Let $C$ be a connected component of $X\subset\mathbb{R}^n$. I want to prove of disprove that $\partial{C}\subset\partial{X}$ (where $\partial{A}$ means the boundary set of $A$).
In metric space, I know a connected component is closed.
In locally connected space(clearly, $\mathbb{R}^n$ is locally connected), I know a connected component is open.
So, in $\mathbb{R}^n$, any components are clopen. Therefore, the boundary set of a given connected component $C$ is exactly the empty set, which trivially is contained in $\partial{X}$
However, it looks very strange to me. There must be something wrong.
I know this question is definitely a newbie topology question. Any explanation will be appreciated.
Best Answer
Let $x \in \mathrm{cl}(C)$. Then, suppose that $x \in \mathrm{int}(X)$. Since $\mathbb{R}^n$ is locally connected, there is a connected set $V \subset X$ which is a neighborhood of $x$.
Since $C$ is connected and $V$ intersects $C$, $C \cup V \subset X$ is connected (why?). Therefore, $V \subset C$. That is, $x \in \mathrm{int}(C)$.
That is, for any $x \in \mathrm{cl}(C)$, $$ x \in \mathrm{int}(X) \Rightarrow x \in \mathrm{int}(C). $$ Therefore, $$ \partial C = \mathrm{cl}(C) \setminus \mathrm{int}(C) \subset \mathrm{cl}(C) \setminus \mathrm{int}(X) \subset \mathrm{cl}(X) \setminus \mathrm{int}(X) = \partial X. $$
Notice that this proof works for any locally connected space in place of $\mathbb{R}^n$.
Edit: Proof made much much simpler (and correct :-P).