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,
- showing that $C$ is connected, and
- 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!