elements U of J.  In other words, Oη : O pt I(OX) ≅ I(OX) → OX is just the familiar ‘union’ map, a.k.a., the sup map.

In the book, I usually write rL for the sup map from I(L) to L, where L is a complete lattice.  For L=OY, where Y is stably compact, which is the situation in Johnstone’s Theorem 9.5.14, rL is a retraction.  Its associated section sL : LI(L) maps every element V of L (an open subset of Y) to ↡V.  One checks easily that not only rL o sL = identity, but also sL o rL ≤ identity.  In particular, sL is left adjoint to rL: sL(V) ⊆ I if and only V ⊆ rL(I).

We are looking for a continuous map g’ : pt I(OX) → Y such that g’ o η = g.  Going to the Stone duals, and since all involved spaces are sober, it is equivalent to find a frame homomorphism f : OY → I(OX) such that Oη o f = Og: there will be a unique continuous map g’ : pt I(OX) → Y determined from f by the adjunction Opt.  We have seen that Oη is the sup map rL, so we are looking for a frame homomorphism f : OY → I(OX) such that Og = rL o f.  By the adjunction sLrL, OgrL o f if and only if sL o Ogf, so any solution f to our problem must be above sL o Og.  But sL o Og is itself a solution!  since rL o  (sL o Og) = (rLsL) o Og = Og.

This finishes the proof.  We have also shown that we could even find g’ minimal, in the sense that g’ is pointwise minimal among all possible solution frame homomorphisms.  ◻︎

Corollary.  If X is T4, then ωX is (isomorphic to) the Stone-Čech compactification ßX.

Proof. We must show the universal property of Stone-Čech compactifications: every continuous map g : XY, where Y is compact T2, has exactly one continuous extension g’ from ωX to Y, where by extension we mean that g’(η(x))=g(x) for every x in X.

By the previous theorem, we know there is at least one such extension from the larger space pt I(OX) to Y.  Its restriction to ωX is then one possible candidate for an extension.  However, g can have at most one continuous extension to ωX, since Im η is dense in ωX.  ◻︎

 

Wrapping up.

If you’ve survived until this point, you may be happy to learn that the story does not end here.  The fact that we only managed to show that the Wallman compactification ωX is the Stone-Čech compactification ßX only when X is T4, is nagging.  The Stone-Čech compactification exists for all T3 1/2 spaces, right?  Gillman and Jerrison showed [3] that a construction that is very similar to ωX yields a Stone-Čech compactification of X for all T3 1/2 spaces X.  Briefly, instead of considering maximal non-trivial filters of closed subsets, you need to consider maximal non-trivial filters of zero sets of X.  (A zero set is the inverse image f-1 {0} of the one-element [closed] subset {0} of R by a continuous map f from X to R, with its usual, metric topology.)  Frink [2] generalized this result to Z-sets, for certain, so-called normal bases of closed sets Z.

Jean Goubault-Larrecqjgl-2011

[1] Henry Wallman.  Lattices and Topological Spaces.  Annals of Mathematics, Second Series 39(1), January 1938, 112-126.  Available on JSTOR.

[2] Orrin Frink.  Compactifications and Semi-Normal Spaces. American Journal of Mathematics 86(3), July 1964, 602-607.  Available on JSTOR.

[3]  L. Gillman and M. Jerison.  Rings of Continuous Functions. Princeton, 1960.

 

 

Leave a Reply