# TD spaces

In any topological space, the closure of any one-element set {x} is also its downward closure ↓x with respect to the specialization preordering. A TD space is a topological space in which, for every point x, ↓x – {x} is closed, too.

The concept is due to Aull and Thron , who also studied half a dozen other separation axioms.

## Preliminary reflections

When I hear of a concept for the first time in mathematics, often I find it weird, useless, and not worthy of interest. But experience has told me I should pay attention nonetheless: among all those ‘uninteresting’ concepts, some will play a decisive role in solving some important problem later.

In contrast, when I hear of a concept for the first time in computer science, I usually know at once whether it is interesting or not.

This is probably because of a difference of approaches between the two scientific communities: in computer science, it is simply unacceptable to present a new notion without a rock-solid motivation for introducing it first. (“Here is the problem that needs to be solved. Here is the difficulty. Here is the notion I need to solve it.”)

In mathematics, I have often seen orators delve immediately into their pet problem without any motivation: either you work in the field and you have a chance of getting something out of the talk, or you’ll have no clue whatsoever.

Anyway, the notion of TD space long remained in the category of useless curiosities to me, but they occur in so many different places nowadays… hence one should probably pay some attention to them, and I will say why.

Perhaps the most convincing argument in favor of the notion is in the correspondence between subspaces and sublocales of a topological space, which I will say a word about below. But I will maintain some suspense, and I will only talk about it near the middle of this post.

Oh, by the way, Aull and Thron did give a reason why they studied TD property (and others), and that was to explore separation axioms between T1 and T0. In passing, that motivation is so weak that you are sure to have you paper rejected to a computer science conference, with a similar motivating statement.

Anyway, before we go to other things, let me say that, indeed, TD lies between T1 and T0.

It is easy to see that every T1 is TD: in a T1 space, ↓x={x}, so ↓x – {x} is empty, hence closed.

It is a bit more difficult to see that every TD space is T0: if xy and yx, but xy in a TD space, then ↓x – {x} contains y (since xy), and since it is closed, it must also contain ↓y; therefore ↓x – {x,y} contains ↓y – {y}, which itself contains x (since xy again); but this implies that x is in ↓x – {x,y}, which is absurd.

## Alternate definitions

There are many alternate definitions of a TD space, and the one I have taken here is the one I find easiest to understand. However, it is probably interesting to know that there are several alternative, equivalent characterizations.

#### Equivalent definition 1.

An isolated point in a subset S of a topological space X is any point x such that {x} is open in S, seen as a subspace. Equivalently, such that there is an open neighborhood of x whose intersection with S is just {x}. A TD space is then a space in which, for every point x, x is isolated in its closure ↓x.

#### Equivalent definition 2.

Given any set S of points, one can form its derived set S’. S’ is the set of limit points of S; and a limit point of S is a limit x of a net of points of S different from x. Because of the latter, S’ is not in general the closure of S. Instead, this is SS’ that is the closure of S.

In general, S is not included in S’: in fact, the points that lie in SS’ are just the isolated points of S. Hence a space is TD if and only if, for every point x, (↓x)’ does not contain x.

#### Equivalent definition 3.

If S=↓x, then every limit point of S must be below x, because S is closed. Also, every point y in ↓x – {x} is the limit of the constant net (y), showing that y is in S’. Hence S’ is included in ↓x, and must contain at least all the points in ↓x – {x}. As a conclusion, a space is TD if and only if, for every point x, (↓x)’ is different from ↓x; it must then be equal to ↓x – {x}.

## Examples and counterexamples

There are three important examples of TD spaces.

#### Example 1.

First… the T1 spaces, since every T1 space is TD.

#### Example 2.

Second, all the T0 scattered spaces are TD.

A scattered space is a space in which every non-empty subset S contains an isolated point (in S). Now, consider any point x in a scattered space X. Then ↓x must contain some isolated point y. Namely, there is an open neighborhood U of y such that U ∩ ↓x = {y}. Using the T0 property, this implies that y=x. Then x is isolated in ↓x; alternatively, U ∩ ↓x = {x}, so ↓x – {x} = ↓xU is closed. Hence, as promised, X is TD.

Oh, by the way, no, the converse implication fails: a TD space need not be scattered. Consider for example any non-empty T1 space without any isolated point, such as [0, 1].

#### Example 3.

Third, all T0 Alexandroff spaces, that is, all the T0 spaces in which every intersection of open sets is open, or equivalently, the spaces whose topology is given by the upwards-closed subsets in some partial ordering ≤. Indeed, for every point x, ↓x – {x} is again downwards-closed, hence closed. (The fact that ≤ is an ordering, not just a preordering, is important. Note that, in that case, ↓x – {x} is also the set of points strictly below x.)

#### Dcpos.

Would dcpos be TD in their Scott topology, too?

Well, very rarely so. If a dcpo X is TD, then by definition for every point x, and every directed family D of points strictly below x, sup D will still be strictly below x. By taking contrapositives, this means that if sup D=x, then x must be in D. In particular, that implies that X has the ascending chain condition: every strictly ascending chain x0 < x1 < … < xn < … is finite.

Conversely, if X has the ascending chain condition, then I claim that X is TD in its Scott topology. Indeed, for every x in X, and every directed family D of points strictly below x, it suffices to show that sup D is in D: this will show that the set of points strictly below x is Scott-closed. If sup D is not in D, that means that no point y of D is largest in D. For every y in D, there is a point z in D such that z is not ≤ y. By directedness, there is a point t in D above both y and z. Then we cannot have ty, since otherwise zty. Hence we have shown that every element of D is strictly below another element of D. Iterating the process, we will find an infinite strictly ascending chain, which is absurd.

We conclude:

Fact. A dcpo is TD 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 TD dcpos are already Alexandroff spaces: dcpos do not provide us with any new example of TD space, compared to the Alexandroff spaces.

#### Sober spaces.

One of the first things you usually learn about TD spaces is that the TD separation axiom, which lies between T0 and T1, is completely incomparable with sobriety. There are TD 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 TD, 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 .

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

Proof. Let us assume that S(X) is TD. We claim that the map η : XS(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 TD, 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 TD space S(X), hence is itself TD.

In the converse direction, if X is sober and TD, then it is homeomorphic to S(X), which is then TD 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 . 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 TD if and only if it is discrete in its Skula topology.

Proof. If X is TD, 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 CU (with C closed, U open in X) must contain x, and then CU is the whole of {x}. Since C contains x and is closed in X, it contains ↓x. It follows that ↓xU = {x}. But that means that x is isolated in ↓x. Therefore X is TD. ☐

## Sublocales

The place where TD 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 , 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 νY(U)=U, where νY(U) denotes the largest open subset of X with the same intersection with Y that U has. (In turn, νY 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 TD 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 TD, then Y=Z.

Proof. The assumption means that: (*) for every pair of open sets U and V, it is equivalent to say that UY=VY or that UZ=VZ. Indeed, UY=VY is equivalent to νY(U)=νY(V), and similarly with Z in place of Y. The assumption means that νYZ, whence (*).

By taking complements, (*) entails that for every pair of closed sets C and D, CY=DY is equivalent to CZ=DZ.

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

And the TD condition is necessary:

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

Proof. Since X is not TD, 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 νZ 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 νY. For every open set U, νY(U) is the largest open set V such that VY=UY. Any such V must be such that VYU. If V contains x, by assumption it intersects ↓x – {x} = Y, say at y; then y is in U, and since yx and U is upwards-closed, x must also be in U; this entails that VZ, which is equal to (VY) ∪ {x}, is also included in U. If V does not contain x, then VY=VZ, so trivially VZU. Whichever is the case, VZU, so V is included in νZ(U). Therefore νY(U) ⊆ νZ(U).

Since Y is included in Z, we see easily that νZ(U) is included in νY(U) for every open set U. Indeed, every open set V such that UZ=VZ will also satisfy UY=VY, hence is included in νY(U).

We have shown that νY = νZ, 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 TD 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 TD property are incomparable, we can also reconstruct X from O(X) when X is a TD space.

Let us see how this can be done. Given any point x in a TD 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 TD 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 TD 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 uv 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 VU. 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 ∩ (UO) = U ∪ (VO). 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 ∩ (UO), hence V = V ∩ (UO), we deduce that V is included in UO. Since V is in F, UO 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, VO is also in F. We consider again W = V ∩ (UO) = U ∪ (VO), and by the same argument W is equal to U or to V. It cannot be equal to U, otherwise U = U ∪ (VO), so VO would be included in U, and that would imply that U is also in F. Therefore W = V, namely V ∩ (UO) = V, meaning that V is included in UO. 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 (). 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 TD and Y is T0, then there is a unique homeomorphism f : XY such that φ=f-1.
In particular, any two lattice equivalent TD 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 TD 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  (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?

1. Charles Edward Aull and Wolfgang Joseph Thron, Separation axioms between T0 and T1. Indagationes Mathematicae 23, pages 26-37, 1962.
2. Jorge Picado and Aleš Pultr.  Frames and locales — topology without points.  Birkhäuser, 2010.
3. Wolfgang Joseph Thron. Lattice-equivalence of topological spaces. Duke Mathematical Journal, 29:671–679, 1962.
4. Weng Kin Ho, Jean Goubault-Larrecq, Achim Jung, and Xiaoyong Xi. The Ho-Zhao problem. Logical Methods in Computer Science, 14(1:7), 1-19, 2018. doi: 10.23638/LMCS-14(1:7)2018