**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.