Confusion about the boundary of connected components 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. 
 A: 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).
A: Suppose $X$ is locally connected, in this case $\mathbb{R}^n$ is locally connected for any $n \in \mathbb{N}$. If $A \subseteq X$ and $C$ is a connected component of $A$ then:

*

*$C^\circ = C \cap A^\circ$.

*$\delta(C) \subseteq \delta(A)$
The inclusion $C^\circ \subseteq C \cap A^\circ$ is trivial since $C$ is a subset of $A$.
We want to prove that $C \cap A^\circ \subset C^\circ$. Let $x \in C \cap A^{\circ}$, since $X$ is locally connected then there must exist $U$ connected neighborhood of $x$ such that $U \subseteq A$. Since $U$ is connected and contains $x$ and $C$ is the maximal connected set that contains $x$ then $U \subset C$, therefore $x \in C^{\circ}$.
$\therefore C^{\circ} = C \cap A^{\circ}$
By definition $\delta(C) = \overline{C} \backslash C^{\circ}$, using what we just proved we have
\begin{align}
\delta(C) &= \overline{C} \backslash (C \cap A^{\circ})\\
&= \overline{C} \cap (C \cap A^{\circ})^c\\
&= \overline{C} \cap (C^c \cup (A^{\circ})^{c})\\
&= (\overline{C} \cap C^c) \cup (\overline{C} \cap A^{\circ})
\end{align}
Notice that $C$ is always a closed set, therefore $C = \overline{C}$, therefore
\begin{align}
\delta(C) &= (C \cap C^c) \cup (C \cap (A^{\circ})^c)\\
&= \varnothing \cup (C \backslash A^{\circ})\\
&= C \backslash A^{\circ}
\end{align}
Lastly, since $C \subseteq A$ it's not hard to see that $C \backslash A^{\circ} \subseteq \overline{A}  \backslash A^{\circ}$.
$\therefore \delta(C) \subseteq \delta(A)$
