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.

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    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Dec 23 '19 at 12:54
  • $\begingroup$ Your definition of strictly descending seems to rule out a chain with only one element. $\endgroup$ – Joel David Hamkins Dec 23 '19 at 12:56
  • $\begingroup$ 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. $\endgroup$ – Mad Hatter Dec 23 '19 at 13:18
  • $\begingroup$ 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. $\endgroup$ – Mad Hatter Dec 23 '19 at 13:21
  • $\begingroup$ Maybe it woube more reasonable to say $\mathcal C$ covers $\mathcal D$ and reserve the coinitiality for mutual covering? $\endgroup$ – Mad Hatter 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.

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  • $\begingroup$ 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. $\endgroup$ – Mad Hatter Jan 5 at 10:35
  • $\begingroup$ 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. $\endgroup$ – KP Hart Jan 5 at 10:44
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    $\begingroup$ 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 $\endgroup$ – KP Hart Jan 5 at 10:57
  • $\begingroup$ 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. $\endgroup$ – Mad Hatter Jan 7 at 9:01
  • $\begingroup$ 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. $\endgroup$ – Mad Hatter Jan 7 at 9:02

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