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A standard method to show that a set $C$ is the connected component of a point $p\in C$ (in some metric space $X$) consists of two parts,

  1. showing that $C$ is connected, and
  2. showing that every $q \notin C$ can be separated from $p$, i.e. there are disjoint open sets $U ,V $ with $p \in U $, $q \in V $, and $X = U \cup V $.

I am trying to prove that given the above two criteria hold, $C$ is a connected component. Here's my attempt:

Suppose $C$ is a connected set in a metric space $X$. We further know that every $q \notin C$ can be separated from $p$, i.e. there are disjoint open sets $U ,V $ with $p \in U $, $q \in V $, and $X = U \cup V ~;~ U \cap V = \phi$. Therefore $U \subseteq C$. $p \in U \cap C~$ and $q \in X\setminus C~ \cap ~V$

Suppose $C \subseteq D$ where $D$ is a connected proper subset of $X$. Then, $D$ clearly includes some points from $V$

EDIT: Then $D = ((X \setminus U) \bigcap D) \bigcup U$ where $(X \setminus U )\bigcap D$ is open in $D$ and $(V \cap D) \bigcap U = \phi$ which means $D$ is disconnected, a contradiction from our earlier assumption. Therefore, there does not exist a super connected set for $C$. Thus, $C$ is a connected component of $X$.

Is my proof correct?

Does the converse also hold true? Here's my attempt:

Suppose $C \ne X$ is a connected component and $p \in C$. Let $q \notin C$. Since, $C \ne X ~\therefore~X$ is a disconnected metric space and there exist two closed disjoint subsets $A,B$ of $X$ such that $X = A \cup B$ and $A \cap B = \phi$. Since, $C$ is a connected component, $C$ is closed in $X$.

Without lose of generality, Suppose both $p,q \in A$. Since, $p \in A \cap C \implies A \cap C \ne \phi$. Then, since $q \notin C~\therefore~q \in A \setminus C \implies A \setminus C \ne \phi$ which would mean $A \cup C$ is also a connected set contradicting the given statement that $C$ is a connected component.

Therefore, $p,q$ must belong to separate open sets in $X$.

Is my proof correct? Thank you for reading through!

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The converse does not hold. If conditions 1 and 2 hold, then $C$ is the connected component of $p$, but it is not necessarily the case that every point outside the component can be separated from $p$ by open sets.

To see that 1 and 2 imply that $C$ is the connected component of $p$, let $K$ be the connected component of $p$ in $X$. We need to show that $K = C$. Since by definition connected components are maximal connected sets, and $C \cup K$ is a connected set (the union of two connected sets with nonempty intersection is connected) containing $K$, it follows that $C \cup K = K$, i.e. $C \subset K$. For the revers inclusion, consider $q \in X \setminus C$. By assumption 2, there are disjoint open $U,V$ such that $p \in U$, $q\in V$, and $X = U \cup V$. Then $(U \cap K)$ and $(V \cap K)$ are disjoint relatively open (in $K$) sets with $K = (U \cap K) \cup (V \cap K)$, and by the connectedness of $K$ it follows that $U \cap K = \varnothing$ or $V \cap K = \varnothing$. Since $p \in U \cap K$ we must have $V \cap K = \varnothing$, in particular $q \notin K$. Since this holds for all $q \in X \setminus C$, the inclusion $K \subset C$, and consequently $K = C$, is proved.

To see that the converse does not hold, an example suffices. Let

$$X = \{(0,0),(0,1)\} \cup \bigcup_{n = 1}^{\infty} \bigl\{ (1/n,t) : 0 < t < 1\bigr\}$$

with the subspace topology inherited from $\mathbb{R}^2$. Then the connected component of $p = (0,0)$ in $X$ is the singleton $\{(0,0)\}$, but $q = (0,1)$ cannot be separated from $p$ by open sets. The quasicomponent of $p$ is the two-element set $\{p,q\}$.

By definition, the quasicomponent of a point is the intersection of all clopen (that is, closed and open) sets containing that point. The quasicomponent of $p$ consists of exactly those points of $X$ that cannot be separated from $p$ by open sets. For if $r$ does not belong to the quasicomponent of $p$, then there is a clopen set $A$ containing $p$ but not $r$, then $U = A$ and $V = X\setminus A$ is a decomposition of $X$ into two disjoint open sets that separates $p$ from $r$. And if $r$ can be separated from $p$ by open sets $U,V$, then these two sets are actually clopen, so $r$ does not belong to the quasicomponent of $p$.

To see that in the above example $q$ belongs to the quasicomponent of $p$, consider a clopen $A$ containing $p$. Since $A$ is open, there is an $N$ such that $A \cap \{(1/n,t) : 0 < t < 1\} \neq \varnothing$ for all $n \geqslant N$. Since the segment $L_n = \{(1/n,t) : 0 < t < 1\}$ is connected (even path-connected), and $L_n \cap A,\, L_n \setminus A$ is a decomposition of $L_n$ into disjoint open sets, one of these must be empty, thus $L_n \subset A$ for all $n \geqslant N$. But $q$ is the limit of the sequence $\bigl((1/n, 1-1/n)\bigr)_{n \geqslant N + 1}$ all of whose points belong to $A$ by what we've seen, and $A$ is closed, so it follows that $q \in A$. Showing that no other point belongs to the quasicomponent of $p$ is easy: $X \setminus L_n$ is a clopen set containing $p$ but none of the points on $L_n$.

And finally, since $\{p,q\}$ is not connected, it follows that the component of $p$ is a proper subset of its quasicomponent.

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  • $\begingroup$ Many many thanks. I have one query though. Would you be able to please pin point where I committed an error in any deduction in the proof I presented while trying to prove the converse? $\endgroup$
    – MathMan
    Commented Mar 8, 2018 at 15:23
  • $\begingroup$ You could only deduce that $A \cup C$ is connected if $A$ is connected, but that need not be the case. $\endgroup$ Commented Mar 8, 2018 at 15:28
  • $\begingroup$ since $A \cap C \ne \phi$, therefore, any continuous function that is constant on $A$ will be constant on $A \cup C$ as well? thus, shouldn't $A \cup C$ be connected as well? and since, $q \in A \setminus C$, we know that $A $ is not a subset of $C$. $\endgroup$
    – MathMan
    Commented Mar 8, 2018 at 15:45
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    $\begingroup$ You're probably thinking of the characterisation of connectedness in terms of continuous functions to the discrete space $Z = \{0,1\}$. Then sure, a continuous $f \colon X \to Z$ that is constant on $A$ will be constant on $A \cup C$. But that's not the point. One needs to check whether every continuous $f \colon A \to Z$ (or $f \colon (A\cup C) \to Z$) is constant. And if $A$ is not connected, that isn't the case. $\endgroup$ Commented Mar 8, 2018 at 15:59
  • $\begingroup$ aah got it. Thank you very much sir. $\endgroup$
    – MathMan
    Commented Mar 8, 2018 at 16:02

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