# The closure of a connected set is connected.

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

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

The contradiction is that $G\cup H = S$ where G and H are separated, not $A\cup B=S$
• Can you be sure that $G$ and $H$ are both not empty? Apr 30 at 19:39
• Yes, if $x \in S$ then $x \in \overline{S}$, and then $x \in A$ or $x \in B$. If $x \in A$ then $x \in A \cap S$, which implies that $A \cap S \neq \emptyset$. Now, since $A \cup B = \overline{S}$ with $A \cap B = \emptyset$, then $B = \overline{S} - A$. $B \cap S = (\overline{S} - A) \cap S = \overline{S} \cap A^{c} \cap S = S - A = \emptyset$ if and only if $S = A$, but $\overline{A} \cap B = \overline{S} \cap B = \emptyset$, which implies that $B = \emptyset$ due to $B \subset \overline{S}$, but $B$ is a non emptyset by definition, which is a contradiction. So, $B \cap S \neq 0$ too. May 1 at 14:56