Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: order structure on open covers of a topological space?

Naso (Sep 04 2022 at 07:20):

Does the set of open covers on a topological space $X$ with the refinement ordering form a lattice? What about if $X$ is finite? The linked page seems to suggest it is always at least a meet-semilattice (?). There is also a 'star' operation mentioned there whose role I'm not sure about. http://www.phy.olemiss.edu/~luca/Topics/top/cover.html

David Egolf (Sep 04 2022 at 16:00):

Page 161 of "Topology" by Dugundji (which I found uploaded as a pdf by googling "Dugundji topology") seems to be relevant, although I would have to look at it more carefully. (Numbered page 161 is page 178 in the pdf).

David Egolf (Sep 04 2022 at 16:03):

In particular, that page claims that refinement gives a preorder, but not a partial order. Apparently you can have two distinct covers, each being a refinement of the other.

Morgan Rogers (he/him) (Sep 04 2022 at 20:51):

Indeed: consider the open cover of the real line by intervals whose endpoints are rationals with an odd denominator and an even numerator vs the cover of the same form with odd numerators.

Morgan Rogers (he/him) (Sep 04 2022 at 20:53):

If we close the covers downward (so that whenever U is in the cover and V is contained in U then V is in the cover) to get "covering sieves" then we get a lattice structure on these, by union and intersection.

Naso (Sep 05 2022 at 00:43):

Good to know!! Thanks a lot for the ref and example. So what about if $X$ is finite? Is it a partial order then? A lattice? If not, maybe I should just consider covering sieves.

David Egolf (Sep 05 2022 at 01:36):

I don't know if this helps, but maybe you can turn the preorder into a partial order? Put two open covers $a$ and $b$ in the same equivalence class if $a$ is a refinement of $b$ and $b$ is a refinement of $a$. Then work with equivalence classes of open covers - maybe these form a partial order?

I saw a related idea in "Seven Sketches in Compositionality", page 13. A preorder can be thought of as a category, where you have a single morphism from $a$ to $b$ exactly when $a \leq b$. Then if $a \leq b$ and $b \leq a$ we have $a \cong b$. So the idea is to switch from a preorder category to an equivalent(?) partial order category where there are no two distinct objects that are isomorphic.

I haven't checked if this works, and I also don't know if it would be helpful to what you want to do.

Naso (Sep 05 2022 at 05:47):

thanks for the suggestion. I think though, for my application it may be simpler to restrict the class of covers instead of quotienting. Also, I really only care about GOOD covers, i.e. where all nonempty intersections are contractible.

Morgan Rogers (he/him) (Sep 05 2022 at 07:54):

My example is still "GOOD" in that sense, so choosing those doesn't guarantee that they'll form a poset

Naso (Sep 05 2022 at 09:36):

Right, I see. Any idea on whether finiteness of $X$ changes anything?

David Egolf (Sep 05 2022 at 17:23):

Consider a two-element set $\{a,b\}$ equipped with the discrete topology $\{∅, \{a\},\{b\}, \{a,b\}\}$. Here are two distinct open covers: $T=\{\{a\}, \{a,b\}\}$ and $S=\{\{b\},\{a,b\}\}$. Each set of $T$ is contained in a set of $S$, and each set of $S$ is contained in a set of $T$. I think that means that $T$ and $S$ are refinements of one another, while being distinct. So, it seems like the open covers of a finite topological space don't always form a partial order under refinement.

That being said, it does feel a little bit like cheating to include $\{a,b\}$ as a set in open covers of $\{a,b\}$. Maybe the answer would change if we didn't allow this sort of thing?

David Egolf (Sep 05 2022 at 17:35):

Actually, if we have the set $\{a,b,c\}$ with the discrete topology, we have open covers with subsets $\{\{a\}, \{b,c\}\}$ and $\{\{a\}, \{b\}, \{c\}, \{b,c\}\}$. Each subset in the first open cover is contained in a subset in the second open cover, and each subset in the second open cover is contained in a subset in the first open cover. So, two open covers can be refinements of one another for a finite topological space even when neither cover contains the whole set (in this case $\{a,b,c\}$) as a subset.

David Egolf (Sep 05 2022 at 17:37):

I wonder if this relates to what Morgan Rogers was saying about closing covers downwards. Notice that $T$ and $S$ become equal when closed downwards, as do the two open covers provided above of $\{a,b,c\}$.

David Egolf (Sep 05 2022 at 17:43):

A question that occurs to me: if one has an open cover $M$ and one closes it downwards "(so that whenever U is in the cover and V is contained in U then V is in the cover)", is the resulting open cover $D(M)$ a refinement of $M$? And is $M$ a refinement of $D(M)$? That is, is $D(M)$ isomorphic to $M$ in the preorder of open covers for some topological space under refinement?
Let's see. Each subset of $M$ is contained in $D(M)$, because $D(M)$ contains every subset of $M$ (plus possible some additional ones). Each subset of $D(M)$ was either in $M$, or it was a subset of one in $M$ (by definition of $D(M)$). So each subset of $D(M)$ is contained in a subset of $M$. It seems that $D(M)$ and $M$ are refinements of one another, if I did this right.

David Egolf (Sep 05 2022 at 17:48):

I think that means that closing open covers downwards and working with those ("covering sieves"?) corresponds to picking a skeleton of the preorder of open covers of a topological space under refinement: we pick one open cover from each isomorphism class to work with.

Naso (Sep 06 2022 at 01:10):

Great analysis---thank you, David!!

Naso (Sep 06 2022 at 01:45):

A more efficient representation may be to consider only maximal covers, i.e. ones that are antichains. I'm not sure if the intersection/union of two maximal covers gives another maximal cover? If not, it may still have (semi-)lattice structure by taking the class corresponding to the intersection/union. But we need to check that satisfies the universal properties of meets/joins.
Relevant link: https://en.wikipedia.org/wiki/Antichain#Join_and_meet_operations

Naso (Sep 06 2022 at 02:29):

If I'm not mistaken, a maximal cover is the same thing as a maximal antichain in the poset $\mathcal{O}(X)$, and according to here, maximal antichains of a partially ordered set form a distributive lattice under the natural ordering (this is wrong!) . I'm happy about this as I'm considering presheaves on these covers, and a presheaf on a cover that isn't maximal gives no additional information anyway.

Morgan Rogers (he/him) (Sep 06 2022 at 16:34):

There was some lack of clarity about what "refinement" meant that had me second-guessing myself about replying again in this thread, but I hope you got the answers you were after @naso!

David Egolf (Sep 06 2022 at 16:44):

I had a thought, and was wondering if it relates to "maximal covers" mentioned by naso above, which I understand to be open covers where no open set in the cover is a proper subset of another. Define an operation $U$ that takes an open cover and removes every set from that cover that is a proper subset of a set in the open cover. If $M$ is an open cover, then so is $U(M)$. Also, every set in $M$ is contained in a set in $U(M)$, and every set in $U(M)$ is contained in a set in $M$. If that's right, then $M$ and $U(M)$ are isomorphic in the preorder on open covers induced by refinement. I think the covers produced by this $U$ operator are the "maximal covers" mentioned above(?).

If that is right, then for every open cover $M$ we have two nice isomorphic covers $D(M)$ and $U(M)$. To my understanding, Morgan Rogers stated above ("If we close the covers downward (so that whenever U is in the cover and V is contained in U then V is in the cover) to get "covering sieves" then we get a lattice structure on these, by union and intersection") that the open covers of the form $D(M)$ as $M$ varies do form a lattice structure.
For each $D(M)$ we have an isomorphic open cover $U(D(M))$. I would hope that this means that the covers of the form $U(M)$ (as $M$ varies) would also form a lattice structure. But I would need to think about that.

Intuitively, though, the skeleton formed by taking the open covers produced by $D$ should be isomorphic as a category to the skeleton formed by taking the open covers produced by $U$. So I would hope that the covers produced by the $U$ operator also form a lattice structure.

Naso (Sep 07 2022 at 03:24):

Morgan Rogers (he/him) said:

There was some lack of clarity about what "refinement" meant that had me second-guessing myself about replying again in this thread, but I hope you got the answers you were after naso!

Thanks, Morgan! but now I'm worried that I've got things mixed up. The refinement definition I was going with is the one on wikipedia which I think coincides with the one in the in the paper above by Ralph Freese (on page 1) on maximal antichains, where we consider the poset $P = \mathcal{O}(X)$ so that maximal covers are maximal antichains in $P$. ...

Looking at that paper more closely, it's not saying what I thought it was! It's talking about maximal SIZED antichains, e.g. antichains on a finite poset whose cardinality is maximal among all antichains, whereas I want to talk about the set of maximal covers, meaning covers that are also antichains...

On the other hand, and I think this is what David may be getting at, there seems to be a 1-1 correspondence between covering sieves and maximal covers, so since you said covering sieves form a (distributive?) lattice, the same should be true for the maximal covers (although Freese's paper doesn't show that)? But did you have a different refinement relation in mind?
And do you know of a reference for the covering sieves forming a lattice? Sorry for the confusion .

(I also asked this on stackexchange (and also my other question)).

David Egolf (Sep 07 2022 at 15:20):

Morgan Rogers stated about that "covering sieves" form a lattice. I would like to check this, and understand it for myself.
A covering sieve is an open cover $M$, so that if $U$ is in the cover and $V$ is a subset of $U$, then $V$ is in the cover.
Let's say we're trying to make a sieve that covers $\{a,b,c\}$ equipped with the discrete topology. Then $C_1=\{\{a,b\},\{c\}\}$ is not a sieve, but $C_2=\{\{\},\{a,b\},\{a\},\{b\},\{c\}\}$ is.

Let us put an ordering on the open subsets of $\{a,b,c\}$ by inclusion. This forms a partial order (even more - a lattice). Then, to turn $C_1$ into $C_2$, we put in every open subset of $\{a,b,c\}$ that is less than or equal to one in $C_1$. This might be where the phrase "closing downwards" Morgan Rogers used came from, for describing how covering sieves are formed.

David Egolf (Sep 07 2022 at 15:20):

Let us now consider the set $S$ of covering sieves of a topological space $X$, ordered under refinement. (If $M$ and $N$ are two open covers of the same topological space then we say $M$ is a refinement of $N$ exactly when every set in $M$ is contained in a set in $N$. Let us write $M \leq N$ to say that $M$ is a refinement of $N$.) We have $M \leq M$ for any covering sieve in $S$. If $M \leq N$ and $N \leq P$ for covering sieves $M,N,P$, do we have $M \leq P$? Every subset of $M$ is contained in a subset of $N$, and every subset of $N$ is contained in a subset of $P$. So, indeed every subset of $M$ is contained in a subset of $P$, so $M \leq P$. So, $S$ forms a preorder with respect to refinement.

David Egolf (Sep 07 2022 at 15:22):

Is $S$ a partial order, so that if $M \leq N$ and $N \leq M$ for covering sieves $M,N$ of $X$, then $M=N$? To show this, consider an open set $A$ in $M$. Because $M \leq N$, $A$ must be contained in a set in $N$. But because $N$ is a covering sieve, it contains every open subset of every set in it. So, it must actually contain $A$. Consequently, any set in $M$ must be in $N$ and similarly any set in $N$ must be in $M$. So if $M \leq N$ and $N \leq M$ then $N=M$ for two covering sieves over a topological space $X$. So, $S$ is indeed a partial order.

David Egolf (Sep 07 2022 at 15:42):

Next, we want to see if $S$ has meets and joins. The meet $M \wedge N$ of two covering sieves $M,N$ would be the largest covering sieve so $M \wedge N \leq M$ and $M \wedge N \leq N$. So, we want a covering sieve that is a refinement of both $M$ and $N$ to start with. We can get one by taking the intersection of $M$ and $N$. Now, we want to show that if $A \leq M$ and $A \leq N$ for some covering sieve $A$, then $A \leq M \cap N$. Let $B$ be a set in $A$. Then $B$ must be contained in a set in $M$, and because $M$ is a covering sieve it must actually be a set in $M$. Similarly, $B$ must be a set in $N$. Therefore, $B$ is a set in $M \cap N$. So, $A \leq M \cap N$.

David Egolf (Sep 07 2022 at 15:46):

The join $M \vee N$ of two covering sieves $M, N$ would be the smallest covering sieve so that $N \leq M \vee N$ and $M \leq M \vee N$. So we want a covering sieve that is refined by both $M$ and $N$ to start with. We can get one by taking the union of $M$ and $N$. Now, we want to show that if $M \leq A$ and $N \leq A$ for some covering sieve $A$, then $M \cup N \leq A$. Let $B$ be a set in $M \cup N$. This set is in both $M$ and $N$ and so it is in $A$, by a similar argument to that we used above for meets. So, indeed $M \cup N \leq A$.

David Egolf (Sep 07 2022 at 15:49):

So, we have at least pair-wise meets and joins. I hope this implies we have finite meets and joins, which would make $S$ into a lattice. Intuitively, it seems that the arguments used above for joins and meets for pairs of covering sieves would work for finite collections of them.

David Egolf (Sep 07 2022 at 16:17):

Now, for a fixed topological space $X$, let us consider the covering sieves $S$ ordered under refinement as a category $C_S$. The objects are covering sieves, and there is a unique morphism from $M$ to $N$ exactly when $M \leq N$. I am assuming that this category forms a lattice, based on the arguments above. Let us similarly also consider the open covers of $X$ ordered under refinement (which is a preorder) as a category $C_O$. The objects are open covers and there is a unique morphism from $M$ to $N$ exactly when $M \leq N$.

We have a map $D$ that sends an open cover $M$ to the corresponding covering sieve $D(M)$, obtained by adding all open subsets of the open sets in $M$. I want to turn this map into a functor. To do this, let us first see if $M \leq N$ implies $D(M) \leq D(N)$. Let $A$ be a set in $D(M)$. Then $A$ is a subset of a set in $M$. Let $A'$ be a set in $M$ that contains it. Then $A'$ is contained in a set in $N$; call this $A''$. Then $A''$ is in $D(N)$. Because $D(N)$ is a covering sieve, all of the subsets of $A''$ are in $D(N)$. In particular, $A'$ is in $D(N)$, and therefore $A$ is contained in $D(N)$. So, indeed $D(M) \leq D(N)$.

Now let us extend $D: C_O \to C_S$ to a map on morphisms as well as objects. If $f$ is the unique morphism from $M$ to $N$, then define $D(f)$ to be the unique morphism from $D(M)$ to $D(N)$. This exists because $M \leq N \implies D(M) \leq D(N)$. I believe that $D$ is a functor. Without checking carefully, it intuitively sends identity morphisms to identity morphisms, it respects sources and targets of morphisms, and it respects composition by the uniqueness of the morphisms between the objects.

David Egolf (Sep 07 2022 at 16:44):

Now, we want to consider a different kind of open cover: where each open set in a cover of this kind contains no proper subsets that are also in that cover. I will call these "maximal" open covers (I hope this is what naso means by a maximal open cover). Let $U$ send an open cover $M$ to its corresponding maximal open cover $U(M)$, by removing sets that are proper subsets of other sets in the cover. (We will see shortly that there is a problem with this!)

Let $C_M$ be the category of maximal open covers ordered by refinement. This is a preorder, and we are interested to see if it has additional structure. Let us first see if $M \leq N$ implies $U(M) \leq U(N)$ for open covers $M$, $N$. Let $A$ be a set in $U(M)$. Then $A$ is a set in $M$, and is therefore contained in a set in $N$. Let $A'$ be the set in $N$ that contains it. Then I would like to conclude that $A'$ is contained in some set $A''$ in $N$ that is not contained in another set also in $N$. However, I could imagine an open cover where this isn't the case. This seems like a problem.

David Egolf (Sep 07 2022 at 16:48):

To avoid this problem, let $C_M'$ be the category with objects given by maximal open covers and morphisms corresponding to the ordering by refinement, with an additional restriction on the open covers allowed as objects. We only allow maximal covers $M$ where every set $A$ in the cover is a subset of some set $A'$ in the cover that is not the proper subset of any other set in the cover. This is still a preoder.
Let us also create a modified form of $C_O$, and call it $C_O'$. The objects of $C_O'$ are open covers of our topological space $X$, with the additional requirement that each open cover $M$ in $C_O'$ satisfies the following: for every set $A$ in $M$, there exists a set $A'$ in $M$ containing $A$ so that $A'$ is not the proper subset of a set in $M$.
Now, define a map $U': C_O' \to C_M'$ that sends open covers to their corresponding maximal cover by removing sets that are the proper subset of sets in the open cover. Notice that the $U: C_O \to C_M$ we define above actually didn't make sense. If $M$ was an open cover consisting of infinitely nested open sets with no largest open set, then $U(M)$ would have been the empty set, which would not have been a cover of (a non-empty) $X$. Now, however, we should avoid this problem, as we are only removing sets that are proper subsets of those we keep.

David Egolf (Sep 07 2022 at 16:59):

Now we can see if $U'$ preserves the ordering of refinement. If $M \leq N$ in $C_O'$, is $U'(M) \leq U'(N)$? Let $A$ be a set in $U'(M)$. Then $A$ is a set in $M$, and is therefore contained in a set $A'$ in $N$. This set $A'$ is contained in some set $A''$ also in $N$ so that $A''$ is not the proper subset of some set in $N$ (because we are working in $C_O'$). Then $A''$ is contained in $U'(N)$ and also contains $A$, which means that an arbitrary set $A$ in $U'(M)$ is contained in some set in $U'(N)$. So $U'(M) \leq U'(N)$. I think this makes $U'$ into a functor $U': C_O' \to C_M'$.

David Egolf (Sep 07 2022 at 17:16):

Next, I would like to relate $C'_M$ and $C_S$. Define a map $F: Obj(C'_M) \to Obj(C_S)$ that sends a maximal open cover $M$ to its corresponding sieve $F(S)$. If $M \leq N$ in $C'_M$, do we have $F(M) \leq F(N)$? Yes - we saw earlier that this holds for $M$ and $N$ any open covers, not just maximal ones. So, I think we have a functor $F: C'_M \to C_S$.

We also have a map $G: Obj(F(C'_M)) \to Obj(C_M')$ that sends a covering sieve to its corresponding maximal cover. To check this makes sense, we need to check that each open cover $M$ in $F(C'_M)$ satisfies the following property: any open set $A$ in $M$ is contained in a open set $A'$ in $M$ where $A'$ is not the proper subset of any set in $M$. This will be satisfied, since this property is satisfied for any open cover $M$ in $C'_M$, and forming the covering sieve only adds subsets of sets already present. Then we can set $G$ to be the restriction of $U':C_O' \to C'_M$ to $F(C'_M)$. Because $U'$ is a functor, I think that means that if $M \leq N$ in $F(C'_M)$, then $G(M) \leq G(N)$, and so we can extend $G$ to be a functor $G: F(C'_M) \to C_M'$.

David Egolf (Sep 07 2022 at 17:29):

Now, I want to know if $(G \circ F)(M) =M$ for a maximal cover $M$. I think this should be the case: we add in extra sets that are subsets of sets in the cover with $F$, and then we remove them with $G$. Do we also have $(F \circ G)(M)=M$ for a covering sieve in $F(C'_M)$? The process of making the maximal open cover works nicely because we are working in $F(C'_M)$, intuitively I think we do have $(F \circ G)(M) = M$. I think this means that $F \circ G$ and $G \circ F$ are both identity functors. If this is right, I think that means that $C'_M \cong F(C'_M)$. So, the category of "nice" maximal open covers of a topological space $X$ is isomorphic to the category of sieves corresponding to these.

David Egolf (Sep 07 2022 at 17:36):

To continue this line of thought, I would then want to see if $F(C'_M)$ somehow inherits a lattice structure from $C_S$. If it did, then $C'_M$ should also be a lattice, and so we would have a lattice structure on the "nice" maximal open covers. However, I need to take a rest here.

To summarize: (1) maximal open covers and sieves seem related, but it's not as close a relationship as I thought; (2) there are some open covers for which the procedure "remove sets that are subsets of other sets in the cover" does not produce maximal open covers

(Other questions: might there be some kind of adjoint relating $C'_M$ and $C_S$? And is $C'_M$ really different from $C_M$?)

General disclaimer: There could be lots of mistakes in the above. I haven't checked it carefully, and may or may not do so in the future. Hopefully there is something here of interest to naso, though!

David Egolf (Sep 08 2022 at 00:13):

Under some assumptions about the "niceness" of the covers we are working with, I now have a guess for what the join of two maximal open covers should be. Let $M$ and $N$ be two maximal open covers. Then maybe $M \vee N = U(D(M) \cup D(N))$ where $D$ maps a maximal open cover to its corresponding covering sieve, and $U$ maps a (nice enough) covering sieve to its corresponding maximal open cover.
I don't know if this actually works. The idea is to try and grab the join from the world of sieves.

Naso (Sep 08 2022 at 07:41):

Wow, thank you, @David Egolf ! This will take me some to digest; I'll need to spend some time carefully going through this...

Just a quick comment for now: on the wikipedia entry for 'antichain', it says 'the family of all antichains in a finite partially ordered set can be given join and meet operations, making them into a distributive lattice.' (the meet and join operations are defined at that link.)

The set of maximal covers $\mathsf{Cov_{max}}(X)$ are the antichains in $\mathcal{O}(X)$ that satisfy $\bigcup M = X$.

So assuming $\mathcal{O}(X)$ is finite, to find out if $\mathsf{Cov_{max}}(X)$ is a (distrib.) lattice we should only need to verify that whenever $M,N$ are maximal covers,

$\bigcup(M \land N) = X$

and also

$\bigcup(M \lor N) = X$

It seems like both of these may be true, but I have not written down a proof yet. What do you think?

Also, I'm not sure why the finite hypothesis is necessary.

Tobias Schmude (Sep 08 2022 at 07:53):

David Egolf said:

Define an operation $U$ that takes an open cover and removes every set from that cover that is a proper subset of a set in the open cover. If $M$ is an open cover, then so is $U(M)$.

I don't think this behaves well though: take for example the open cover $M$ of the real line that consists of the bounded open sets. Each of them is contained in a bigger one, so $U(M)$ is empty and thus not an open cover.

Tobias Schmude (Sep 08 2022 at 07:53):

So overall each open cover is equivalent to a unique sieve, but not every open cover is equivalent to a maximal one. If it is, this is still unique though.

John Baez (Sep 08 2022 at 07:54):

naso said:

Also, I'm not sure why the finite hypothesis is necessary.

It might not be.

Jens Hemelaer (Sep 08 2022 at 08:40):

In general, the sieves form a distributive lattice. Antichains can be identified with sieves by taking the downwards closure, but not every sieve is the downwards closure of an antichain as @Tobias Schmude showed. A sieve $S$ comes from an antichain if and only if for every element $x \in S$ there is a maximal element $y \in S$ such that $x \leq y$.

If $S$ and $S'$ come from an antichain, then so does $S \cup S'$. The problem is with intersections. Take for example the set $\mathbb{N} \cup \{\infty_a,\infty_b\}$ with partial order extending the one on $\mathbb{N}$, such that $\infty_a$ and $\infty_b$ are incomparable but both larger than any natural number. Every singleton is an antichain, and the only other antichain is the set $\{\infty_a,\infty_b\}$. The antichains do not form a distributive lattice here because the meet of antichains $\{\infty_a\} \wedge \{\infty_b\}$ does not exist.

David Egolf (Sep 08 2022 at 15:03):

Tobias Schmude said:

David Egolf said:

Define an operation $U$ that takes an open cover and removes every set from that cover that is a proper subset of a set in the open cover. If $M$ is an open cover, then so is $U(M)$.

I don't think this behaves well though: take for example the open cover $M$ of the real line that consists of the bounded open sets. Each of them is contained in a bigger one, so $U(M)$ is empty and thus not an open cover.

Yes, I noticed this above! Admittedly, it's rather buried in my lengthy response.
Above, I end up trying to avoid this problem by restricting attention to open covers satisfying the following property: each open set in the open cover is contained in an open set that is "maximal" in that cover; by this I mean, each open set in the cover is contained in an open set also in that cover which is not contained in another set in that open cover.

David Egolf (Sep 08 2022 at 15:04):

The maximal open covers we are interested in I believe are those where every open set in the cover is maximal (is not contained as a proper subset of another open set in the cover). So, all the maximal open covers satisfy this property.

David Egolf (Sep 08 2022 at 15:07):

To avoid keeping on saying "this property", I've been in my own notes calling open covers that satisfy this property "restrained open covers". The big questions to me at this point are: is the union of two restrained open covers a restrained open cover? is the intersection of two restrained open covers a restrained open cover?

David Egolf (Sep 08 2022 at 15:15):

naso has mentioned above some interest in finite topological spaces.
I am hoping that all open covers of a finite topological space are restrained.
Let $M$ be an open cover of a finite topological space $(X, \tau)$. Then, let $A$ be an open set in $M$. It could be that $A$ is contained as a proper subset of some other open set $A'$ in $M$. And $A'$ could be contained as a proper subset of some other open set $A''$ in $M$. We can't keep finding larger sets in this way forever though, as each set in this sequence of nested sets has more elements than the previous one, and the maximum number of elements in a set is bounded above by the number of elements in $X$, because $X$ is finite. So, eventually there is a largest open set in $M$ containing $A$.
So, in a finite topological space, any open cover is restrained (and so can be associated with a maximal cover).

Tobias Schmude (Sep 08 2022 at 15:18):

I think this should hold for compact spaces already?

Tobias Schmude (Sep 08 2022 at 15:21):

Yes, I noticed this above! Admittedly, it's rather buried in my lengthy response.

Ah, sorry! I skimmed your response, I just didn't catch it.

David Egolf (Sep 08 2022 at 15:23):

Tobias Schmude said:

I think this should hold for compact spaces already?

That sounds cool to me! I'm not immediately seeing how to prove this (that any open cover of a compact space is a restrained open cover), but I also don't know many tricks or patterns from topology yet.

David Egolf (Sep 08 2022 at 15:31):

Before I forget, I also wanted to try out the join operation that I think might work for nice enough maximal open covers, and I am very hopeful it will work for finite maximal open covers.
Let our topological space $X$ for the moment be $\{a,b,c\}$ equipped with the discrete topology.
Let $M$ be the maximal open cover $\{\{a\},\{b,c\}\}$ and let $N$ be the maximal open cover $\{\{a,b\},\{c\}\}$. I want to find a join for these two: the smallest maximal open cover larger than these two.
To find such a join, I think we can try $U(D(M) \cup D(N))$. That is, we compute the covering sieves corresponding to $N$ and $M$, take the union, and then compute the corresponding maximal open cover.
We have $D(M) = \{\{\}, \{a\}, \{b\}, \{c\}, \{b,c\}\}$ and $D(N)=\{\{\},\{a\},\{b\},\{a,b\},\{c\}\}$.
The union is: $D(M) \cup D(N) = \{\{\}, \{a\}, \{b\}, \{c\}, \{b,c\}, \{a,b\}\}$.
The corresponding maximal open cover from this union is $U(D(M) \cup D(N)) = \{\{a,b\},\{b,c\}\}$, obtained by taking the maximal elements.

David Egolf (Sep 08 2022 at 15:36):

Notice that, at least in this test case, $M \leq U(D(M) \cup D(N))$ and $N \leq U(D(M) \cup D(N))$. And I think this set will indeed refine any other set that is refined by both $N$ and $M$, although I haven't checked carefully.

David Egolf (Sep 08 2022 at 15:39):

This procedure may get into trouble if $D(M) \cup D(N)$ isn't restrained any more, but otherwise I am hopeful it will work based on the correspondence (preoder isomorphism?) between the maximal covers and the restrained covering sieves of a topological space. Whether it works in general or not, I thought it was incredibly cool how a way of combining sieves can give at least an idea for how to combine maximal covers! This whole strategy of grabbing concepts from related categories seems quite powerful to me.

David Egolf (Sep 08 2022 at 15:51):

Angling towards (hopefully) showing that open covers of compact spaces are restrained, let us consider finite open covers. That is, open covers that only have a finite number of open sets. Are finite open covers restrained? Yes, because we can not infinitely keep fitting open sets from such a cover inside other sets inside that cover - eventually we run out of sets in the open cover. So, every open set in a finite open cover is contained in a maximal open set in that cover, which is not contained in any other open set in the cover.

David Egolf (Sep 08 2022 at 15:58):

I think I maybe though of a counterexample - a compact topological space having an open cover that is not restrained. Consider the closed interval $[0,1]$ under the subspace topology. This is a compact topological space (https://ncatlab.org/nlab/show/closed+intervals+are+compact+topological+spaces). Let us construct an open cover from the sets $[0,0.6)$ and $(0.4, 1]$, plus some other less nice sets. Let us also add an infinite nested family of open intervals centered on 0.5 and expanding outwards so that the open intervals approach (but never quite reach) the open interval (0,1). I think most of the sets in this infinite nested family will be not be contained inside a maximal set in the open cover. So, I think this open cover is not restrained, even though the topological space is compact.

Interestingly though, the maximal elements of this open cover do form a maximal open cover. So in some cases we can get a maximal open cover from a non-restrained open cover by taking maximal elements! It seems like an interesting question: Which open covers produce a maximal open cover when we take their maximal elements?

Naso (Sep 09 2022 at 07:58):

David Egolf said:

Yes, I noticed this above! Admittedly, it's rather buried in my lengthy response.
Above, I end up trying to avoid this problem by restricting attention to open covers satisfying the following property: each open set in the open cover is contained in an open set that is "maximal" in that cover; by this I mean, each open set in the cover is contained in an open set also in that cover which is not contained in another set in that open cover.

David could you please rephrase this or explain again the motivation for this definition? I'm having trouble connecting it to what I called 'maximal covers', which are simply covers that are antichains in $\mathcal{O}(X)$. In other words, a maximal cover $M$ is a cover ($\bigcup M = X$) such that distinct elements of $M$ are incomparable, i.e. if $U, V \in M$ then $U \leq V$ or $V \leq U$ implies $U = V$.

(Note that a maximal cover is not a maximal antichain like I mistakenly suggested before, since we can have two maximal covers $M$ and $N$ where $M$ is a proper subset of $N$, e.g. take $M = \{\{a,b\}, \{b,c\}\}$ and $N = \{\{a,b\}, \{b,c\}, \{a,c\}\}$ on the discrete space $X=\{a,b,c\}$.)

Naso (Sep 09 2022 at 08:00):

Jens Hemelaer said:

In general, the sieves form a distributive lattice. Antichains can be identified with sieves by taking the downwards closure, but not every sieve is the downwards closure of an antichain as Tobias Schmude showed. A sieve $S$ comes from an antichain if and only if for every element $x \in S$ there is a maximal element $y \in S$ such that $x \leq y$.

If $S$ and $S'$ come from an antichain, then so does $S \cup S'$. The problem is with intersections. Take for example the set $\mathbb{N} \cup \{\infty_a,\infty_b\}$ with partial order extending the one on $\mathbb{N}$, such that $\infty_a$ and $\infty_b$ are incomparable but both larger than any natural number. Every singleton is an antichain, and the only other antichain is the set $\{\infty_a,\infty_b\}$. The antichains do not form a distributive lattice here because the meet of antichains $\{\infty_a\} \wedge \{\infty_b\}$ does not exist.

Can this example be adapted to show that two maximal covers of a space may not have a meet?

David Egolf (Sep 09 2022 at 16:08):

naso said:

David Egolf said:

Yes, I noticed this above! Admittedly, it's rather buried in my lengthy response.
Above, I end up trying to avoid this problem by restricting attention to open covers satisfying the following property: each open set in the open cover is contained in an open set that is "maximal" in that cover; by this I mean, each open set in the cover is contained in an open set also in that cover which is not contained in another set in that open cover.

David could you please rephrase this or explain again the motivation for this definition? I'm having trouble connecting it to what I called 'maximal covers', which are simply covers that are antichains in $\mathcal{O}(X)$. In other words, a maximal cover $M$ is a cover ($\bigcup M = X$) such that distinct elements of $M$ are incomparable, i.e. if $U, V \in M$ then $U \leq V$ or $V \leq U$ implies $U = V$.

(Note that a maximal cover is not a maximal antichain like I mistakenly suggested before, since we can have two maximal covers $M$ and $N$ where $M$ is a proper subset of $N$, e.g. take $M = \{\{a,b\}, \{b,c\}\}$ and $N = \{\{a,b\}, \{b,c\}, \{a,c\}\}$ on the discrete space $X=\{a,b,c\}$.)

Sure, naso!
First, let me explain what I mean by a "maximal" element of an open cover. Each open cover $M$ can be considered as a preorder under inclusion of subsets. For a set $U$ in $M$, define $\uparrow U$ by $\uparrow U = \{V \in M | U \leq V\}$. This is the set of all elements at least as large as $U$ in $M$. An open set $U$ in an open cover $M$ is maximal exactly $\uparrow U = \{U\}$. In other words, a maximal element in an open cover $M$ is one that is not contained as a proper subset of a set in $M$.

Now we can connect the concept of "maximal element" to that of "maximal cover". As you define above, maximal covers $M$ are those where distinct open sets are incomporable. In particular, this means that if $U \in M$, then the only open set in $M$ that contains $U$ is $U$. Therefore, there is no open set in $M$ that contains $U$ as a proper subset. Therefore, each open set $U$ in a maximal cover is itself a "maximal element" with respect to the preorder of open sets of that maximal cover.

To summarize, I think the maximal open covers are exactly those where each element in the open cover is maximal.

David Egolf (Sep 09 2022 at 16:16):

Now, there are some open covers that can't be nicely made into maximal open covers, by the process of removing sets from the cover if they are proper subsets of other sets in the cover. Note that this process is equivalent of taking an open cover, and keeping only its maximal elements. Let $U$ be a map that sends each open cover to the set containing only the maximal elements of it. Then, for some open covers $M$, $U(M)$ is not a maximal open cover.

For example, consider an open cover $M$ of the real line provided by an infinite family of nested open intervals centred about the origin. Each open set in this open cover is of the form $(-a,a)$ for $a \in \mathbb{R}$. This is an open cover, as each set in it is open, and each point on the real line is contained in some $(-a,a)$. However, $U(M)$ is the empty set, as this cover has no maximal elements! I want to be able to associate to each open cover a maximal open cover, but this example show that at least this doesn't work with the operation $U$.

David Egolf (Sep 09 2022 at 16:24):

Consider now open covers where each set in the open cover is a subset of a maximal element in that open cover. (This is the property/definition you were asking about). I call these open covers "restrained" open covers. If we take the maximal elements of a restrained open cover, I believe we do get a maximal open cover.

Let $M$ be a restrained open cover, and let us consider $U(M)$. Let $a$ be a point in the topological space we are hoping to cover with $U(M)$. Then $a$ is contained in some open set $A$ in $M$, because $M$ is an open cover. Because $M$ is restrained, $A$ is a subset of a maximal element $A'$ of $M$. By definition of $U(M)$, we must have $A'$ in $U(M)$. So, for any point $a$, $U(M)$ contains an open set that contains $a$. So, $U(M)$ is indeed a (maximal) open cover if $M$ is a restrained open cover.

David Egolf (Sep 09 2022 at 16:26):

In summary, the concept you asked about (what I called "restrained open covers") is relevant, because it provides a type of open cover that is guaranteed to make a maximal open cover when we take its maximal elements.

David Egolf (Sep 09 2022 at 16:30):

I hope this helps!

David Egolf (Sep 09 2022 at 21:47):

I also asked on stack exchange: "Is the union of two restrained open covers a restrained open cover?"
The answer appears to be "yes", see here.
Using that result, I think the operation $U(D(M) \cup D(N))$ is well defined for maximal open covers $M$ and $N$. (This is the operation that I hope gives a join of maximal open covers, where $D(M)$ is the covering sieve induced by an open cover $M$ and $U(A)$ is the maximal cover induced by a restrained open cover $A$).

Naso (Sep 10 2022 at 03:47):

thanks David, that clears it up! I am more interested in the case where we start with the set of maximal covers $\mathsf{Cov_{max}}(X)$ as given, and then whether this forms a distributive lattice. I feel fairly satisfied that it does when $\mathcal{O}(X)$ is finite, and even if $\mathcal{O}(X)$ is infinite I don't see a problem. But I'd feel better about it if someone can agree with me ...

Naso (Sep 10 2022 at 05:07):

My reasoning for when $\mathcal{O}(X)$ is finite, is that the antichains on a finite poset are known to form a distributive lattice, so if $\mathsf{Cov_{mat}}(X)$ is a lattice under the ordering it must have the same meet and join and we only need to check it is closed under those operations.

If $A,B$ are maximal covers and $t \in X$ we know there is a $U \in A, V \in B$ with $t \in U$ and $t \in V$.

For the join, if $U,V$ are incomparable then $U \in A \lor B$ and so $t \in \bigcup{(A \lor B)}$, so we have $X = \bigcup{(A \lor B)}$ since $t$ was arbitrary.
If $U, V$ are comparable, then the larger one is in $A \lor B$ and again $X = \bigcup{(A \lor B)}$.

For the meet, we have $t \in U \cap V$ and $U \cap V \in L_A \cap L_B$.
If there is no $Z >U \cap V$ in $L_A \cap L_B$, then we have $U \cap V \in A \land B$ and therefore $t \in \bigcup{(A \land B)}$ and $X = \bigcup{(A \land B)}$.
Otherwise let $Z$ be the maximal element of $L_A \cap L_B$ above $U \cap V$, which must exist because $L_A \cap L_B$ is a finite set. Then $Z \in A \land B$ and so $t \in \bigcup{(A \land B)}$ and $X = \bigcup{(A \land B)}$.

However, if $\mathcal{O}(X)$ is infinite, we cannot rely on $L_A \cap L_B$ being a finite set, and we also can't start with the fact that the antichains on $\mathcal{O}(X)$ form a distributive lattice (Jens's counterexample).

Actually in my application I'm happy to assume $\mathcal{O}(X)$ is finite, but it would be interesting to know what exactly changes in the infinite case.

David Egolf (Sep 10 2022 at 17:45):

When $\mathcal{O}(X)$ is finite, then every open cover is finite, and therefore every open cover is restrained. In this case, I guess we can go back and forth between maximal covers and covering sieves in a functorial way with no problem, and so the maximal open covers I expect will inherit lattice structure from the covering sieves. I think what we've been discussing above can be cleaned up to show that, hopefully.

The case when $\mathcal{O}(X)$ is infinite seems harder. The union of restrained open covers is restrained, which I hope means that the maximal open covers inherit the join operation from the restrained covering sieves, as discussed above. With respect to the meet, one approach would be to try and figure out if the intersection of restrained covering sieves is also a restrained covering sieve. If this is the case, then I am thinking that the meet of maximal open covers in this infinite case will exist. If it isn't, I am currently thinking that the meet won't exist.

Incidentally, my guess for the meet of maximal covers $M$ and $N$ would be $U(D(M) \cap D(N))$. The question is if the intersection $D(M) \cap D(N)$ of the resulting (restrained) covering sieves is restrained, so that it can be made into a maximal open cover by $U$.

Jens Hemelaer (Sep 12 2022 at 18:56):

naso said:

Jens Hemelaer said:

In general, the sieves form a distributive lattice. Antichains can be identified with sieves by taking the downwards closure, but not every sieve is the downwards closure of an antichain as Tobias Schmude showed. A sieve $S$ comes from an antichain if and only if for every element $x \in S$ there is a maximal element $y \in S$ such that $x \leq y$.

If $S$ and $S'$ come from an antichain, then so does $S \cup S'$. The problem is with intersections. Take for example the set $\mathbb{N} \cup \{\infty_a,\infty_b\}$ with partial order extending the one on $\mathbb{N}$, such that $\infty_a$ and $\infty_b$ are incomparable but both larger than any natural number. Every singleton is an antichain, and the only other antichain is the set $\{\infty_a,\infty_b\}$. The antichains do not form a distributive lattice here because the meet of antichains $\{\infty_a\} \wedge \{\infty_b\}$ does not exist.

Can this example be adapted to show that two maximal covers of a space may not have a meet?

EDIT: Originally, I wrote a different example below, but that one did not work. I believe the one below works, but it seems likely that a much simpler example exists.

Consider the set $\mathbb{R}-\{0\}$ with the standard Euclidean topology. There is a maximal cover given by $(0,+\infty)$ and $(-\infty,0)$. Another maximal cover is given by the open sets $(-e^{-n},e^n)$ for $n$ an integer.

The intersection $\mathcal{U}$ of the two maximal covers above consists of the sets $(0,e^n)$ and $(-e^n,0)$, for $n$ going over the integers. As a covering sieve, this consists precisely of all open sets that are bounded and
contained in either $(0,+\infty)$ or $(-\infty,0)$. This is not a maximal cover.

We have to prove something stronger: that the set of all maximal covers refining $\mathcal{U}$ does not have a largest element. Let $\mathcal{V}$ be this maximal element, we want to find a contradiction.

Take any maximal cover $\mathcal{W}$ that refines $\mathcal{U}$ and contains the open sets $(-1,0)$ and $(0,1)$. Because $\mathcal{W}$ is a refinement of $\mathcal{U}$, it is by definition also a refinement of $\mathcal{V}$. So $\mathcal{V}$ contains an open set $X_1$ with $(0,1) \subseteq X_1$.

Now we notice that $\mathcal{U}$ as a covering sieve is invariant under any homeomorphism
$f _\lambda: \mathbb{R} - \{0\} \longrightarrow \mathbb{R} - \{0\},~ x \mapsto \lambda\, x$
for $\lambda > 0$ a real number.

It follows that then also $\mathcal{V}$ is invariant under the $f_\lambda$. The open set $f_\lambda(X_1) \in \mathcal{V}$ contains the interval $(0,\lambda)$. Similarly, we can for any $\lambda$ find an open set in $\mathcal{V}$ that contains the interval $(-\lambda,0)$. It then follows that, as a covering sieve, $\mathcal{V}$ agrees with $\mathcal{U}$. But this means it is not a maximal cover, a contradiction.

David Egolf (Sep 12 2022 at 21:15):

I'm trying to understand the example above given by @Jens Hemelaer .
Let $A = \{(0, +\infty), (-\infty,0)\}$. Let $B = \{(-e^{-n}, e^n) | n \in \mathbb{Z}\}$. Every set in $B$ is finite in its length, while each set in $A$ is infinite in its length. I would then think that $A \cap B = \emptyset$. There are no elements in common between these two sets as far as I can tell.

However, the example above seems to be indicating that the intersection of these two sets is nonempty. So I am confused! I guess some kind of fancier pair-wise intersection of elements is being used? (I don't know if this really matters, but it confused me!)

Jens Hemelaer (Sep 13 2022 at 06:52):

Yes, I should have mentioned that I identify each maximal cover with the covering sieve it generates. For example, the maximal cover $A$ is identified with the covering sieve consisting of all open sets $U$ with either $U \subseteq (-\infty,0)$ or $U \subseteq (0,\infty)$. The covering sieve generated by a maximal cover is a restrained open covering in your terminology above. So you can recover the original maximal cover from the covering sieve generated by it. In this way, the maximal covers form a subposet of the lattice of covering sieves.

By "intersection of maximal covers" $A \cap B$ I meant the meet/intersection as covering sieves. As you guessed, it is generated as a covering sieve by the open sets $a \cap b$ with $a \in A$ and $b \in B$.

The idea is now that the covering sieve $A \cap B$ is not restrained, meaning it does not come from a maximal cover. Moreover, if the meet of $A$ and $B$ as maximal covers would exist, then it would be the largest maximal cover refining $A \cap B$. This then gives a contradiction in the example.

David Egolf (Sep 13 2022 at 15:26):

Thanks, that helps clarify things!
To rephrase part of the argument, for my own understanding:
Let us have the two maximal open covers $A = \{(0, +\infty), (-\infty,0)\}$ and $B = \{(-e^{-n}, e^n) | n \in \mathbb{Z}\}$.
If $D$ is the operation that sends a maximal open cover to its corresponding covering sieve, then $D(A) \cap D(B)$ is not a restrained covering sieve, even though $D(A)$ and $D(B)$ are.
That means $U(D(A) \cap D(B))$ does not produce a maximal cover. That was my candidate for the join of two maximal open covers, and we can now see it will not work.

I think I can follow the rest of Jens Hemelaer's argument now as well. It's awesome to have an answer to the question "Do meets of maximal open covers exist?"!