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I just proved this statement like the below. Is this valid or solid proof?

Thank you!

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It wouldn't hurt if you could try and write that in $\TeX$. It usually looks nicer and takes up less space. –  Pedro Tamaroff Jul 16 '13 at 2:28
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It’s fine, but as @Peter says, you’re working harder than necessary: for each $x\in A$, $A$ is an open nbhd of $x$ disjoint from $B$, so $x\notin\operatorname{cl}B$, and vice versa. And this simpler approach has the extra virtue of not relying an a metric: the result is true whether the space is metrizable or not. –  Brian M. Scott Jul 16 '13 at 2:30

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up vote 2 down vote accepted

Your idea looks fine. The proof that $\overline A\cap B=\varnothing$ can be direct, and by symmetry (as you say), we also get $A\cap\overline B=\varnothing$.

Take $b\in B$. Since $B$ is open, there is some open neighborhood $N$ with $b\in N\subseteq B$. But then from $A\cap B=\varnothing$ we get $N\cap A=\varnothing$; so $b\notin\overline A$. (Yes, we can even take $B=N$!)

ADD Recall that $x\in \overline A$ if and only if for each open set $O$ contaning $x$, we have $A\cap O\neq \varnothing$.

In fact, we can say more:

Let $(X,\mathscr T)$ be a (topological, metric) space. Then the following definitions of connectedness are equivalent:

$(1)$ There exist two non-empty disjoint open sets $A,B$ with $A\cup B=X$.

$(2)$ There exist two non-empty disjoint closed sets $A',B'$ with $A'\cup B'=X$.

$(3)$ There exist two non-empty separated sets $A'',B''$ with $A''\cup B''=X$.

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