Proving $\overline{A}$ is connected

Question

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!

Possible duplicate of The closure of a connected set in a topological space is connected — freakish, Nov 07 '18 at 19:03
@freakish I am asking if my proof is right or wrong not if there are analogous proofs. — Pedro Gomes, Nov 07 '18 at 19:06

score 4 · Accepted Answer · edited Nov 07 '18 at 20:21

4

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.

edited Nov 07 '18 at 20:21

Matematleta

29,139

answered Nov 07 '18 at 19:16

Mariah

3,760

score 0 · Answer 2 · answered Nov 07 '18 at 19:05

0

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.

answered Nov 07 '18 at 19:05

Arthur

199,419

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 07 '18 at 19:10
I guess $U,V$ are clopen. – Pedro Gomes Nov 07 '18 at 19:14

Proving $\overline{A}$ is connected

2 Answers2

Linked