{"id":2626,"date":"2020-07-20T18:29:52","date_gmt":"2020-07-20T16:29:52","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=2626"},"modified":"2022-11-19T15:02:20","modified_gmt":"2022-11-19T14:02:20","slug":"td-spaces","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=2626","title":{"rendered":"TD spaces"},"content":{"rendered":"\n

In any topological space, the closure of any one-element set {x<\/em>} is also its downward closure \u2193x<\/em> with respect to the specialization preordering. A TD<\/em><\/sub> space is a topological space in which, for every point x<\/em>, \u2193x<\/em> \u2013 {x<\/em>} is closed, too.<\/p>\n\n\n\n

The concept is due to Aull and Thron [1], who also studied half a dozen other separation axioms.<\/p>\n\n\n\n

Preliminary reflections<\/h2>\n\n\n\n

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.<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

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.”)<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

Anyway, the notion of TD<\/em><\/sub> 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. <\/p>\n\n\n\n

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.<\/p>\n\n\n\n

Oh, by the way, Aull and Thron did<\/em> give a reason why they studied TD<\/em><\/sub> property (and others), and that was to explore separation axioms between T1<\/sub> and T0<\/sub>. 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.<\/p>\n\n\n\n

Anyway, before we go to other things, let me say that, indeed, TD<\/em><\/sub> lies between T1<\/sub> and T0<\/sub>.<\/p>\n\n\n\n

It is easy to see that every T1<\/sub> space is TD<\/em><\/sub>: in a T1<\/sub> space, \u2193x<\/em>={x<\/em>}, so \u2193x<\/em> \u2013 {x<\/em>} is empty, hence closed.<\/p>\n\n\n\n

It is a bit more difficult to see that every TD<\/em><\/sub> space is T0<\/sub>: if x<\/em>\u2264y<\/em> and y<\/em>\u2264x<\/em>, but x<\/em>\u2260y<\/em> in a TD<\/em><\/sub> space, then \u2193x<\/em> \u2013 {x<\/em>} contains y<\/em> (since x<\/em>\u2260y<\/em>), and since it is closed, it must also contain \u2193y<\/em>; therefore \u2193x<\/em> \u2013 {x<\/em>,y<\/em>} contains \u2193y<\/em> \u2013 {y<\/em>}, which itself contains x<\/em> (since x<\/em>\u2260y<\/em> again); but this implies that x<\/em> is in \u2193x<\/em> \u2013 {x<\/em>,y<\/em>}, which is absurd.<\/p>\n\n\n\n

Alternate definitions<\/h2>\n\n\n\n

There are many alternate definitions of a TD<\/em><\/sub> 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.<\/p>\n\n\n\n

Equivalent definition 1.<\/h4>\n\n\n\n

An isolated point<\/em> in a subset S<\/em> of a topological space X<\/em> is any point x<\/em> such that {x<\/em>} is open in S<\/em>, seen as a subspace. Equivalently, such that there is an open neighborhood of x<\/em> whose intersection with S<\/em> is just {x<\/em>}. A TD<\/em><\/sub> space is then a space in which, for every point x<\/em>, x<\/em> is isolated in its closure \u2193x<\/em>.<\/p>\n\n\n\n

Equivalent definition 2.<\/h4>\n\n\n\n

Given any set S<\/em> of points, one can form its derived set<\/em> S’<\/em>. S’<\/em> is the set of limit points<\/em> of S<\/em>; and a limit point of S<\/em> is a limit x<\/em> of a net of points of S<\/em> different from x<\/em>. Because of the latter, S’<\/em> is not in general the closure of S<\/em>. Instead, this is S<\/em> \u222a S’<\/em> that is the closure of S<\/em>.<\/p>\n\n\n\n

In general, S<\/em> is not included in S’<\/em>: in fact, the points that lie in S<\/em>\u2013S’<\/em> are just the isolated points of S<\/em>. Hence a space is TD<\/em><\/sub> if and only if, for every point x<\/em>, (\u2193x<\/em>)’ does not contain x<\/em>.<\/p>\n\n\n\n

Equivalent definition 3.<\/h4>\n\n\n\n

If S<\/em>=\u2193x<\/em>, then every limit point of S<\/em> must be below x<\/em>, because S<\/em> is closed. Also, every point y<\/em> in \u2193x<\/em> \u2013 {x<\/em>} is the limit of the constant net (y<\/em>), showing that y<\/em> is in S’<\/em>. Hence S’<\/em> is included in \u2193x<\/em>, and must contain at least all the points in \u2193x<\/em> \u2013 {x<\/em>}. As a conclusion, a space is TD<\/em><\/sub> if and only if, for every point x<\/em>, (\u2193x<\/em>)’ is different from \u2193x<\/em>; it must then be equal to \u2193x<\/em> \u2013 {x<\/em>}.<\/p>\n\n\n\n

Examples and counterexamples<\/h2>\n\n\n\n

There are three important examples of TD<\/em><\/sub> spaces.<\/p>\n\n\n\n

Example 1.<\/h4>\n\n\n\n

First… the T1<\/sub> spaces, since every T1<\/sub> space is TD<\/em><\/sub>.<\/p>\n\n\n\n

Example 2.<\/h4>\n\n\n\n

Second, all the T0<\/sub> scattered spaces are TD<\/em><\/sub>.<\/p>\n\n\n\n

A scattered space<\/em> is a space in which every non-empty subset S<\/em> contains an isolated point (in S<\/em>). Now, consider any point x<\/em> in a scattered space X<\/em>. Then \u2193x<\/em> must contain some isolated point y<\/em>. Namely, there is an open neighborhood U<\/em> of y<\/em> such that U<\/em> \u2229 \u2193x<\/em> = {y<\/em>}. Using the T0<\/sub> property, this implies that y<\/em>=x<\/em>. Then x<\/em> is isolated in \u2193x<\/em>; alternatively, U<\/em> \u2229 \u2193x<\/em> = {x<\/em>}, so \u2193x<\/em> \u2013 {x<\/em>} = \u2193x<\/em> \u2013 U<\/em> is closed. Hence, as promised, X<\/em> is TD<\/em><\/sub>.<\/p>\n\n\n\n

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

Example 3.<\/h4>\n\n\n\n

Third, all T0<\/sub> Alexandroff spaces, that is, all the T0<\/sub> 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 \u2264. Indeed, for every point x<\/em>, \u2193x<\/em> \u2013 {x<\/em>} is again downwards-closed, hence closed. (The fact that \u2264 is an ordering, not just a preordering, is important. Note that, in that case, \u2193x<\/em> \u2013 {x<\/em>} is also the set of points strictly below x<\/em>.)<\/p>\n\n\n\n

Dcpos.<\/h4>\n\n\n\n

Would dcpos be TD<\/em><\/sub> in their Scott topology, too?<\/p>\n\n\n\n

Well, very rarely so. If a dcpo X<\/em> is TD<\/em><\/sub>, then by definition for every point x<\/em>, and every directed family D<\/em> of points strictly below x<\/em>, sup D<\/em> will still be strictly below x<\/em>. By taking contrapositives, this means that if sup D<\/em>=x<\/em>, then x<\/em> must be in D<\/em>. In particular, that implies that X<\/em> has the ascending chain condition<\/em>: every strictly ascending chain x<\/em>0<\/sub> < x<\/em>1<\/sub> < … < xn<\/sub><\/em> < … is finite.<\/p>\n\n\n\n

Conversely, if X<\/em> has the ascending chain condition, then I claim that X<\/em> is TD<\/em><\/sub> in its Scott topology. Indeed, for every x<\/em> in X<\/em>, and every directed family D<\/em> of points strictly below x<\/em>, it suffices to show that sup D<\/em> is in D<\/em>: this will show that the set of points strictly below x<\/em> is Scott-closed. If sup D<\/em> is not in D<\/em>, that means that no point y<\/em> of D<\/em> is largest in D<\/em>. For every y<\/em> in D<\/em>, there is a point z<\/em> in D<\/em> such that z<\/em> is not \u2264 y<\/em>. By directedness, there is a point t<\/em> in D<\/em> above both y<\/em> and z<\/em>. Then we cannot have t<\/em>\u2264y<\/em>, since otherwise z<\/em>\u2264t<\/em>\u2264y<\/em>. Hence we have shown that every element of D<\/em> is strictly below another element of D<\/em>. Iterating the process, we will find an infinite strictly ascending chain, which is absurd.<\/p>\n\n\n\n

We conclude:<\/p>\n\n\n\n

\n

Fact.<\/strong> A dcpo is TD<\/em><\/sub> in its Scott topology if and only if it satisfies the ascending chain condition.<\/p>\n\n\n\n

Note that, in a dcpo with the ascending chain condition, the Scott topology coincides with the Alexandroff topology… hence all the TD<\/em><\/sub> dcpos are already Alexandroff spaces: dcpos do not provide us with any<\/em> new example of TD<\/em><\/sub> space, compared to the Alexandroff spaces.<\/p>\n\n\n\n

Sober spaces.<\/h4>\n\n\n\n

One of the first things you usually learn about TD<\/em><\/sub> spaces is that the TD<\/em><\/sub> separation axiom, which lies between T0<\/sub> and T1<\/sub>, is completely incomparable with sobriety. There are TD<\/em><\/sub> spaces that are not sober, simply because there are T1<\/sub> spaces that are not sober, for example N<\/strong> with the cofinite topology (see Exercise 8.2.13 in the book<\/a>). And there are sober spaces that are not TD<\/em><\/sub>, for example any continuous dcpo that does not satisfy the ascending chain condition, such as N<\/strong> \u222a {\u221e} in its usual ordering.<\/p>\n\n\n\n

The situation is even worse than that. The following is Note VI-2.3.2 in [2].<\/p>\n\n\n\n

Proposition.<\/strong> Let X<\/em> be a T0<\/sub> space. The sobrification S<\/strong>(X<\/em>) of a space X<\/em> is never TD<\/em><\/sub>, unless X<\/em> is already sober and TD<\/em><\/sub>.<\/p>\n\n\n\n

Proof.<\/em> Let us assume that S<\/strong>(X<\/em>) is TD<\/em><\/sub>. We claim that the map \u03b7 : X<\/em> \u2192 S<\/strong>(X<\/em>), which sends x<\/em> to \u2193x<\/em>, is surjective. Let C<\/em> be any element of S<\/strong>(X<\/em>), namely any irreducible closed subset of X<\/em>. Since S<\/strong>(X<\/em>) is TD<\/em><\/sub>, C<\/em> is isolated in \u2193C<\/em> (where downward closure is taken in S<\/strong>(X<\/em>)). That is, there is an open neighborhood of C<\/em> in S<\/strong>(X<\/em>), necessarily of the form \u2662U<\/em> with U<\/em> open in X<\/em>, which intersects \u2193C<\/em> at C<\/em> only. (\u2662U<\/em> is the set of irreducible closed subsets of X<\/em> that intersect U<\/em>.) That \u2662U<\/em> is an open neighborhood of C<\/em> means that C<\/em> and U<\/em> intersect, say at x<\/em> in X<\/em>. Then \u2193x<\/em> is in \u2193C<\/em> since \u2193x<\/em> is included in C<\/em>, and is also in \u2662U<\/em>, so \u2193x<\/em> is also in the intersection \u2193C<\/em> \u2229 \u2662U<\/em>, namely in {C<\/em>} since C<\/em> is isolated in \u2193C<\/em>. This shows that C<\/em> must be equal to \u2193x<\/em>.<\/p>\n\n\n\n

We have just shown that \u03b7 is surjective. Since X<\/em> is T0<\/sub>, \u03b7 is injective. It is then a homeomorphism (Proposition 8.2.1 in the book<\/a>). Therefore X<\/em> is sober, and homeomorphic to the TD<\/em><\/sub> space S<\/strong>(X<\/em>), hence is itself TD<\/em><\/sub>.<\/p>\n\n\n\n

In the converse direction, if X<\/em> is sober and TD<\/em><\/sub>, then it is homeomorphic to S<\/strong>(X<\/em>), which is then TD<\/em><\/sub> as well. \u2610<\/p>\n\n\n\n

The Skula topology<\/h2>\n\n\n\n

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!<\/p>\n\n\n\n

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

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<\/em> is the coarsest topology that contains all the open subsets and all the closed subsets of X<\/em> (as new open subsets).<\/p>\n\n\n\n

Proposition.<\/strong> A space X<\/em> is TD<\/em><\/sub> if and only if it is discrete in its Skula topology.<\/p>\n\n\n\n

Proof.<\/em> If X<\/em> is TD<\/em><\/sub>, then for every point x<\/em>, \u2193x<\/em> \u2013 {x<\/em>} is closed, so {x<\/em>} can be expressed as the difference \u2193x<\/em> \u2013 (\u2193x<\/em> \u2013 {x<\/em>}) of two closed sets, namely as the intersection of a closed set and of an open set (a crescent<\/em>); in particular, {x<\/em>} is open in the Skula topology.<\/p>\n\n\n\n

Conversely, if X<\/em> is discrete in its Skula topology, let us fix a point x<\/em> in X<\/em>. By discreteness, {x<\/em>} is open in the Skula topology, hence a union of crescents. One of those crescents C<\/em> \u2229 U<\/em> (with C<\/em> closed, U<\/em> open in X<\/em>) must contain x<\/em>, and then C<\/em> \u2229 U<\/em> is the whole of {x<\/em>}. Since C<\/em> contains x<\/em> and is closed in X<\/em>, it contains \u2193x<\/em>. It follows that \u2193x<\/em> \u2229 U<\/em> = {x<\/em>}. But that means that x<\/em> is isolated in \u2193x<\/em>. Therefore X<\/em> is TD<\/em><\/sub>. \u2610<\/p>\n\n\n\n

Sublocales<\/h2>\n\n\n\n

The place where TD<\/em><\/sub> spaces perhaps have a more prominent role is in the theory of locales. I have already talked about sublocales quite a few times<\/a>.<\/p>\n\n\n\n

You probably remember that sublocales are a kind of localic analogue of the notion of subspace, but one should beware of the differences.<\/p>\n\n\n\n

Let me define a sublocale of a frame \u03a9, as in [2], as a subset L<\/em> of \u03a9 that is closed under arbitrary infima (taken in \u03a9), and such that \u03c9 \u27f9 x<\/em> is in L<\/em> for every x<\/em> in L<\/em> and every \u03c9 in \u03a9.  Every subspace Y<\/em> of a topological space X<\/em> defines a sublocale of the frame O<\/strong>(X<\/em>) of open subsets of X<\/em>, as the family of open subsets U<\/em> of X<\/em> such that \u03bdY<\/em><\/sub>(U<\/em>)=U<\/em>, where \u03bdY<\/em><\/sub>(U<\/em>) denotes the largest open subset of X<\/em> with the same intersection with Y<\/em> that U<\/em> has. (In turn, \u03bdY<\/em><\/sub> is the nucleus<\/em> associated with Y<\/em>.)<\/p>\n\n\n\n

In general, there are many more sublocales than there are subspaces. For example, let us recall Isbell’s density theorem<\/a>: any frame contains a smallest dense sublocale. Moreover, the smallest dense sublocale of O<\/strong>(X<\/em>) 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<\/em> is dense-in-itself, namely if it has no isolated point.<\/p>\n\n\n\n

We claim that the smallest dense sublocale of O<\/strong>(X<\/em>), where X<\/em> is dense-in-itself and non-empty, cannot be the sublocale associated with any subspace of X<\/em>. Indeed, since X<\/em> has no isolated point, X<\/em>\u2013{x<\/em>} is dense for every point x<\/em>, so that if the sublocale were associated with a subspace Y<\/em>, that Y<\/em> would have to be included in X<\/em>\u2013{x<\/em>} for every x<\/em>. Hence Y<\/em> would have to be empty. But then Y<\/em> would be dense in X<\/em>, which is non-empty, and that is absurd.<\/p>\n\n\n\n

But there is more. It may be that a given sublocale is associated with two or more <\/em>subspaces!<\/p>\n\n\n\n

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<\/em><\/sub> spaces, as we now show.<\/p>\n\n\n\n

Proposition.<\/strong> Let Y<\/em> and Z<\/em> be two subspaces of the same topological space X<\/em>, and assume that they define the same sublocale of O<\/strong>(X<\/em>). If X<\/em> is TD<\/em><\/sub>, then Y<\/em>=Z<\/em>.<\/p>\n\n\n\n

Proof.<\/em> The assumption means that: (*) for every pair of open sets U<\/em> and V<\/em>, it is equivalent to say that U<\/em> \u2229 Y<\/em>=V<\/em> \u2229 Y<\/em> or that U<\/em> \u2229 Z<\/em>=V<\/em> \u2229 Z<\/em>. Indeed, U<\/em> \u2229 Y<\/em>=V<\/em> \u2229 Y<\/em> is equivalent to \u03bdY<\/em><\/sub>(U<\/em>)=\u03bdY<\/em><\/sub>(V<\/em>), and similarly with Z<\/em> in place of Y<\/em>. The assumption means that \u03bdY<\/em><\/sub>=\u03bdZ<\/em><\/sub>, whence (*).<\/p>\n\n\n\n

By taking complements, (*) entails that for every pair of closed sets C<\/em> and D<\/em>, C<\/em> \u2229 Y<\/em>=D<\/em> \u2229 Y<\/em> is equivalent to C<\/em> \u2229 Z<\/em>=D<\/em> \u2229 Z<\/em>.<\/p>\n\n\n\n

Let us assume that Y<\/em>\u2260Z<\/em>. Without loss of generality, there is a point x<\/em> in Y<\/em> that is not in Z<\/em>. Since X<\/em> is TD<\/em><\/sub>, C<\/em> = \u2193x<\/em> \u2013 {x<\/em>} is closed. Let D<\/em> = \u2193x<\/em>. Then C<\/em> \u2229 Z<\/em>=D<\/em> \u2229 Z<\/em>, since x<\/em> is not in Z<\/em>. Hence C<\/em> \u2229 Y<\/em>=D<\/em> \u2229 Y<\/em>. However, x<\/em> is in D<\/em> \u2229 Y<\/em> but not in C<\/em> \u2229 Y<\/em>, which is absurd. \u2610<\/p>\n\n\n\n

And the TD<\/em><\/sub> condition is necessary:<\/p>\n\n\n\n

Proposition.<\/strong> If X<\/em> is not TD<\/em><\/sub>, then it has two distinct subspaces that define the same sublocale.<\/p>\n\n\n\n

Proof.<\/em> Since X<\/em> is not TD<\/em><\/sub>, it contains a point x<\/em> that is not isolated in \u2193x<\/em>. This means that every open neighborhood U<\/em> of x<\/em> intersects \u2193x<\/em> \u2013 {x<\/em>}. We let Y<\/em> = \u2193x<\/em> \u2013 {x<\/em>}, Z<\/em> = \u2193x<\/em>. The nucleus \u03bdZ<\/em><\/sub> maps every open set U<\/em> to U<\/em> \u222a (X<\/em> \u2013 \u2193x<\/em>): this is the closed nucleus c<\/strong>(X<\/em> \u2013 \u2193x<\/em>), as defined at the end of this post<\/a>.<\/p>\n\n\n\n

Let us compute \u03bdY<\/em><\/sub>. For every open set U<\/em>, \u03bdY<\/em><\/sub>(U<\/em>) is the largest open set V<\/em> such that V<\/em> \u2229 Y<\/em>=U<\/em> \u2229 Y<\/em>. Any such V<\/em> must be such that V<\/em> \u2229 Y<\/em> \u2286 U<\/em>. If V<\/em> contains x<\/em>, by assumption it intersects \u2193x<\/em> \u2013 {x<\/em>} = Y<\/em>, say at y<\/em>; then y<\/em> is in U<\/em>, and since y<\/em>\u2264x<\/em> and U<\/em> is upwards-closed, x<\/em> must also be in U<\/em>; this entails that V<\/em> \u2229 Z<\/em>, which is equal to (V<\/em> \u2229 Y<\/em>) \u222a {x<\/em>}, is also included in U<\/em>. If V<\/em> does not contain x<\/em>, then V<\/em> \u2229 Y<\/em>=V<\/em> \u2229 Z<\/em>, so trivially V<\/em> \u2229 Z<\/em> \u2286 U<\/em>. Whichever is the case, V<\/em> \u2229 Z<\/em> \u2286 U<\/em>, so V<\/em> is included in \u03bdZ<\/em><\/sub>(U<\/em>). Therefore \u03bdY<\/em><\/sub>(U<\/em>) \u2286 \u03bdZ<\/em><\/sub>(U<\/em>).<\/p>\n\n\n\n

Since Y<\/em> is included in Z<\/em>, we see easily that \u03bdZ<\/em><\/sub>(U<\/em>) is included in \u03bdY<\/em><\/sub>(U<\/em>) for every open set U<\/em>. Indeed, every open set V<\/em> such that U<\/em> \u2229 Z<\/em>=V<\/em> \u2229 Z<\/em> will also satisfy U<\/em> \u2229 Y<\/em>=V<\/em> \u2229 Y<\/em>, hence is included in \u03bdY<\/em><\/sub>(U<\/em>).<\/p>\n\n\n\n

We have shown that \u03bdY<\/em><\/sub> = \u03bdZ<\/em><\/sub>, hence that Y<\/em> and Z<\/em> define the same sublocale. \u2610 <\/p>\n\n\n\n

In other words:<\/p>\n\n\n\n

A space in which every sublocale corresponds to at most one subspace is the same thing as a TD<\/sub> space.<\/em><\/p>\n\n\n\n

Lattice equivalence<\/h2>\n\n\n\n

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<\/em><\/sub> property are incomparable, we can also reconstruct X<\/em> from O<\/strong>(X<\/em>) when X<\/em> is a TD<\/em><\/sub> space.<\/p>\n\n\n\n

Let us see how this can be done. Given any point x<\/em> in a TD<\/em><\/sub> space X<\/em>, and as in every space, the family N<\/em>(x<\/em>) of open neighborhoods of x<\/em> is a completely prime filter in O<\/strong>(X<\/em>). If X<\/em> were sober, we could recover a unique point from any completely prime filter of open sets, but in a non-sober TD<\/em><\/sub> space (such as N<\/strong> 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<\/strong>…) that are not of the form N<\/em>(x<\/em>) for any point x<\/em>.<\/p>\n\n\n\n

In fact, in a TD<\/em><\/sub> space, N<\/em>(x<\/em>) has an additional property: there are two open sets, namely the complement U<\/em> of \u2193x<\/em> and the complement V<\/em> of \u2193x<\/em> \u2013 {x<\/em>}, such that V<\/em> is in N<\/em>(x<\/em>), U<\/em> is not in N<\/em>(x<\/em>), and U<\/em> is immediately below<\/em> V<\/em>. The latter means that U<\/em> is strictly below (strictly included in) V<\/em>, but there is absolutely no other open set between U<\/em> and V<\/em>.<\/p>\n\n\n\n

Picado and Pultr [2, page 5] call slicing<\/em> any prime filter with that property. In general, a filter F<\/em><\/strong> of elements of a frame \u03a9 is slicing if and only if it is prime<\/em> (not containing the bottom element and such that if it contains u<\/em> \u22c1 v<\/em> then it contains u<\/em> or it contains v<\/em>), and there is an element u<\/em> not in F<\/em><\/strong>, an element v<\/em> in F<\/em><\/strong> such that u<\/em> is immediately below v<\/em>, in the sense that there is no element of \u03a9 strictly above u<\/em> and strictly below v<\/em>. Note that we do not require F<\/em><\/strong> to be completely<\/em> prime, only prime, by the way.<\/p>\n\n\n\n

Lemma.<\/strong> In a T0<\/sub> space X<\/em>, each slicing filter is of the form N<\/em>(x<\/em>) for a unique point x<\/em>.<\/p>\n\n\n\n

Proof.<\/em> Let F<\/em><\/strong> be a slicing filter of open sets. There is an open set U<\/em> outside F<\/em><\/strong> and immediately below some other open set V<\/em> inside F<\/em><\/strong>. Since U<\/em> is strictly contained in V<\/em>, there is a point x<\/em> in V<\/em> \u2013 U<\/em>. We will show that F<\/em><\/strong>=N<\/em>(x<\/em>). The uniqueness of x<\/em> will follow from the fact that X<\/em> is T0<\/sub>: in a T0<\/sub> space, any two points with the same set of open neighborhoods must be equal.<\/p>\n\n\n\n

For every open neighborhood O<\/em> of x<\/em>, let us consider the open set W<\/em> = V<\/em> \u2229 (U<\/em> \u222a O<\/em>) = U<\/em> \u222a (V<\/em> \u2229 O<\/em>). This lies between U<\/em> and V<\/em>, hence is equal to one of them. But x<\/em> is in W<\/em> (since it is both in V<\/em> and in O<\/em>), and not in U<\/em>, so W<\/em> cannot be equal to U<\/em>. It follows that W<\/em> = V<\/em>. From W<\/em> = V<\/em> \u2229 (U<\/em> \u222a O<\/em>), hence V<\/em> = V<\/em> \u2229 (U<\/em> \u222a O<\/em>), we deduce that V<\/em> is included in U<\/em> \u222a O<\/em>. Since V<\/em> is in F<\/em><\/strong>, U<\/em> \u222a O<\/em> is in F<\/em><\/strong>, and because F<\/em><\/strong> is prime, U<\/em> or O<\/em> is in F<\/em><\/strong>. However, U<\/em> is not in F<\/em><\/strong>, so O<\/em> is. It follows that N<\/em>(x<\/em>) is included in F<\/em><\/strong>.<\/p>\n\n\n\n

Conversely, we claim that every element O<\/em> of F<\/em><\/strong> contains x<\/em>. Since F<\/em><\/strong> is a filter, V<\/em> \u2229 O<\/em> is also in F<\/em><\/strong>. We consider again W<\/em> = V<\/em> \u2229 (U<\/em> \u222a O<\/em>) = U<\/em> \u222a (V<\/em> \u2229 O<\/em>), and by the same argument W<\/em> is equal to U<\/em> or to V<\/em>. It cannot be equal to U<\/em>, otherwise U<\/em> = U<\/em> \u222a (V<\/em> \u2229 O<\/em>), so V<\/em> \u2229 O<\/em> would be included in U<\/em>, and that would imply that U<\/em> is also in F<\/em><\/strong>. Therefore W<\/em> = V<\/em>, namely V<\/em> \u2229 (U<\/em> \u222a O<\/em>) = V<\/em>, meaning that V<\/em> is included in U<\/em> \u222a O<\/em>. Since V<\/em> contains x<\/em> but U<\/em> does not, O<\/em> must contain x<\/em>. This shows that F<\/em><\/strong> is included in N<\/em>(x<\/em>). \u2610<\/p>\n\n\n\n

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

Theorem ([3]).<\/strong> Let X<\/em> and Y<\/em> be two topological spaces with isomorphic lattices of open sets, and let \u03c6 : O<\/strong>(Y<\/em>) \u2192 O<\/strong>(X<\/em>) be such an isomorphism. If X<\/em> is TD<\/em><\/sub> and Y<\/em> is T0<\/sub>, then there is a unique homeomorphism f<\/em> : X<\/em> \u2192 Y<\/em> such that \u03c6=f<\/em>-1<\/sup>.
In particular, any two lattice equivalent TD<\/em><\/sub> spaces are homeomorphic.<\/p>\n\n\n\n

Proof. The key is to observe that for every slicing filter F<\/em><\/strong> of open sets of Y<\/em>, \u03c6-1<\/sup>(F<\/em><\/strong>) is also a slicing filter. (I will always write \u03c6-1<\/sup> for “the inverse image by \u03c6”, and I will call \u03c8 the inverse \u03c6. Of course \u03c6-1<\/sup>(F<\/em><\/strong>) is also the directed image of F<\/em><\/strong> by \u03c8.) This is easy, but funnily, that would not work if we had only assumed \u03c6 to be a frame homomorphism: in that case, we would be able to prove that given any prime filter F<\/em><\/strong> of open sets of Y<\/em>, \u03c6-1<\/sup>(F<\/em><\/strong>) is a prime filter, but slicing may fail to be preserved. Given that \u03c6 is an isomorphism, there is no difficulty in showing that \u03c6-1<\/sup> preserves slicing, however.<\/p>\n\n\n\n

If f<\/em> exists as specified, then f<\/em>-1<\/sup>(N<\/em>(f<\/em>(x<\/em>))) should be equal to N<\/em>(x<\/em>), so one must have N<\/em>(f<\/em>(x<\/em>))=\u03c8(N<\/em>(x<\/em>))=\u03c6-1<\/sup>(N<\/em>(x<\/em>)). This defines f<\/em>(x<\/em>) uniquely since Y<\/em> is T0<\/sub>, where any point is defined uniquely (if at all) by its set of open neighborhoods.<\/p>\n\n\n\n

In order to show the existence of f<\/em>, we observe that for every x<\/em> in X<\/em>, N<\/em>(x<\/em>) is a slicing filter of open sets, hence also \u03c6-1<\/sup>(N<\/em>(x<\/em>)), and by the previous lemma it is equal to N<\/em>(f<\/em>(x<\/em>)) for some unique point f<\/em>(x<\/em>).<\/p>\n\n\n\n

We claim that \u03c6=f<\/em>-1<\/sup>, namely that for every open subset V<\/em> of Y<\/em>, \u03c6(V<\/em>)=f<\/em>-1<\/sup>(V<\/em>). For every x<\/em> in X<\/em>, x<\/em> is in f<\/em>-1<\/sup>(V<\/em>) if and only if f<\/em>(x<\/em>) is in V<\/em>, if and only if V<\/em> is in N<\/em>(f<\/em>(x<\/em>)). Since the latter is equal to \u03c6-1<\/sup>(N<\/em>(x<\/em>)), this is equivalent to \u03c6(V<\/em>) \u2208 N<\/em>(x<\/em>), hence to x<\/em> \u2208 \u03c6(V<\/em>).<\/p>\n\n\n\n

In particular, f<\/em>-1<\/sup>(V<\/em>) = \u03c6(V<\/em>) is open for every open subset V<\/em> of Y<\/em>, so f<\/em> is continuous.<\/p>\n\n\n\n

For every open subset U<\/em> of X<\/em>, U<\/em> is equal to \u03c6(V<\/em>) for some open subset V<\/em> of Y<\/em>, since \u03c6 is surjective; hence to f<\/em>-1<\/sup>(V<\/em>). This shows that f<\/em> is almost open.<\/p>\n\n\n\n

For every point y<\/em> of Y<\/em>, N<\/em>(y<\/em>) is a slicing filter of open subsets of Y<\/em>. Recall that \u03c8 is the inverse of \u03c6. Reasoning with \u03c8 instead of \u03c6, \u03c8-1<\/sup>(N<\/em><\/em>(y<\/em><\/em>)) is also a slicing filter, hence a filter of the form N<\/em>(x<\/em>) for some point x<\/em> of X<\/em>. Then N<\/em><\/em>(y<\/em><\/em>)=\u03c6-1<\/sup>(N<\/em>(x<\/em>)), which is equal to N<\/em>(f<\/em>(x<\/em>)). It follows that y<\/em>=f<\/em>(x<\/em>). This shows that f<\/em> is surjective.<\/p>\n\n\n\n

Finally, imagine that f<\/em>(x<\/em>)=f<\/em>(x’<\/em>). Then N<\/em>(f<\/em>(x<\/em>))=N<\/em>(f<\/em>(x<\/em>‘)), so \u03c6-1<\/sup>(N<\/em>(x<\/em>))=\u03c6-1<\/sup>(N<\/em>(x<\/em>‘)), and since \u03c6 is bijective, it follows that N<\/em>(x<\/em>)=N<\/em>(x<\/em>‘). Since X<\/em> is TD<\/em><\/sub> hence T0<\/sub>, x<\/em>=x’<\/em>. Therefore f<\/em> is injective. \u2610<\/p>\n\n\n\n

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<\/em> is a non-sober space, then X<\/em> and its sobrification S<\/strong>(X<\/em>) are lattice equivalent, but cannot be homeomorphic.<\/p>\n\n\n\n

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<\/strong> functor of Stone duality, and recalling that for every sober space X<\/em>, pt<\/strong>(O<\/strong>(X<\/em>)) and X<\/em> are homeomorphic.<\/p>\n\n\n\n

There are several variants of this question. For example, the following is the Ho-Zhao problem<\/em>: 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?<\/p>\n<\/div><\/div>\n\n\n\n

    \n
  1. Charles Edward Aull and Wolfgang Joseph Thron, Separation axioms between T0<\/sub> and T1<\/sub><\/a>. Indagationes Mathematicae 23, pages 26-37, 1962.<\/li>\n\n\n\n
  2. Jorge Picado and Ale\u0161 Pultr.  Frames and locales \u2014 topology without points.  Birkh\u00e4user, 2010.<\/li>\n\n\n\n
  3. Wolfgang Joseph Thron. Lattice-equivalence of topological spaces. Duke Mathematical Journal, 29:671\u2013679, 1962.<\/li>\n\n\n\n
  4. Weng Kin Ho, Jean Goubault-Larrecq, Achim Jung, and Xiaoyong Xi. The Ho-Zhao problem<\/a>. Logical Methods in Computer Science, 14(1:7), 1-19, 2018. doi: 10.23638\/LMCS-14(1:7)2018<\/li>\n<\/ol>\n\n\n
    \n
    \"jgl-2011\"<\/figure>\n<\/div>\n\n\n

    \u2014 Jean Goubault-Larrecq<\/a> (July 20th, 2020)<\/p>\n","protected":false},"excerpt":{"rendered":"

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