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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.

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The conclusion $\partial C = \emptyset$ is certainly false. For an easy counter-example, take $X = [1,2] \cup [3,4] \subseteq \mathbb R$ and $C = [1,2]$. – Srivatsan Nov 8 '11 at 16:48
Hint: You are mixing up the topological spaces $\mathbb{R}^n$ and $X$. $C$ is clopen in $X$ but not in $\mathbb{R}^n$. – Nate Eldredge Nov 8 '11 at 16:51
Try to prove $\partial C\subseteq \partial X$. Also for any topological space, any connected component is closed. – user18119 Nov 8 '11 at 21:26
@NateEldredge Thanks!! You are right! – Y. Fan Nov 9 '11 at 5:09
@Y.Fan: you seem to be contradicting yourself when you say that in a mtric space a component is closed and then saying that in (a locally-connected space like) $\mathbb R^n$ a component is closed. – gary Nov 10 '11 at 2:42
up vote 4 down vote accepted

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).

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Excellent proof!Thanks! – Y. Fan Nov 9 '11 at 5:09
Why $X-int(X)=\partial{X}$? I thought the definition of $\partial{X}$ is $\partial{X}=cl(X)-int(X)$. Right? Is $X$ closed? Or connectness implies close? – Y. Fan Nov 10 '11 at 1:20
@Y.Fan: It should be $X - int(X) \subset \partial X$. I will fix it. – André Caldas Nov 10 '11 at 2:15
@Y.Fan: Actually, it should be totally different. I have corrected the proof. – André Caldas Nov 10 '11 at 2:37
I like your simpler proof. Thank you, again! – Y. Fan Nov 10 '11 at 5:25

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