# Proving $\overline{A}$ is connected

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?

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.
Either one of $$A\cap U$$ and $$A\cap V$$ may be empty. You have to show that they are not.
I would perhaps begin with $$U,V$$ being closed instead of open. Since we're dealing with the closure of $$A$$, that feels more natural.
• They cannot be empty since they are in the closure right? If either $U$ or $V$ does not intersect $A$ then $A'=\emptyset$,right? – Pedro Gomes Nov 7 '18 at 19:10
• I guess $U,V$ are clopen. – Pedro Gomes Nov 7 '18 at 19:14