Definition. A closed subset $S$ of a topological space $X$ is called

$\bullet$ a separator of $X$ if $X\setminus S$ is disconnected;

$\bullet$ an irreducible separator if $S$ is a separator of $X$ and $S$ coincides with each closed subset $C\subset S$ that separates $X$.

Problem. Let $X$ be a compact connected $n$-manifold without boundary.

1) Does each closed separator of $X$ contain a closed irreducible separator of $X$?

2) Is it true that each irreducible separator of $X$ has only finitely many connected components?

Remark. Using the Mayer-Vietoris sequence, I can prove that each irreducible separator in a manifold $X$ with trivial homology group $H_1(X)$ is connected. So, I suggest that if the homology group $H_1(X)$ of a manifold $X$ is finitely generated, then each irreducible separator of $X$ has only finitely many connected components. But I am no so fluent in Algebraic Topology, so cannot catch a proof of this (I hope true) fact. I admit that the answer could follow from a version of Mayer-Vietoris exact sequence for unions of arbitrary finite family of sets (something a la van Kampen Theorem), but I cannot find such a modification of the Mayer-Vietoris in the literature.

  • $\begingroup$ Try duality (it's stronger). $\endgroup$ – Wlod AA Sep 3 '18 at 19:04
  • $\begingroup$ Q1 seems a simple consequence of Kuratowski-Zorn (answer: yes). It's so for arbitrary Hausdorff compact spaces. (Even for arbitrary Hausdorff spaces for compact separators). $\endgroup$ – Wlod AA Sep 3 '18 at 19:07
  • $\begingroup$ @WlodAA The problem with Kuratowski-Zorn is that a decreasing sequence of separators can separate different pairs of points and then for the intersection it will be not clear why it does separate the space. Just think of the Hawai earings: each small piece of a Hawai earing does separate the plane but their intersection is a singleton, which is not a separator. $\endgroup$ – Taras Banakh Sep 3 '18 at 19:16
  • $\begingroup$ @WlodAA Do you mean Poincare or Alexander duality? $\endgroup$ – Taras Banakh Sep 3 '18 at 19:17
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    $\begingroup$ In fact, I have already a proof that an irreducible separator between a pair of points in a triangulable manifold has countably many connected components and this is sufficient for my purposes. So, now these question are just of academic interest. But anyway irreducible separators turn to be quite interesting creatures. $\endgroup$ – Taras Banakh Sep 3 '18 at 19:20

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