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Let $X$ be a compact metric space satisfying the following condition: for any given positive number $\delta>0$, only finitely many components of $X$ have diameter larger than $\delta$.

For a given component $P$ of $X$, let us remove a connected compact subset $C$ of $P$ such that $P\setminus C$ is disconnected. Consider a disjoint partition $P\setminus C=A\cup B$, with $A$ and $B$ at a positive distance. Suppose that $x\in A$ and $y\in B$.

Are then $x$ and $y$ in different quasi-components of $X\setminus C$? or can we find a separation of $X\setminus C=M\cup N$ such that $x\in M$ and $y\in N$?

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  • $\begingroup$ Can you clarify whether we can assume $x\in A$ and $y\in B$? Or is it only that a partition $A\cup B$ of $P\setminus C$ exists, and $x,y$ might both be in $A$? $\endgroup$
    – Geva Yashfe
    Commented Jul 30, 2020 at 10:09
  • $\begingroup$ sorry for the confuse, actually, I want to express that $x\in A$ and $y\in B$. $\endgroup$
    – Yee Neil
    Commented Jul 30, 2020 at 10:29

2 Answers 2

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(This answers the previous version of the question, for components but not quasicomponents).

I think, it us true in every topological space: if $X$ is a topological space with connected components $P$ and $Q_j$: $X=P\sqcup (\sqcup_j Q_j)$, and you remove a set $C$ from a connected component $P$ so that it becomes disconnected: $P\setminus C=\sqcup A_i$, where $A_i$ are the connected components of $P\setminus C$, then the connected components of $X\setminus C$ are $A_i$'s and $Q_j$'s. The reason is that the connected components are the inclusion-maximal connected subsets, and each connected subset of $X\setminus C$ lies inside a single connected component of $X$.

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  • $\begingroup$ what are the sets A_i'? $\endgroup$
    – Yee Neil
    Commented Jul 30, 2020 at 7:55
  • $\begingroup$ connected components of $P\setminus C$ $\endgroup$
    – Fedor Petrov
    Commented Jul 30, 2020 at 7:56
  • $\begingroup$ If P is a segment with length 1, and there is a sequence of segmenst with length 1 which are all parallel to $P$ converging to $P$ , $X$ is the union of these segmetns. Then we remove the midpoint $p$ of $P$, now $P\setminus p$ is disconnected, but $P\setminus p$ is thecomponent of the $X$. $\endgroup$
    – Yee Neil
    Commented Jul 30, 2020 at 8:02
  • $\begingroup$ You probably mean that $P\setminus p$ is the component of $X\setminus p$, but it is not the case: the components must be connected, and $P\setminus p$ is not. $\endgroup$
    – Fedor Petrov
    Commented Jul 30, 2020 at 8:08
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    $\begingroup$ @YeeNeil I think you might be talking about quasi-components. Your example is example 2 in this wikipedia page: en.wikipedia.org/wiki/Locally_connected_space#Quasicomponents However, connected components really are always connected. They are precisely the connected subsets which are maximal under the inclusion ordering. $\endgroup$
    – Geva Yashfe
    Commented Jul 30, 2020 at 9:05
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Your compactum $X$ is finitely Suslinian (by defnition). As such, every connected subset of $X$ is locally connected (see this paper), and this implies that the components of $X\setminus C$ are the same as the quasi-components of $X\setminus C$. See this paper, Theorem 2.1. @Fedor Petrov already argued that $x$ and $y$ are in different components of $X\setminus C$, therefore they are also in different quasi-components.

Note that $C$ does not have to be connected or compact. It can be any subset of $P$ such that $P\setminus C$ is disconnected.

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  • $\begingroup$ Yeah, as far as I can see, the idea is that finitely Suslinian implies hereditarily locally connected, and then hereditarily locally connected implies (q=c) according to Theorem 2.1. But I wonder, by the definition in my question, can we say that $X$ is finitely Suslinian, note that we only requirement that the components with diameter than a given positive number $\delta>0$ is finitely, and this of course different with that a sequence a pairwise disjoint subcontinua form a null sequence. $\endgroup$
    – Yee Neil
    Commented Jul 31, 2020 at 0:55
  • $\begingroup$ @YeeNeil Your objection is perfectly valid. But I think that a more elementary approach can be taken to show that $x$ and $y$ are in different quasi-components. I will update my answer when I get the chance. $\endgroup$
    – D.S. Lipham
    Commented Aug 5, 2020 at 22:15
  • $\begingroup$ that sounds good, thanks for your works! $\endgroup$
    – Yee Neil
    Commented Aug 8, 2020 at 23:42

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