TD spaces

Fact. A dcpo is T_D in its Scott topology if and only if it satisfies the ascending chain condition.

Note that, in a dcpo with the ascending chain condition, the Scott topology coincides with the Alexandroff topology… hence all the T_D dcpos are already Alexandroff spaces: dcpos do not provide us with any new example of T_D space, compared to the Alexandroff spaces.

Sober spaces.

One of the first things you usually learn about T_D spaces is that the T_D separation axiom, which lies between T₀ and T₁, is completely incomparable with sobriety. There are T_D spaces that are not sober, simply because there are T₁ spaces that are not sober, for example N with the cofinite topology (see Exercise 8.2.13 in the book). And there are sober spaces that are not T_D, for example any continuous dcpo that does not satisfy the ascending chain condition, such as N ∪ {∞} in its usual ordering.

The situation is even worse than that. The following is Note VI-2.3.2 in [2].

Proposition. Let X be a T₀ space. The sobrification S(X) of a space X is never T_D, unless X is already sober and T_D.

Proof. Let us assume that S(X) is T_D. We claim that the map η : X → S(X), which sends x to ↓x, is surjective. Let C be any element of S(X), namely any irreducible closed subset of X. Since S(X) is T_D, C is isolated in ↓C (where downward closure is taken in S(X)). That is, there is an open neighborhood of C in S(X), necessarily of the form ♢U with U open in X, which intersects ↓C at C only. (♢U is the set of irreducible closed subsets of X that intersect U.) That ♢U is an open neighborhood of C means that C and U intersect, say at x in X. Then ↓x is in ↓C since ↓x is included in C, and is also in ♢U, so ↓x is also in the intersection ↓C ∩ ♢U, namely in {C} since C is isolated in ↓C. This shows that C must be equal to ↓x.

We have just shown that η is surjective. Since X is T₀, η is injective. It is then a homeomorphism (Proposition 8.2.1 in the book). Therefore X is sober, and homeomorphic to the T_D space S(X), hence is itself T_D.

In the converse direction, if X is sober and T_D, then it is homeomorphic to S(X), which is then T_D as well. ☐

The Skula topology

Another concept which I had found almost completely uninteresting at first is the Skula topology. But is has uses almost everywhere I can see now!

For example, the sober Noetherian spaces are exactly those that are compact in their Skula topology (Exercise 9.7.16 in the book, R.-E. Hoffmann’s theorem). Also, the sobrification of a subspace Y of a sober space X is exactly its Skula-closure.

Here is a nifty theorem, which one can find as part of I-4.2 in [2]. I should recall that the Skula topology on X is the coarsest topology that contains all the open subsets and all the closed subsets of X (as new open subsets).

Proposition. A space X is T_D if and only if it is discrete in its Skula topology.

Proof. If X is T_D, then for every point x, ↓x – {x} is closed, so {x} can be expressed as the difference ↓x – (↓x – {x}) of two closed sets, namely as the intersection of a closed set and of an open set (a crescent); in particular, {x} is open in the Skula topology.

Conversely, if X is discrete in its Skula topology, let us fix a point x in X. By discreteness, {x} is open in the Skula topology, hence a union of crescents. One of those crescents C ∩ U (with C closed, U open in X) must contain x, and then C ∩ U is the whole of {x}. Since C contains x and is closed in X, it contains ↓x. It follows that ↓x ∩ U = {x}. But that means that x is isolated in ↓x. Therefore X is T_D. ☐

Sublocales

The place where T_D spaces perhaps have a more prominent role is in the theory of locales. I have already talked about sublocales quite a few times.

You probably remember that sublocales are a kind of localic analogue of the notion of subspace, but one should beware of the differences.

Let me define a sublocale of a frame Ω, as in [2], as a subset L of Ω that is closed under arbitrary infima (taken in Ω), and such that ω ⟹ x is in L for every x in L and every ω in Ω.  Every subspace Y of a topological space X defines a sublocale of the frame O(X) of open subsets of X, as the family of open subsets U of X such that ν_Y(U)=U, where ν_Y(U) denotes the largest open subset of X with the same intersection with Y that U has. (In turn, ν_Y is the nucleus associated with Y.)

In general, there are many more sublocales than there are subspaces. For example, let us recall Isbell’s density theorem: any frame contains a smallest dense sublocale. Moreover, the smallest dense sublocale of O(X) is the lattice of regular open subsets, namely those open subsets that are equal to the interior of their closure. Now look at the case where X is dense-in-itself, namely if it has no isolated point.

We claim that the smallest dense sublocale of O(X), where X is dense-in-itself and non-empty, cannot be the sublocale associated with any subspace of X. Indeed, since X has no isolated point, X–{x} is dense for every point x, so that if the sublocale were associated with a subspace Y, that Y would have to be included in X–{x} for every x. Hence Y would have to be empty. But then Y would be dense in X, which is non-empty, and that is absurd.

But there is more. It may be that a given sublocale is associated with two or more subspaces!

In other words, if I give you the sublocale, and even it comes from a subspace, that subspace may fail to be unique. But that does not happen with T_D spaces, as we now show.

Proposition. Let Y and Z be two subspaces of the same topological space X, and assume that they define the same sublocale of O(X). If X is T_D, then Y=Z.

Proof. The assumption means that: (*) for every pair of open sets U and V, it is equivalent to say that U ∩ Y=V ∩ Y or that U ∩ Z=V ∩ Z. Indeed, U ∩ Y=V ∩ Y is equivalent to ν_Y(U)=ν_Y(V), and similarly with Z in place of Y. The assumption means that ν_Y=ν_Z, whence (*).

By taking complements, (*) entails that for every pair of closed sets C and D, C ∩ Y=D ∩ Y is equivalent to C ∩ Z=D ∩ Z.

Let us assume that Y≠Z. Without loss of generality, there is a point x in Y that is not in Z. Since X is T_D, C = ↓x – {x} is closed. Let D = ↓x. Then C ∩ Z=D ∩ Z, since x is not in Z. Hence C ∩ Y=D ∩ Y. However, x is in D ∩ Y but not in C ∩ Y, which is absurd. ☐

And the T_D condition is necessary:

Proposition. If X is not T_D, then it has two distinct subspaces that define the same sublocale.

Proof. Since X is not T_D, it contains a point x that is not isolated in ↓x. This means that every open neighborhood U of x intersects ↓x – {x}. We let Y = ↓x – {x}, Z = ↓x. The nucleus ν_Z maps every open set U to U ∪ (X – ↓x): this is the closed nucleus c(X – ↓x), as defined at the end of this post.

Let us compute ν_Y. For every open set U, ν_Y(U) is the largest open set V such that V ∩ Y=U ∩ Y. Any such V must be such that V ∩ Y ⊆ U. If V contains x, by assumption it intersects ↓x – {x} = Y, say at y; then y is in U, and since y≤x and U is upwards-closed, x must also be in U; this entails that V ∩ Z, which is equal to (V ∩ Y) ∪ {x}, is also included in U. If V does not contain x, then V ∩ Y=V ∩ Z, so trivially V ∩ Z ⊆ U. Whichever is the case, V ∩ Z ⊆ U, so V is included in ν_Z(U). Therefore ν_Y(U) ⊆ ν_Z(U).

Since Y is included in Z, we see easily that ν_Z(U) is included in ν_Y(U) for every open set U. Indeed, every open set V such that U ∩ Z=V ∩ Z will also satisfy U ∩ Y=V ∩ Y, hence is included in ν_Y(U).

We have shown that ν_Y = ν_Z, hence that Y and Z define the same sublocale. ☐

In other words:

A space in which every sublocale corresponds to at most one subspace is the same thing as a T_D space.

Lattice equivalence

One of the nice things about sober spaces is that you can reconstruct the space from its lattice of open sets. While sobriety and the T_D property are incomparable, we can also reconstruct X from O(X) when X is a T_D space.

Let us see how this can be done. Given any point x in a T_D space X, and as in every space, the family N(x) of open neighborhoods of x is a completely prime filter in O(X). If X were sober, we could recover a unique point from any completely prime filter of open sets, but in a non-sober T_D space (such as N with its Alexandroff topology) there are completely prime filters of opens sets (such as the set of all the non-empty upwards-closed subsets of N…) that are not of the form N(x) for any point x.

In fact, in a T_D space, N(x) has an additional property: there are two open sets, namely the complement U of ↓x and the complement V of ↓x – {x}, such that V is in N(x), U is not in N(x), and U is immediately below V. The latter means that U is strictly below (strictly included in) V, but there is absolutely no other open set between U and V.

Picado and Pultr [2, page 5] call slicing any prime filter with that property. In general, a filter F of elements of a frame Ω is slicing if and only if it is prime (not containing the bottom element and such that if it contains u ⋁ v then it contains u or it contains v), and there is an element u not in F, an element v in F such that u is immediately below v, in the sense that there is no element of Ω strictly above u and strictly below v. Note that we do not require F to be completely prime, only prime, by the way.

Lemma. In a T₀ space X, each slicing filter is of the form N(x) for a unique point x.

Proof. Let F be a slicing filter of open sets. There is an open set U outside F and immediately below some other open set V inside F. Since U is strictly contained in V, there is a point x in V – U. We will show that F=N(x). The uniqueness of x will follow from the fact that X is T₀: in a T₀ space, any two points with the same set of open neighborhoods must be equal.

For every open neighborhood O of x, let us consider the open set W = V ∩ (U ∪ O) = U ∪ (V ∩ O). This lies between U and V, hence is equal to one of them. But x is in W (since it is both in V and in O), and not in U, so W cannot be equal to U. It follows that W = V. From W = V ∩ (U ∪ O), hence V = V ∩ (U ∪ O), we deduce that V is included in U ∪ O. Since V is in F, U ∪ O is in F, and because F is prime, U or O is in F. However, U is not in F, so O is. It follows that N(x) is included in F.

Conversely, we claim that every element O of F contains x. Since F is a filter, V ∩ O is also in F. We consider again W = V ∩ (U ∪ O) = U ∪ (V ∩ O), and by the same argument W is equal to U or to V. It cannot be equal to U, otherwise U = U ∪ (V ∩ O), so V ∩ O would be included in U, and that would imply that U is also in F. Therefore W = V, namely V ∩ (U ∪ O) = V, meaning that V is included in U ∪ O. Since V contains x but U does not, O must contain x. This shows that F is included in N(x). ☐

This entails the following interesting observation of W. J. Thron, who calls two spaces with isomorphic lattices of open sets “lattice equivalent”.

Theorem ([3]). Let X and Y be two topological spaces with isomorphic lattices of open sets, and let φ : O(Y) → O(X) be such an isomorphism. If X is T_D and Y is T₀, then there is a unique homeomorphism f : X → Y such that φ=f^-1.
In particular, any two lattice equivalent T_D spaces are homeomorphic.

Proof. The key is to observe that for every slicing filter F of open sets of Y, φ^-1(F) is also a slicing filter. (I will always write φ^-1 for “the inverse image by φ”, and I will call ψ the inverse φ. Of course φ^-1(F) is also the directed image of F by ψ.) This is easy, but funnily, that would not work if we had only assumed φ to be a frame homomorphism: in that case, we would be able to prove that given any prime filter F of open sets of Y, φ^-1(F) is a prime filter, but slicing may fail to be preserved. Given that φ is an isomorphism, there is no difficulty in showing that φ^-1 preserves slicing, however.

If f exists as specified, then f^-1(N(f(x))) should be equal to N(x), so one must have N(f(x))=ψ(N(x))=φ^-1(N(x)). This defines f(x) uniquely since Y is T₀, where any point is defined uniquely (if at all) by its set of open neighborhoods.

In order to show the existence of f, we observe that for every x in X, N(x) is a slicing filter of open sets, hence also φ^-1(N(x)), and by the previous lemma it is equal to N(f(x)) for some unique point f(x).

We claim that φ=f^-1, namely that for every open subset V of Y, φ(V)=f^-1(V). For every x in X, x is in f^-1(V) if and only if f(x) is in V, if and only if V is in N(f(x)). Since the latter is equal to φ^-1(N(x)), this is equivalent to φ(V) ∈ N(x), hence to x ∈ φ(V).

In particular, f^-1(V) = φ(V) is open for every open subset V of Y, so f is continuous.

For every open subset U of X, U is equal to φ(V) for some open subset V of Y, since φ is surjective; hence to f^-1(V). This shows that f is almost open.

For every point y of Y, N(y) is a slicing filter of open subsets of Y. Recall that ψ is the inverse of φ. Reasoning with ψ instead of φ, ψ^-1(N(y)) is also a slicing filter, hence a filter of the form N(x) for some point x of X. Then N(y)=φ^-1(N(x)), which is equal to N(f(x)). It follows that y=f(x). This shows that f is surjective.

Finally, imagine that f(x)=f(x’). Then N(f(x))=N(f(x‘)), so φ^-1(N(x))=φ^-1(N(x‘)), and since φ is bijective, it follows that N(x)=N(x‘). Since X is T_D hence T₀, x=x’. Therefore f is injective. ☐

This is rather remarkable, and similar properties fail for lots of other classes of spaces. For example, it fails for the class of all topological spaces: if X is a non-sober space, then X and its sobrification S(X) are lattice equivalent, but cannot be homeomorphic.

But it holds for the class of sober spaces: any two lattice equivalent sober spaces must be homeomorphic. This is a simple instance of applying the pt functor of Stone duality, and recalling that for every sober space X, pt(O(X)) and X are homeomorphic.

There are several variants of this question. For example, the following is the Ho-Zhao problem: given any two dcpos with isomorphic lattices of open sets, and those two dcpos homeomorphic in their Scott topology, or (equivalently) are they order-isormophic? This was solved in the negative in [4] (and also, in the positive, for the rather intriguing class of so-called dominated dcpos, a very rich class of dcpos). I may talk about that in a later post, who knows?