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 T_{D} space is a topological space in which, for every point *x*, ↓*x* – {*x*} is closed, too.

The concept is due to Aull and Thron [1], 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 T_{D} 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 T_{D} property (and others), and that was to explore separation axioms between T_{1} and T_{0}. 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, T_{D} lies between T_{1} and T_{0}.

It is easy to see that every T_{1} space is T_{D}: in a T_{1} space, ↓*x*={*x*}, so ↓*x* – {*x*} is empty, hence closed.

It is a bit more difficult to see that every T_{D} space is T_{0}: if *x*≤*y* and *y*≤*x*, but *x*≠*y* in a T_{D} space, then ↓*x* – {*x*} contains *y* (since *x*≠*y*), and since it is closed, it must also contain ↓*y*; therefore ↓*x* – {*x*,*y*} contains ↓*y* – {*y*}, which itself contains *x* (since *x*≠*y* again); but this implies that *x* is in ↓*x* – {*x*,*y*}, which is absurd.

## Alternate definitions

There are many alternate definitions of a T_{D} 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 T_{D} 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 *S* ∪ *S’* that is the closure of *S*.

In general, *S* is not included in *S’*: in fact, the points that lie in *S*–*S’* are just the isolated points of *S*. Hence a space is T_{D} 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 T_{D} 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 T_{D} spaces.

#### Example 1.

First… the T_{1} spaces, since every T_{1} space is T_{D}.

#### Example 2.

Second, all the T_{0} scattered spaces are T_{D}.

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 T_{0} property, this implies that *y*=*x*. Then *x* is isolated in ↓*x*; alternatively, *U* ∩ ↓*x* = {*x*}, so ↓*x* – {*x*} = ↓*x* – *U* is closed. Hence, as promised, *X* is T_{D}.

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

#### Example 3.

Third, all T_{0} Alexandroff spaces, that is, all the T_{0} 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 T_{D} in their Scott topology, too?

Well, very rarely so. If a dcpo *X* is T_{D}, 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 *x*_{0} < *x*_{1} < … < *x _{n}* < … is finite.

Conversely, if *X* has the ascending chain condition, then I claim that *X* is T_{D} 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 *t*≤*y*, since otherwise *z*≤*t*≤*y*. 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 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 T_{0} and T_{1}, is completely incomparable with sobriety. There are T_{D} spaces that are not sober, simply because there are T_{1} 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 T_{0} 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 T_{0}, 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

**, an element**

*F**v*in

**such that**

*F**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

**to be**

*F**completely*prime, only prime, by the way.

**Lemma.** In a T_{0} 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

**and immediately below some other open set**

*F**V*inside

**. Since**

*F**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 T

_{0}: in a T

_{0}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

**, and because**

*F***is prime,**

*F**U*or

*O*is in

**. However,**

*F**U*is not in

**, so**

*F**O*is. It follows that

*N*(

*x*) is included in

**.**

*F*Conversely, we claim that every element *O* of ** F** contains

*x*. Since

**is a filter,**

*F**V*∩

*O*is also in

**. We consider again**

*F**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

**. Therefore**

*F**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

**is included in**

*F**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 T_{0}, 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}(

**) is also a slicing filter. (I will always write φ**

*F*^{-1}for “the inverse image by φ”, and I will call ψ the inverse φ. Of course φ

^{-1}(

**) 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**

*F**Y*, φ

^{-1}(

**) 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**

*F*^{-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 T_{0}, 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*(

*)) is also a slicing filter, hence a filter of the form*

*y**N*(

*x*) for some point

*x*of

*X*. Then

*(*

*N**)=φ*

*y*^{-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 T_{0}, *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?

- Charles Edward Aull and Wolfgang Joseph Thron, Separation axioms between T
_{0}and T_{1}. Indagationes Mathematicae 23, pages 26-37, 1962. - Jorge Picado and Aleš Pultr. Frames and locales — topology without points. Birkhäuser, 2010.
- Wolfgang Joseph Thron. Lattice-equivalence of topological spaces. Duke Mathematical Journal, 29:671–679, 1962.
- 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