# The Closure of a connected set is a connected set

If i suppose that $\overline{A}$ is not connected then $\begin{cases} \overline{A}=U\cup V\\ U\neq\emptyset, V\neq\emptyset\\U\cap V=\emptyset\end{cases}$ where $U,V$ are open in $\overline{A}$

We know that $A$ is connected and $A\subset \overline{A}$ then $A\subset U$ or $A\subset V$

How to continue?

Thank you

• Since $U,V$ are nonempty, choose points in them... – Lee Mosher Jan 31 '16 at 18:18
• Ok let $x\in U$ and $y\in V$ wht i must do after ? – Vrouvrou Jan 31 '16 at 18:22
• This is a "follow your nose" proof. What you should be doing is to look at the next thing in front of your nose. Since $x \in U$, and $U$ is open in $\overline A$, and $\overline A$ is the closure of $A$, ... – Lee Mosher Jan 31 '16 at 18:24
• Rather than continually posting and deleting questions when they are downvoted, please edit your questions to improve them. I've seen this happen quite a few times. – user296602 Jan 31 '16 at 18:29
• @LeeMosher if $x\in U$ open then $U\cap A\neq \emptyset,$ i don't see how to find a contradiction withe the fact that $A$ is connected – Vrouvrou Jan 31 '16 at 20:13

Let $f:\bar A\rightarrow \{0,1\}$ be a continuous map. Since $A$ is connected, $f(A)=0$ or $f(A)=1$. Suppose $f(A)=0$, then $f^{-1}(0)$ is a closed subset which contains $A$, so $\bar A\subset f^{-1}(0)$. Same argument if $f(A)=1$. We deduce that $f$ is constant. Thus $\bar A$ is connected.
Since $U,V$ are nonempty and open in $\overline A$, and since $\overline A$ is the closure of $A$, each of $U,V$ has nonempty intersection with $A$. The sets $U'=U \cap A$, $V'=V \cap A$ therefore have the following properties: $A = U' \cup V'$; $U' \ne \emptyset$ and $V' \ne \emptyset$; $U',V'$ are open in $A$; $U' \cap V' = \emptyset$. It follows that $A$ is disconnected.
• so we don't use the fact that $A\subset U$ or $A\subset V$? – Vrouvrou Jan 31 '16 at 20:40
• I am continuing the proof by contradiction that you started in your question: starting from the assumption that $\overline A$ is disconnected, one proves that $A$ is disconnected. – Lee Mosher Jan 31 '16 at 21:04
• in my proof i used the fact that $A$ is connected then $A\subset U$ or $A\subset V$ can i continue with this and found a contradiction ? – Vrouvrou Jan 31 '16 at 21:07
• Every open subset of a topological space (in this case $\overline A$) has nonempty intersection with every dense subset of that space (in this case $A$) @lorenzo – Lee Mosher Sep 4 '17 at 18:02