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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

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    $\begingroup$ Since $U,V$ are nonempty, choose points in them... $\endgroup$ – Lee Mosher Jan 31 '16 at 18:18
  • $\begingroup$ Ok let $x\in U$ and $y\in V$ wht i must do after ? $\endgroup$ – Vrouvrou Jan 31 '16 at 18:22
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    $\begingroup$ 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 $A$, ... $\endgroup$ – Lee Mosher Jan 31 '16 at 18:24
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    $\begingroup$ 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. $\endgroup$ – user296602 Jan 31 '16 at 18:29
  • $\begingroup$ @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 $\endgroup$ – Vrouvrou Jan 31 '16 at 20:13
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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.

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  • $\begingroup$ I don't understand how you deduce thar f is constant ? $\endgroup$ – Vrouvrou Jan 31 '16 at 19:43
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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.

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  • $\begingroup$ so we don't use the fact that $A\subset U$ or $A\subset V$? $\endgroup$ – Vrouvrou Jan 31 '16 at 20:40
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    $\begingroup$ 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. $\endgroup$ – Lee Mosher Jan 31 '16 at 21:04
  • $\begingroup$ 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 ? $\endgroup$ – Vrouvrou Jan 31 '16 at 21:07
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    $\begingroup$ 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 $\endgroup$ – Lee Mosher Sep 4 '17 at 18:02
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    $\begingroup$ My comment with the words "topological space" remains true if you replace those words with "metric space" (when you learn topology, you will learn that a metric space is a special kind of topological space). If you have any further followup questions, I suggest that you post your own question on math.stackexchange. $\endgroup$ – Lee Mosher Sep 4 '17 at 18:43

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