# Connectedness of compact metric space

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 $$y\in N$$?

• 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 $A$? Commented Jul 30, 2020 at 10:09
• sorry for the confuse, actually, I want to express that $x\in A$ and $y\in B$. Commented Jul 30, 2020 at 10:29

(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 $$X$$.

• what are the sets A_i'? Commented Jul 30, 2020 at 7:55
• connected components of $P\setminus C$ Commented Jul 30, 2020 at 7:56
• 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 $X$. Commented Jul 30, 2020 at 8:02
• 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. Commented Jul 30, 2020 at 8:08
• @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. Commented Jul 30, 2020 at 9:05

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.

• 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. Commented Jul 31, 2020 at 0:55
• @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. Commented Aug 5, 2020 at 22:15
• that sounds good, thanks for your works! Commented Aug 8, 2020 at 23:42