# A problem of non-emptiness of intersections of certain chains of regular open sets

Let $$X$$ be a topological space and $$\mathrm{RO}(X)$$ its complete boolean algebra of regular opens. Define well inside relation: $$U\prec V\iff\overline{U}\subseteq V.$$ Let $$\mathcal C\subseteq\mathrm{RO}(X)$$ be a chain such that (a) any two distinct elements of $$\mathcal C$$ are comparable w.r.t. $$\prec$$ and (b) $$\mathcal C$$ is strictly descending, i.e. for every $$U$$ in the chain there is $$V$$ in the same chain with $$V\subsetneq U$$.

Of course, the intersection of $$\mathcal C$$ does not have to be non-empty (e.g. take $$\{(n,+\infty)\mid n\in\omega\}\subseteq\mathrm{RO}(\mathbf{R})$$, with $$\mathbf{R}$$ the set of reals), but it is always non-empty if the space is compact.

Say that a chain $$\mathcal C$$ covers $$\mathcal D$$ iff for every $$U\in\mathcal C$$ there is $$D\in\mathcal D$$ such that $$D\subseteq U$$. Further, call $$\mathcal C$$ c-minimal if for every $$\mathcal D$$ covered by $$\mathcal C$$, $$\mathcal C$$ is also covered by $$\mathcal D$$.

My problem is whether for every non-empty c-minimal chain $$\mathcal C$$ (in an atomless $$\mathrm{RO}(X)$$) which satisfies (a) and (b), $$\bigcap\mathcal C$$ is non-empty.

EDIT: Changed the terminology since, as pointed in the comments, the one I used originally was a bit misleading.

• Your definition of co-initial seems strange to me, since it would be that $\mathcal{C}$ is a chain converging to a much smaller intersection than $\mathcal{D}$, but this isn't usually what is meant by "co-initial". Usually, co-initial would imply that the limit is the same. Dec 23 '19 at 12:54
• Your definition of strictly descending seems to rule out a chain with only one element. Dec 23 '19 at 12:56
• Yes, according to my definition one of the chains may happen to converge a larger (smaller) intersection than the other. And one-element chains are excluded. By the definition and by dependent choices every such chain is infinite. Dec 23 '19 at 13:18
• My use of coinitial is the same as this one: en.m.wiktionary.org/wiki/coinitial To be honest I have never encountered a different notion, but this may be my ignorance of course. Dec 23 '19 at 13:21
• Maybe it woube more reasonable to say $\mathcal C$ covers $\mathcal D$ and reserve the coinitiality for mutual covering? Dec 23 '19 at 13:27

Here is a provisional negative answer. If $$\mathcal{C}$$ is a c-minimal chain then $$N=\bigcap\mathcal{C}$$ is closed and nowhere dense (if nonempty): if the interior is nonempty take a nonempty regular open set $$O$$ such that $$\overline{O}\subseteq \operatorname{int}N$$. Then the chain $$\mathcal{C}\cup\{O\}$$ is covered by $$\mathcal{C}$$ but it does not cover it. This shows that a c-minimal chain does not have a minimum and that $$\bigcap\mathcal{C}=\bigcap\{\overline{C}:C\in\mathcal{C}\}$$ is closed. Now the chain $$\{C\setminus N:C\in\mathcal{C}\}$$ is c-minimal in $$X\setminus N$$ and its intersection is empty. This is provisional in the sense that I could not think of a c-minimal chain. For example, in the real line every chain is countable and by diagonalising a co-initial sequence one can construct a strictly smaller chain. Correction: every well-ordered (up or down) chain is countable; every chain still has a co-initial sequence.
• Take the chain $\mathcal C=\{(-\frac{1}{n}, \frac{1}{n})\colon n\in\omega^+\}$. Then $\bigcap\mathcal C=\{0\}$, but $\{C\setminus N\colon C\in\mathcal C\}$ is not c-minimal in $X\setminus{N}$ , since it covers the chain $(-\frac{1}{n},0)$ but not vice versa. I believe that $\mathcal C$ is an example of a c-minimal chain. Jan 5 '20 at 10:35
• It is not c-minimal, let $O=\bigcup_{n=1}^\infty (\frac1{2n+1},\frac1{2n})$; The chain $\{O\cap C:C\in\mathcal{C}\}$ is covered by $\mathcal{C}$ but not vice versa. Jan 5 '20 at 10:44
• There was an error in the commenst above: the new family is not a a chain in the sense of the question. Here is a better chain: let $O_n=\bigcup_{k=n}^\infty (2^{-k}-2^{-k-n-2}, 2^{-k}+2^{-k-n-2})$; The chain $\{O_n:n\in\mathbb{N}\}$ is covered by $\mathcal{C}$ but not vice versa. add a comment Jan 5 '20 at 10:57
• I think there is a misunderstanding between us (which is my fault). Take $\mathcal C$ as above and $\mathcal D\subseteq\mathrm{RO}(\mathbf{R})$ which satisfies (a) (b), and is covered by $\mathcal C$. Every element of $\mathcal D$ must contain 0. For if $D\in\mathcal D$ is such that $0\notin D$, there is $E\in\mathcal D$ such that $Cl E\subseteq D$, $0\notin Cl E$ and for sufficiently big $n\in\omega$, $(-\frac{1}{n},\frac{1}{n})\cap E=\emptyset$, a contradiction. t.b.c. Jan 7 '20 at 9:01
• continued: So every element of $\mathcal D$ contains 0 (and is open), and so $\mathcal D$ covers $\mathcal C$, as $\mathcal C$ is a local basis at 0. I am only interested in chains that satisfy both (a) and (b), and I think my the way I formulated the question causes ambiguity. Thus the conclusion is that $\mathcal C$ is minimal among all chains which have (a) and (b) from my question. Jan 7 '20 at 9:02