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Show that a subspace $T$ of a topological space $S$ is disconnected iff there are nonempty sets $A,B \subset T$ such that $T= A\cup B$ and $\overline{A} \cap B = A \cap \overline{B} = \emptyset$. Where the closure is taken in $S$.

I've used this relatively simple proof for many of these slightly different types of questions so I was wondering if it's the right method. It seems pretty good, except for the 'where the closure is taken in $S$ part'.

$T$ is disconnected if and only if there exists a partition $A,B \subset T$ such that $T = A \cup B$ and $A \cap B = \emptyset$. Also, $A$ and $B$ are both open and closed therefore $\overline{A} = A$ and $\overline{B} = B$. The result follows.

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  • $\begingroup$ This is fine. In fact, it’s Theorem 26.5 in Stephen Willard, General Topology. $\endgroup$ – Brian M. Scott May 19 '12 at 19:16
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It looks fine to me, in particular because as $\,A\subset \overline{A}\Longrightarrow A\cap B\subset \overline{A}\cap B\,$ , so if the rightmost intersection is empty then also the leftmost one is, which is the usual definition

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  • $\begingroup$ So there's nothing funny with the fact that the closure is in $S$ in the question and in the proof the closure is obviously in $T$? $\endgroup$ – user26069 May 19 '12 at 18:58

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