Let $\mathrm{r}\mathscr{O}$ be the family of open domains (regular open sets) of a topological space $\langle X,\mathscr{O}\rangle$, that is: $$A\in\mathrm{r}\mathscr{O}\iff A=\mathrm{int(\mathrm{cl(A)})}.$$

Define $A$ to be well inside $B$: $$\tag{df $\Subset$} A\Subset B\iff \mathrm{cl}(A)\subseteq B, $$

Call a set $X$ of regular open sets contracting iff $\forall_{A\in X}\exists_{B\in X}\,B\Subset A$. Let $C$ be a maximal contracting chain with respect to well inside relation and consider the filter $\mathscr{F}$ generated from $C$. $\mathscr{F}$ is obviously contracting. My question is:

Does $\mathscr{F}$ have to be a maximal contracting filter?
  • $\begingroup$ It should always be 'contracting' (sorry, my mistake - I have corrected it). By a contracting filter I mean a filter (defined standardly as closed on the Boolean product and upward closed) in the Boolean algebra which is contracting as a set. $\endgroup$ Sep 8, 2015 at 13:50

2 Answers 2


The answer is no. Consider the case of a discrete space $X$, where every set is open and hence also regular open. In this case, the well-inside relation coincides with the ordinary inclusion relation. Your question asks whether the filter generated by an inclusion chain in $P(X)$ is maximal, and the answer is that this can fail.

For example, if $X$ is infinite, then there is a maximal inclusion chain containing no finite sets, and so the filter is not principal. Let $A_\alpha$ be a sequence that is coinitial (downward cofinal) in the chain, with $A_0=X$ the whole space, and let $E=\bigsqcup_{\alpha\text{ even}}A_\alpha-A_{\alpha+1}$ and $O=\bigsqcup_{\alpha\text{ odd}}A_\alpha-A_{\alpha+1}$. That is, you chop the chain into pieces, and let $E$ consist of the union of the even differences, and $O$ the odd differences. So $E\sqcup O=X$, but neither contains an element of the chain. So the filter generated by the chain is not an ultrafilter, and so it isn't a maximal filter.

Essentially the same argument shows that a nonprincipal ultrafilter in a complete Boolean algebra is never generated by a linearly ordered set, and so one can make such examples even in any non-discrete spaces. (See lemma 43 in my paper Well-founded Boolean ultrapowers as large cardinal embeddings.) This result shows that in any complete Boolean algebra (which can always be viewed as the regular open sets of some topological space), a nonprincipal ultrafilter is never generated by a linearly ordered set.

  • $\begingroup$ Good example! Since I was typing out a different one when you posted this, I'll just share it here: the Tychonoff Plank. No maximal contracting chain of neighborhoods of the corner point can be a maximal contracting filter (proof now left as an exercise!). $\endgroup$
    – Will Brian
    Sep 8, 2015 at 14:09
  • $\begingroup$ Will, go ahead and post your example as an answer! $\endgroup$ Sep 8, 2015 at 14:15
  • $\begingroup$ Could you give in few words purely algebraic description of that lemma 34? How are ultrafilters $U$ and $U_A$ related, without referring to ultrapowers of $V$? $\endgroup$ Sep 8, 2015 at 18:03
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    $\begingroup$ @მამუკაჯიბლაძე I'm sorry, I had meant to refer to lemma 43! I think that will be more readily understood. (I am also happy to explain lemma 34, but it is not directly related to this question.) $\endgroup$ Sep 8, 2015 at 22:39
  • $\begingroup$ And I too believe that having a separate answer with the example by @WillBrian would be interesting as it seemingly gives an example involving a compact space which is moreover not extremally disconnected, so that well-insideness does not coincide with inclusion... $\endgroup$ Sep 9, 2015 at 3:31

Although Joel has already given an example showing why the answer to this question is no, I'm going to offer a few more examples, and some comments about the question.

First, here's a really easy example showing that a maximal contracting chain does not need to generate a maximal contracting filter. In the unit interval $[0,1]$, let $$C = \{[0,a) \cup (1-a,1] : 0 < a < 1/2\} \cup \{[0,1]\}.$$ It's not too hard to check that this is a maximal contracting chain, but it obviously does not generate a maximal contracting filter.

The reason we can come up with such an easy counterexample is that "maximal contracting chain" is too weak a condition to hope realistically for a maximal filter. I'll venture to guess, though, that this isn't the sort of counterexample you're looking for -- it feels too much like cheating.

In order to do away with easy counterexamples like this one, I suggest we modify the question as follows:

If $C$ is a maximal contracting chain and $\mathcal F$ is a maximal contracting filter containing $C$, is there a (possibly different) maximal contracting chain $C'$ such that $C'$ generates $\mathcal F$?

Perhaps I'm taking too much license with your question here, and if so I apologize. But I like this version of the question because (1) it does away with the idea behind my easy counterexample, (2) the answer can depend on what space we're working in (i.e., this question is getting at some topological property).

An example where the answer is no: Let $T$ denote the Tychonoff Plank, $T = [0,\omega] \times [0,\omega_1]$. Let $$C = \{(n,\omega] \times [0,\omega_1] \cup \{n\} \times [\alpha,\omega_1] : n < \omega, \alpha < \omega_1\}$$ It's not too hard to check that this is a maximal contracting chain in $T$. Let $\mathcal F$ be the neighborhood filter on the point $(\omega,\omega_1)$. This is a maximal contracting filter, but I claim that it cannot be generated by any chain $C'$. The reason is that the chain $C'$ has a certain cofinality $\kappa$. If $\kappa \neq \omega$, then for some fixed $n$ there are cofinally many elements of $C'$ containing sets of the form $[n,\omega] \times [\alpha,\omega_1]$. Since this happens cofinally often and $C'$ is a chain, every element of $C'$ contains a set of this form. This means that $C'$ cannot generate $\mathcal F$. We get a similar contradiction by supposing $\kappa \neq \omega_1$.

A class of spaces where the answer is always yes: If $X$ is a compact metric space then the answer (to my modified version of your question) is always yes. More generally, suppose $X$ is a compact space that is well-based, meaning that every point has a well-ordered neighborhood basis. Every metric space is well-based, but there are other spaces too (the order topology on $\omega_1+1$, for example). For any space in this class, I claim that every maximal contracting filter is generated by a chain (which is more than enough for an affirmative answer to our question). To see this, use compactness to say that a maximal contracting filter must just be the neighborhood filter of some point. Then use the well-based property to conclude that every maximal contracting filter is generated by a chain.

Another example where the answer is yes: Although every well-based space gives you an affirmative answer, I want to point out that there are other spaces that also work. For example, consider the space obtained by gluing the final point of $\omega_1+1$ to the final point of $\omega+1$ (where $\omega_1+1$ and $\omega+1$ are each given the order topology). This space is not well-based, because the "pit point" where the two spaces were glued together cannot have a well-ordered neighborhood basis (after all, what would its cofinality be?). However, this space still gives us an affirmative answer to our question. To see this, just notice that no maximal contracting chain contains the pit point (supposing it did, use a cofinality-matching argument like the one above to show that the chain is not maximal after all). Therefore every maximal contracting chain $C$ can be thought of as a maximal contracting chain in one of the subspaces $\omega+1$ or $\omega_1+1$. Since each of these is compact and well-based, any maximal contracting filter containing $C$ is generated by a chain.

A few final comments: Notice that I'm only working with compact spaces. The question still makes sense for non-compact spaces, though. I think that, given a space $X$, the answer to our question will be the same for $X$ and for $\beta X$ (the Stone-Cech compactification of $X$). I also think it should be possible to show that every non-compact metric space gives a negative answer to our question.

The class of well-based spaces has a few well-studied generalizations (radial, pseudo-radial, etc.). It could be the case that the property "our question has an affirmative answer for $X$" is equivalent to one of these other properties. I don't really know much about these generalizations, so I can't say, but perhaps it's worth looking into.

  • $\begingroup$ Will, I like your version of the question since it is very closely related to the one I had already asked before: mathoverflow.net/questions/202280/… I do not mention well-inside relation there, but motivations for asking derived from the need to understand some of well-insideness properties in chains. Your answer, for which I am very grateful, has delivered a lot of food for my thought. $\endgroup$ Sep 10, 2015 at 22:13

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