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This is a proof by contradiction, unsing the definition of connectedness from Rudin, Principles of Mathematical Anaylsis. Can someone please check if it is any good?

Suppose S is a connected set, where its closure $ \bar{S} $ is not connected. Therefore there exist two nonempty sets A and B, such that $ \bar{A} \cap B = \bar{B} \cap A = \emptyset $ and $ A \cup B = \bar{S}$. Define $ G:= A \cap S $ and $ H:= B \cap S $. Because G is a subset of A and H is a subset of B, it is clear that $ \bar{G} \cap H = \bar{H} \cap G = \emptyset $. and $A\cup B = S$. So that S is not connected, contrary to our first assumption. q.e.d.

Thank you

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    $\begingroup$ You seem to have changed your $S$ into an $E$ at the end there. $\endgroup$ – user259242 Nov 4 '16 at 11:57
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    $\begingroup$ $G$ and $H$ are the same set. This is probably not what you want. $\endgroup$ – user259242 Nov 4 '16 at 11:59
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    $\begingroup$ Shouldn't $A\cap\overline{S}$ and $B\cap\overline{S}$ be assumed to be nonempty, for the contradiction? $\endgroup$ – Michael Burr Nov 4 '16 at 12:01
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    $\begingroup$ Something is not quite right. You have not used the assumption that $\overline{S}$ is not connected. $\endgroup$ – user259242 Nov 4 '16 at 12:02
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    $\begingroup$ The current version is correct. $\endgroup$ – Brian M. Scott Nov 4 '16 at 14:59
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The contradiction is that $G\cup H = S$ where G and H are separated, not $A\cup B=S$

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