Irreducible elements and irredundant families
If you are familiar with sober spaces, you know that an irreducible closed subset of a topological space Z is a non-empty closed subset C such that, for all closed subsets C1 and C2, if C is included in C1 ∪ C2, then C is included in C1 or in C2.
In general, given any family L of subsets of a set Z, let me say that E ∈ L is irreducible if and only if E is non-empty and for all elements E1 and E2 of L, if E is included in E1 ∪ E2, then E is included in E1 or in E2.
Let me call a family L of subsets of Z irredundant if and only if all its non-empty elements are irreducible. More specifically, I will be interested in irredundant ∩-semilattices, where a ∩-semilattice is simply a collection of sets that is closed under binary intersections. That may look like a stupidly overconstrained notion, to the point that one may legitimitely ask whether there is any non-trivial example.
This is a very interesting notion, as I will attempt to demonstrate. It was introduced in [1], and the credit is entirely due to the late Klaus Keimel. Also, it is at the root of a clever argument due to Zhenchao Lyu and Xiadong Jia [2], which exactly characterizes when the Smyth powerdomain Q(X) of a topological space X is core-compact.
Two notes, before I start:
- The result of [2] was found by Zhenchao Lyu (and perhaps Xiaodong Jia) while he and Xiaodong Jia were postdocs in my lab in 2018-2019. (Update, January 21, 2022: X. Jia tells me his role in this was only in asking the question, and that the solution is entirely due to Zh. Lyu.) I was expecting them to be able to publish this quickly, but somehow I can only guess that something went wrong. I usually have the policy of not talking about a paper before it is published (except for my own ideas), but this is so neat (and available from arXiv) that I have decided to talk about it without waiting any further. (Update, January 21, 2022: also, it is cited as reference 38 in a recent survey paper by Xiaoquan Xu and Dongsheng Zhao.)
- There is some ambiguity in the definition of “irredundant” in [1], which may lead to some mistakes. I have given a clarified definition above. For more details on the ambiguity, see Appendix A at the end of this post.
The main example: the irredundant base of the upper Vietoris topology on Q(X)
There are at least two examples of an irredundant ∩-semilattice in [1], and here is the one which will be the center of our attention in this post.
Fact. For every topological space X, the base of open sets ☐U of the Smyth hyperspace Q(X), where U ranges over the open subsets of X, is an irredundant ∩-semilattice.
Let me recall that Q(X) is the set of non-empty compact saturated subsets of X. (Omitting “non-empty” would not change much in the sequel.) The collection of sets ☐U, U ∈ OX, is a base for a topology called the upper Vietoris topology, and Q(X) with that topology is the Smyth hyperspace of X.
The proof of the above Fact is easy. The sets ☐U form a ∩-semilattice because ☐U ∩ ☐V = ☐(U ∩ V). Let now ☐U be non-empty and included in ☐V1 ∪ ☐V2. For the sake of contradiction, let us assume that ☐U is not included in ☐V1, and not included in ☐V2 either. Then there is a non-empty compact saturated subset Q1 of X that is included in U but not in V1, and there is a non-empty compact saturated subset Q2 of X that is included in U but not in V2. We form Q1 ∪ Q2, which is a non-empty compact saturated subset of X that is included in U, but is included neither in V1 nor in V2, contradicting ☐U ⊆ ☐V1 ∪ ☐V2.
The other example is the collection of crescents of the form ☐U–♢V in the space of compact lenses (also known as the Plotkin powerdomain) with the Vietoris topology. I will not expand on that.
When is Q(X) core-compact? The Lyu-Jia theorem
First, let me write ↑Q Q for the the upward closure of a single point Q of Q(X). That is taken with respect to the specialization ordering of Q(X), which is reverse inclusion ⊇. Hence ↑Q Q is the set of non-empty compact saturated sets Q’ that are included in Q. I know, this may sound confusing at first.
Extending the notation ☐U, I could have just written ↑Q Q as ☐Q, the set of non-empty compact saturated sets included in Q. The ☐ operator, defined by ☐A ≝ {Q ∈ Q(X) | Q ⊆ A} for every subset A of X, is monotonic and commutes with finite intersections.
It is well-known that, for every locally compact space X, Q(X) is locally compact. Indeed, let Q ∈ Q(X), and let U be an open neighborhood of Q. U contains a basic open set ☐U, where U is open in X, such that Q is in ☐U, namely such that Q is included in U. By the interpolation property in locally compact spaces (Proposition 4.8.14 in the book), there is (necessarily non-empty) compact saturated subset Q’ of X such that Q ⊆ U’ ⊆ Q’ ⊆ U, where U’ is the interior of Q’. Therefore Q ∈ ☐U‘ ⊆ ☐Q’ ⊆ ☐U. Since ↑Q Q‘ = ☐Q’, we can interpret the latter as saying that Q is in the interior of ↑Q Q‘, which is itself included in ☐U, hence in U.
We have in fact just shown more: for every locally compact space X, Q(X) is a c-space. (See Section 5.1.2 in the book. A locally compact space is a space Z in which for every point z, and for every open neighborhood W of z, we can find a compact saturated neighborhood K of z included in W. In a c-space, we require that K be the upward closure of a single point, just like ↑Q Q‘ in the previous paragraph.)
It is also known that Q(X) is locally compact if and only if X is locally compact. (We will reprove this as a consequence of a more general result below.)
What Zhenchao was interested in was whether one could prove a similar theorem with “core-compact” replacing “locally compact”. The perhaps surprising outcome is that… no, you cannot.
But let us not move too fast. Here we are right now. We know of the following implications:
X loc. compact ⇒ Q(X) c-space ⇒ Q(X) loc. compact ⇒ Q(X) core-compact,
where the last two implications hold with any space in place of Q(X); the main result of [2] is to show that this is actually a string of equivalences.
In order to see this, we assume that Q(X) is core-compact, and we will show that X is locally compact.
Although not strictly needed, let us extend the notion of irreducibility and of irredundancy to a more abstract setting. (I am saying that this is not strictly needed, because I will always apply that to the concrete cases of irredundant ∩-semilattices as above.) Given a sup-semilattice Ω, a subset L of Ω is irredundant if and only if every element u of L different from ⊥ is irreducible in the following sense: u≠⊥ and for all v1, v2 in L such that u ≤ v1 ⋁ v2, we have u ≤ v1 or u ≤ v2.
A prime-continuous lattice is a complete lattice in which every element is the supremum of the elements that are way-way-below it. The prime-continuous lattices are exactly the completely distributive, complete lattices by Raney’s Theorem (Exercise 8.3.16 in the book).
The key point is the following, which we will apply to the case where Ω is the lattice of open subsets of Q(X). (Let me recall that a space is core-compact if and only if its open set lattice is continuous, see Definition 5.2.3 in the book.)
Lemma. Let Ω be a complete lattice, and L be a family of elements of Ω. If Ω is continuous, if L generates Ω (in the sense that every element of Ω is the supremum of a family of elements of L), and if L is irredundant, then Ω is prime-continuous.
Proof. Let us assume that every element of Ω is the supremum of elements of L. We will require the other assumptions later. The main idea is to use irredundancy in order to prove the following:
- For every irreducible element p of L, for every v in Ω, p is way-way-below v (p⋘v) if and only if p is way-below v (p≪v).
Such a property is well-known if p is coprime in Ω (namely, p≠⊥ and for all v1, v2 in Ω such that u ≤ v1 ⋁ v2, we have u ≤ v1 or u ≤ v2, see Exercise 8.3.47 of the book), but we only require that p be irreducible in L, which is a much weaker property, and also an easier property to verify. (Have you seen the difference, by the way? This is subtle: in the definition of coprime, v1, v2 vary in the whole of Ω, while they range over the smaller set L in the definition of irreducible.)
Anyway, here is how we prove that p⋘v if and only if p≪v, when p is irreducible in L.
Let me recall that u⋘v if and only if every family (not necessarily directed) whose supremum lies above v contains an element that is already above u. Notably, u⋘v implies u≪v.
Conversely, if p≪v and if p is an irreducible element of L, let us consider any family F of elements of Ω such that v≤sup F. Every element w of F is a supremum of a family Fw of elements of L, so sup F = sup (∪w ∈ F Fw). Since sup (∪w ∈ F Fw) can also be written as the directed supremum of all suprema of finite subfamilies of ∪w ∈ F Fw, and using the definition of the way-below relation ≪, p ≤ u1 ⋁ … ⋁ un for some finite subfamily {u1, …, un} of ∪w ∈ F Fw. Since p is irreducible (and since u1, …, un are all in L), we must have p ≤ ui for some i; ui is in Fw for some w ∈ F, so ui ≤ w, and therefore p ≤ w. This shows that p⋘v.
We now assume that L is irredundant. What we have just shown simplifies to: for every p ∈ L, for every v in Ω, p⋘v if and only if p≪v.
Finally, we also assume that Ω is continuous. For every v ∈ Ω, we can write v as the supremum of a directed family of elements vi≪v, where i ranges over some indexing set I. Each vi is a supremum of a family Fi ⊆ L, so v is the supremum of the family ∪i ∈ I Fi. That family is a family of elements p of L, and each of them is not just way-below, but way-way-below v, as we have just seen. Hence Ω is prime-continuous. ☐
As an application, and as promised, we specialize that to the case where Ω=O(Q(X)) and L is the base of the upper Vietoris topology consisting of sets of the form ☐U, U ∈ O(X), which is our primary example of an irredundant ∩-semilattice:
Corollary. For every topological space X, if Q(X) is core-compact then it is a c-space.
Proof. If Q(X) is core-compact, then O(Q(X)) is continuous, hence prime-continuous by the previous Lemma. But every space whose lattice of open sets is prime-continuous is a c-space. Lemma 8.3.42 in the book almost proves this… but assuming that the space is sober. This is really silly: it is pretty easy to see that this assumption is not necessary, and the easiest way to see this is as follows. (Update, January 22nd, 2022: or see the elementary proof in Appendix B.)
Let Y be a space such that O(Y) is prime-continuous, and let S(Y) be the sobrification of Y. Let me recall that the elements of S(Y) are the irreducible closed subsets of Y, and that the open subsets of S(Y) are the sets ♢U ≝ {C ∈ S(Y) | C intersects U}, for every open subset U of Y. The latter defines an order-isomorphism U ↦ ♢U of O(Y) onto O(S(Y)). Then O(S(Y))≅O(Y) is prime-continuous, so S(Y) is a c-space, using Lemma 8.3.42.
It remains to see that every space Y such that S(Y) is a c-space is itself a c-space. In order to see this, let y be any point in Y and let V be any open neighborhood of y in Y. Then ↓y is a point of S(Y), and ♢V is an open neighborhood of ↓y. Since S(Y) is a c-space, there is an element C of ♢V and an open set ♢U such that ↓y ∈ ♢U ⊆ ↑S C, where ↑S C denotes the upward closure {C’ ∈ S(Y) | C ⊆ C’} of C in S(Y). Since C is in ♢V, it intersects V, say at z. The inclusion ♢U ⊆ ↑S C means that every irreducible closed set that intersects U contains C, and therefore that for every point x in U, ↓x contains C; in particular, z≤x. Since x is arbitrary in U, this shows that U ⊆ ↑z. Finally, ↓y ∈ ♢U means that y is in U. In summary, we have obtained a point z in V such that y ∈ U ⊆ ↑z, and this concludes the proof that Y is a c-space. ☐
By the way, Lemma 8.3.41 in the book states that every c-space Y is such that O(Y) is prime-continuous, and therefore the c-spaces are exactly the spaces whose lattice of open sets is prime-continuous. This had been known for a long time. As far as I can tell, this appears as the equivalence between (a) and (e) in Proposition 2.2.C of Marcel Erné’s 1991 paper on a-spaces, b-spaces, and c-spaces [3].
We finally have:
Fact. If Q(X) is a c-space, then X is locally compact.
Indeed, let x be any point of X and U be any open neighborhood of x in X. Then ↑x is in ☐U, and since Q(X) is a c-space, ↑x is in the interior of ↑Q Q = ☐Q for some element Q of ☐U. Namely, there is a basic open set ☐V such that ↑x ∈ ☐V ⊆ ☐Q ⊆ ☐U. Observing that for all sets A, B, the inclusion ☐A ⊆ ☐B implies A ⊆ B (for every z in A, ↑z is in ☐A, hence in ☐B, and therefore z is in B), it follows that x ∈ V ⊆ Q ⊆ U.
Combining this with the previous corollary, if Q(X) is core-compact then X is locally compact. We have obtained the promised chain of equivalences.
Theorem [2, Theorem 3.1]. For every topological space X, the following equivalences hold:
X loc. compact ⇔ Q(X) c-space ⇔ Q(X) loc. compact ⇔ Q(X) core-compact.
It follows that the Q functor does not preserve core-compactness in general: given any space X that is core-compact but not locally compact, such as this one, Q(X) is not core-compact. This is Corollary 3.4 of [2].
The baby Groemer theorem
The original application that Klaus found of irredundant ∩-semilattices [1] was the following problem.
A lattice of sets is a collection Ω of subsets of a given set X that is closed under finite unions and finite intersections. Among the finite intersections, one finds the intersection of the empty family, so X itself is in X. Among the finite unions, one finds the union of the empty family, so the empty set ∅ is in Ω. A lattice of sets is of course a bounded lattice, consisting of subsets of a set X, but the definition also requires that finite infima are intersections and that finite suprema are unions.
Any topology, any σ-algebra is a lattice of sets.
Let us called signed valuation on a lattice of sets Ω any map μ : Ω → R that is:
- strict: μ(∅)=0
- modular: for all U, V in Ω, μ(U ∪ V)=μ(U)+μ(V)–μ(U ∩ V).
For example, any signed measure on a measurable set is a signed valuation on its σ-algebra. The main differences between a signed measure and a measure is that: 1. a signed measure is not required to give non-negative values to sets 2. (less easy to see) a signed measure cannot give infinite measure to any measurable set.
I have already talked briefly about valuations (not signed valuations) in various posts. Valuations are a close cousin of measures, much as signed valuations are a closed cousin of signed measures. You can have a look here, for a nifty idea of Alex Simpson’s about a formalization of the notion of random elements of a space; or the appendix of this post, where I build the so-called Lebesgue valuation on R and on the Sorgenfrey line by a purely domain-theoretic method.
One of the main, basic problems in measure or valuation theory is showing that we can build a measure (or valuation) satisfying certain constraints. For example, building the Lebesgue measure on R reduces to showing that there is a measure on R that maps every interval [a, b] to b–a. There are many theorems that allow you do such things. In measure theory, one usually builds the Lebesgue measure by using the so-called Carathéodory extension theorem.
In [1], one of our basic problems was the following. Let X be a topological space, and let us consider the lattice of sets Ofin(Q(X)) obtained as finite unions of basic open subsets ☐U of Q(X). Ofin(Q(X)) is not the full topology O(Q(X)), and the reason I am considering this here is because this will make the argument simpler, and it will allow me to concentrate on the core of the more complex arguments of [1].
Now imagine you are given values ν(U) for each open subset U of X, and you wish to build a signed valuation μ on Ofin(Q(X)) such that μ(☐U)=ν(U) for every U. Is that possible? Under which conditions?
Naturally, when U is empty, we require ν(∅) to be equal to 0, otherwise there would be no solution for μ. The nice thing is that ν(∅)=0 is the only constraint that ever has to be satisfied for μ to exist; and, in fact, μ is then uniquely determined from the knowledge of ν.
Klaus realized that this followed from a theorem due to Groemer [4], which he found by randomly browsing through a few books on his shelf, and in particular a book by Klain and Rota [5], where this theorem is stated and proved. Klaus also realized that we only needed a very special case of that theorem, and that this very special case had an amazingly simple proof. (I had proved all the main theorems of [1] before Klaus found all that. My proofs were extremely laborious, and Klaus showed how to simplify them drastically using those wonderful theorems, sometimes reducing my original proofs by a factor between 20 and 40.)
I will call that very special case the baby Groemer theorem.
So here is how we proceed. We fix a set X. For every subset A of X, we can build the characteristic function χA, which maps every element of A to 1, and all other elements to 0. The following is an easy exercise, and should remind you of the modularity requirement of signed valuations.
Fact. For all sets U, V, χU ∪ V=χU+χV–χU ∩ V.
The collection of all maps from X to R is a real vector space under pointwise addition and scalar multiplication. For example, every characteristic map χU is a vector in that vector space, as is any linear combination such as χU–2χV, for example.
The key ingredient that will lead us to the baby Groemer theorem is the following.
Lemma. Let L be any irredundant ∩-semilattice of subsets of X. The vectors χU, where U ranges over the non-empty sets in L, are linearly independent.
Proof. Let us assume that ∑U aU . χU = 0, where U ranges over a non-empty finite subset E of L–{∅} and each aU is non-zero. Let U0 be a maximal element of E with respect to inclusion, and U1, …, Un be the other elements of E (namely, Ui≠U0 for every i, 1≤i≤n). If U0 were included in U1 ∪ … ∪ Un, then by irredundancy U0 would be included in some Ui, hence equal to it by maximality. That is impossible since Ui≠U0. Hence U0 is not included in U1 ∪ … ∪ Un. Let x be a point in U0 that is not in U1 ∪ … ∪ Un. Evaluating ∑U aU . χU on x, all the terms except the term U=U0 vanish, and we obtain the non-zero value aU0: this contradicts the fact that ∑U aU . χU = 0. ☐
And here is the baby Groemer theorem. The lattice of subsets L∪ generated by L is the smallest lattice of subsets containing L, namely the smallest collection of subsets containing L and closed under finite unions and finite intersections. When L is a ∩-semilattice, the elements of L∪ are juste the finite unions of elements of L. The Boolean algebra of subsets A(L) generated by L is the smallest family of subsets of X that contains L and is closed under finite unions, finite intersections, and complements. Its elements are the finite unions of finite intersections of L-literals, where an L-literal is an element of L or a complement of an element of L. I will also call L-clause any finite intersection of L-literals. It is relatively easy to check that the elements of A(L) are also the disjoint finite unions of L∪-crescents, where an L∪-crescent is a set of the form A–B where A and B are in L∪ and B is included in A.
Theorem. Let L be any irredundant ∩-semilattice of subsets of X. Given any map ν from L–{∅} to R, there is a unique signed valuation μ on L∪, and in fact on A(L), that extends ν.
Proof. Let F(L) be the linear space of functions from X to R generated by the functions χU, where U ranges over L–{∅}. By the previous Lemma, those functions form a basis of F(L). Hence there is a unique linear map f from F(L) to R such that f(χU)=ν(U) for every U in L–{∅}.
We verify that for every A in L∪, χA is in F(L). This is easy. A is a finite union U1 ∪ … ∪ Un of elements of L, and we prove this by induction on n. If n=0, then χ∅=0 is in F(L). If n=1, this is by definition of F(L). Otherwise, let B ≝ U2 ∪ … ∪ Un. By the Fact seen above, χA = χU1+χB–χU1∩B; χU1 is in F(L) by definition, χB is in F(L) by induction hypothesis, and χU1∩B = χ(U1 ∩ U2) ∪ … ∪ (U1 ∩ Un) is also in F(L) by induction hypothesis, since U1 ∩ U2, …, U1 ∩ Un are all in L, since L is a ∩-semilattice of subsets.
It follows that for every L∪-crescent C ≝ A–B, χC is also in F(L), since χC=χA–χB.
We then obtain that for every E in A(L), χE is in F(L). Indeed, E is a finite disjoint union of L∪-crescents Ci, 1≤i≤n, and then χE = Σi=1n χCi.
We now define μ by μ(E) ≝ f(χE) for every E in A(L). This is legitimate, since we have just proved that χE is in F(L). This map μ is:
- strict, because μ(∅) = f(χ∅) = f(0) = 0;
- modular, because for all E, F in A(L), μ(E ∪ F) = f(χE ∪ F) = f(χE)+f(χF)–f(χE ∩ F)=μ(E)+μ(F)–μ(E ∩ F), by the Fact once again, and the linearity of f.
- Finally, μ extends ν, since for every U in L, μ(U)=f(χU)=ν(U), by definition of f.
That was the essential construction of the proof: just a bit of elementary linear algebra, resting on the irredundancy of L in order to obtain a base of F(L).
In order to see that μ is unique (as a valuation), we reason as follows. The value of μ on elements A ≝ U1 ∪ … ∪ Un of L∪ is uniquely determined from the values it takes on individual elements U1, …, Un, by induction on n, using (strictness and) modularity. (Let me leave the details to you.) For each L-crescent C ≝ A–B, we must have μ(A)=μ(B ∪ C)=μ(B)+μ(C)–μ(B ∩ C)=μ(B)+μ(C)–μ(∅)=μ(B)+μ(C), so μ(C) must be equal to μ(A)–μ(B). This argument shows more generally that if E and F (instead of B and C) are disjoint elements of A(L), then μ(E ∪ F)=μ(E)+μ(F). Using this and strictness, this shows that for every element E of A(L), written as a finite disjoint union of L∪-crescents Ci, 1≤i≤n, μ(E) is uniquely determined as Σi=1n μ(Ci). ☐
We apply this to the case where L is the irredundant ∩-semilattice of subsets of Q(X) of the form ☐U, where U ranges over the open subsets of a space X, and we obtain that for any function ν from OX–{∅} to R, there is a unique signed valuation μ on the lattice of sets L∪=Ofin(Q(X)), such that μ(☐U)=ν(U) for every non-empty open subset U of X, and we are done.
A final word. The problem we were really interested in was whether there exists a continuous valuation μ such that μ(☐U)=ν(U) for every open subset U of X. The obvious difference is that we now require continuity, and to handle this, we simply extend a valuation μ on Ofin(Q(X)) such that μ(☐U)=ν(U) for every open set U to the whole of O(Q(X)), using Scott’s formula; this assumes that X is locally compact (and that ν(∅)=0). There is a less visible difference: a valuation (not a signed valuation) takes its values in the non-negative reals. In order to ensure that μ is indeed a valuation, a necessary and sufficient condition is that ν satisfy the inequality:
ν(U) ≥ ∑I (-1)|I|+1 ν(∩i∈I Ui)
for all open sets U, U1, …, Un such that U contains U1 ∪ … ∪ Un, and where the summation extends over all non-empty subsets I of {1, …, n}. (This is an inequational form of the so-called inclusion-exclusion formula in probability theory.)
You may read [1] if you are interested. I will not explain, sorry… My point was only to show how the use of the (apparently) overconstrained notion of irredundant ∩-semilattices can be put to good use in order to obtain simple proofs of not completely trivial results!
- Jean Goubault-Larrecq and Klaus Keimel. Choquet-Kendall-Matheron theorems for non-Hausdorff spaces. Mathematical Structures in Computer Science 21(3), 2011, pages 511-561.
- Zhenchao Lyu and Xiaodong Jia. Core-compactness of Smyth powerspaces. arXiv:1907.04715, July 2019.
- Marcel Erné. The ABC of order and topology. Pages 57–83 of Category Theory at Work, Proceedings of a Workshop. Research and Exposition in Mathematics 18. Heldermann Verlag, 1991. H. Herrlich and H.-E. Porst, editors.
- Helmut Groemer. On the extension of additive functionals on classes of convex sets. Pacific Journal of Mathematics 75(2):397–410, 1978.
- Daniel A. Klain and Gian-Carlo Rota. Introduction to Geometric Probability. Cambridge University Press, Lezioni Lincee series, 1997.
Appendix A: the ambiguity in [1]
The definition of “irreducible” in [1] is: in an ∩-semilattice L, E is irreducible if and only if one cannot write E as the union of finitely many proper closed subsets still in L. But that is stated in such a way that the reader cannot decide whether we mean “finitely many, possibly 0”, or “finitely many, and at least 1”.
I have decided that we meant “finitely many, possibly 0”. That definition is then equivalent to the definition I gave at the beginning of this post. In particular, that forces E to be non-empty, since the empty set can be written as the union of finitely many (namely, zero) proper closed subsets in L.
Then we defined L as irredundant if and only if all its elements are irreducible. In particular, an irredundant L cannot contain the empty set. This is unfortunate, since our primary example, the collection of sets of the form ☐U, where U ranges over the open subsets of a topological space X, does contain the empty set (as ☐∅).
The easiest fix is what I did in this post: require all non-empty elements of L to be irreducible. With that fix, all theorems of [1], and for that matter, of [2] as well, go through unchanged.
Appendix B: if OX is prime-continuous, then X is a c-space
(Added January 22nd, 2022.) Let X be a topological space, and let us assume that OX is prime-continuous, or equivalently, completely distributive. (See Exercise 8.3.16.)
We claim that, given any family (Di)i ∈ I of downwards closed subsets of X, the closure cl(∩i ∈ I Di) is equal to ∩i ∈ I cl (Di). We write each Di as the union of the sets ↓x, where x ranges over Di. This is possible because each Di is downwards closed. Next, we recall that the powerset of X is completely distributive, so ∩i ∈ I Di = ∩i ∈ I ∪x ∈ Di ↓x = ∪f ∩i ∈ I ↓f(i), where f ranges over Πi ∈ I Di, the set of functions that map each element i of I to an element of Di. The sets ↓f(i) are closed, ∩i ∈ I ↓f(i) is their infimum in the lattice HX of closed subsets of X, and the closure cl(∩i ∈ I Di)=cl(∪f ∩i ∈ I ↓f(i)) of their union is then equal to supf infi ∈ I ↓f(i), where “sup” and “inf” are understood in HX. Since HX is completely distributive, we also have infi ∈ I supx ∈ Di ↓x=supf infi ∈ I ↓f(i). Therefore cl(∩i ∈ I Di)=infi ∈ I supx ∈ Di ↓x. For each i in I, supx ∈ Di ↓x is the smallest closed set that contains every element of Di, and is therefore equal to cl (Di). The outer “inf” is just an intersection, so cl(∩i ∈ I Di)=∩i ∈ I cl(Di).
By taking complements, we obtain that for every family (Ai)i ∈ I of upwards closed subsets of X, the interior int(∪i ∈ I Ai) is equal to ∪i ∈ I int (Ai). By Exercise 5.1.38 of the book, X is a c-space. (Explicitly, let x be a point of X and U be an open neighborhood of x. Then U is the union of all the upwards closed sets ↑y where y ranges over U. In particular, U=int(∪y ∈ U ↑y). Since the sets ↑y form a family of upwards closed sets, we have just seen that U=∪y ∈ U int(↑y). Therefore x is in int(↑y) for some y in U.)
— Jean Goubault-Larrecq (January 20th, 2022)
