Let $A$ be a connected subspace of a topological space $(X,\tau)$. Prove that $\overline{A}$ is also connected.
If $\overline{A}$ is disconnected then there exists $U,V\subset\overline{A}$ open in the subspace topology that $U\cup V=\overline{A}$ and $U\cap V=\emptyset$.
As $A\subseteq\overline{A}$ the $U\cap A$ and $V\cap A$ are two open sets in the subspace topology related to $A$ so that $(U\cap A)\cap(V\cap A)=(U\cap V)\cap A=\emptyset\cap A=\emptyset$ and $A=(A\cap V)\cup (A\cap U)$. So $A$ would be disconnected contradicting the assumption. Then $\overline{A}$ must be connected.
Question:
Is my proof right? If not. Why?
Thanks in advance!
Best Answer
Let $U, V$ be a separation of $\overline{A}$.
Then since $A \subset \overline{A}$ and $A$ is connected, either $A \subset U$ or $A \subset V$. Assume $A \subset U$.
But now since $\overline{A} \cap V \neq \emptyset$, take any point there, $v$. Then $v\in V$ implies that $V \cap A \neq \emptyset$. Then $V \cap U$ isn't empty.