The closure of a connected set in a topological space is connected

Solution 1:

Suppose that $E$ is connected. Let $A,B\subseteq X$ be separated sets (that is, $\overline{A}\cap B=A\cap\overline{B}=\varnothing$) such that $\overline{E}=A\cup B$, and suppose that $A\neq\varnothing$. Let us prove that $B=\varnothing$.

Let $a\in A$. Since $A\cap \overline{B}=\varnothing$, there exists a neighborhood $U$ of $a$ such that $U\cap B=\varnothing$. Since $a\in\overline{E}$, then there exists some point $x\in E\cap U$, so $x\not\in B$, hence $x\in E\cap A$. Therefore, $E\cap A\neq\varnothing$.

Notice that $E=(A\cap E)\cup (B\cap E)$, and $A\cap E$ and $B\cap E$ are obviously separated. As $A\cap E\neq\varnothing$, from the previous paragraph, and $E$ is connected, then $B\cap E=\varnothing$.

(See PS below for an alternative end to the proof without the argument by contradiction)

Finally, suppose, in order to obtain a contradiction, that $B\neq\varnothing$, and take $b\in B$. By the same arguments as those used in the second paragraph above, switching $A$ and $B$ and $a$ by $b$, we would conclude that $B\cap E\neq\varnothing$, contradicting what we have just proved.

Therefore, $B=\varnothing$. This proves that $\overline{E}$ is connected.

PS: As $E\subseteq A\cup B$ and $E\cap B=\varnothing$, then $E\subseteq A$, so $\overline{E}\subseteq\overline{A}$. It follows that $$B=B\cap\overline{E}\subseteq B\cap\overline{A}=\varnothing.$$

Solution 2:

I believe this can be made even more concise: Suppose $\overline{E}=A\cup B$ for disjoint, nonempty, and open $A,B$.

$E$ connected and $E=(A\cap E)\cup (B\cap E)$, so wlog, $A\cap E=E$. Then $B$ is an open set containing a limit point of $E$, and so it must intersect $E\subseteq A$ nontrivially - contradiction, as $A\cap B=\emptyset$.