Return to Revisions

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.