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Definition. A closed subset $S$ of a topological space $X$ is called a separator between points $x,y\in X\setminus S$ if the points $x$ and $y$ belong to different connected components of $X\setminus S$. A separator $S$ is called an irreducible separator between $x$ and $y$ is $S$ coincides with each closed separator between $x$ and $y$ that is contained in $S$.

Using Kuratowski-Zorn Lemma it is easy to prove that each closed separator in a locally path-connected (or more generally locally continuum-connected) space contains a closed irreducible separator. Is the case result true without the local path connectedness?

Question. Does each closed separator between two points $a,b$ of a metrizable continuum contain an irreducible closed separator between $a$ and $b$?

I hope that the answer to this question should be know but somehow I cannot find in the books of Kuratowski and Nadler.

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No. Consider the subset $X$ of $\mathbb{R}^2$ consisting of the union of line segments beginning at $(0,0)$ and ending at $(1,2^{-n})$ for $n\geq 0$ or $(1,0)$. Let $x=(0,0)$ and $y=(1,0)$ and consider the separator $S$ consisting of points of the form $(\frac{1}{2},2^{-n-1})$ or $(\frac{1}{2},0)$. I claim that $S$ is a separator between $x$ and $y$ with no irreducible sub-separator. Clearly $S$ is a separator between $x$ and $y$ since $X\setminus S$ can be partitioned into two relatively clopen sets containing $x$ and $y$. To see that $S$ has no irreducible sub-separator, note that first any set $S^\prime \subseteq S$ that separates $x$ and $y$ must contain the point $(\frac{1}{2},0)$, otherwise the line segment between $(0,0)$ and $(1,0)$ connects $x$ and $y$. Furthermore notice that for any $S^\prime \subseteq S$ that is a separator between $x$ and $y$ and for any point $z$ of the form $(\frac{1}{2},2^{-n-1})$, $S^\prime \setminus \{z\}$ is also a separator between $x$ and $y$. This implies that the only subset of $S$ that can possibly be an irreducible separator is the singleton $\{(\frac{1}{2},0)\}$, but $X \setminus \{(\frac{1}{2},0)\}$ is connected. Therefore $S$ has no irreducible sub-separator.

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  • $\begingroup$ $(0,1)$ should be $(1,0)$, twice - right? $\endgroup$ – GNiklasch Sep 4 '18 at 16:54
  • $\begingroup$ Yes. Thank you for pointing that out. $\endgroup$ – James Hanson Sep 4 '18 at 16:56

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