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I have to prove that if $E$ is a connected space, then so is $\overline{E}$ (the closure of $E$) a connected space. I tried to prove the contrapositive. So suppose that $\overline{E}$ is not connected in a metric space $X$. This implies there are $A\subset X$ and $B\subset X$ such that $A\cap B=\emptyset$, $A\cap \overline{E}\neq \emptyset$, $B\cap \overline{E}\neq \emptyset$ and $\overline{E}\subset A\cup B$. I will try to prove that this $A$ and $B$ will also work for $E$. Suppose that $A\cap E=\emptyset$ and $B\cap E=\emptyset$, this implies that $A\cup B$ is a subset of the limit points of $E$, because $A\cap \overline{E}\neq \emptyset$ and $B\cap \overline{E}\neq \emptyset$. This contradicts the fact that $\overline{E}\subset A\cup B$, so we have to conclude that $A\cap E\neq \emptyset\neq B\cap E$. Since $E\subset\overline{E}\subset A\cup B$, we have that $E$ is also not connected.

My question is whether my proof is correct, because I have a little bit doubt. If it's not correct, what is the best I can do? Thanks in advance!

Edit1: The definition we use for a connected space $E$ in a metric space $X$ is that if $E$ is connected, then there are no open $A,B\subset X$ such that $A\cap B=\emptyset$, $A\cap E\neq \emptyset\neq B\cap E$ and $E\subset A\cup B$.

Edit2: $A$ and $B$ must be open.

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    $\begingroup$ By connected space, do you mean connected set in a given ambient space? (Closure is relative.) And is the ambient space assumed to be a metric space? $\endgroup$ – Trevor Wilson Oct 3 '13 at 17:05
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    $\begingroup$ A subset $E\subseteq X$ is connected iff the only continuous functions $f:E\to \{0,1\}$ are the constant ones. A continuous function $g:X\to Y$ where $Y$ is Hausdorff is uniquely determined by a dense subset of $X$. $E$ is dense in $\overline E$, and continuous functions $f:E\to \{0,1\}$ are constant by assumption. $\endgroup$ – Pedro Tamaroff Oct 3 '13 at 17:06
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    $\begingroup$ Can you say what exactly the characterization of connectedness is that you are trying to use? It seems to be wrong to me. $\endgroup$ – Carsten S Oct 3 '13 at 17:08
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    $\begingroup$ I believe that the $A,B$ should be open subsets of $X.$ Otherwise, we can prove that no space with more than one point is connected. $\endgroup$ – Cameron Buie Oct 3 '13 at 17:11
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    $\begingroup$ @Pedro: That's a very nice proof! $\endgroup$ – Cameron Buie Oct 3 '13 at 17:12
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As other people have already noted, you need to add $A$ and $B$ are open to your definition. Moving into the second part of your proof, the assumption for contradiction is

$A \cap E = \emptyset$ or $B \cap E = \emptyset$.

You can assume, without loss of generality, that the first condition holds, i.e. $A \cap E = \emptyset$.

Now, your conclusion about limit points is true - $A \cap \overline E$ must consist only of limit points of $E$, i.e. $A$ is an open set that contains limit points of $E$, but no points of $E$ itself. Why can't this happen?

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  • $\begingroup$ well, A is open so suppose you have a limit point x in A. then there has to be radius $\epsilon$ such that the open ball around x is a subset of A, but x is a limit point so in that ball there is going to be element of E, which is a contradiction. $\endgroup$ – Badshah Oct 3 '13 at 21:01
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A subset $E⊆X$ is connected iff the only continuous functions $f:E→\{0,1\}$ are the constant ones. A continuous function $g:X→Y$ where $Y$ is Hausdorff is uniquely determined by a dense subset of $X$, and $\{0,1\}$ with the discrete metric is evidently Hausdorff. $E$ is dense in $\overline E$ and continuous functions $f:E→\{0,1\}$ are constant by assumption.

Another proof my proceed by contraposition. Suppose that $\overline E$ is disconencted by $U,V$. Then $E$ being dense has nonempty intersection with both $U,V$, so it is disconnected by $U'=E\cap U\;,\; V'=V\cap E$, for $U',V'$ are disjoint nonempty and relatively open in $E$, and $E=V'\cup U'$ by $\bar E=V\cup U$.

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A bit more general statement which i think worth saying at this time is :

If $A$ is everywhere dense connected subset of $X$ then $X$ is connected.

Suppose not, then, you have $X=U\cup V$ be a separation for $X$ by open sets (non empty, disjoint).

Then, $A=(U\cap A)\cup(V\cap A)$ would then be separation for $A$

But, $A$ is connected.... So, either $U\cap A$ or $V\cap A$ is empty...

Without loss of generality, we assume $U\cap A$ is empty.

But, as $A$ is every where dense, $A$ has non empty intersection with every non empty open set.

Thus, $U$ should be empty and so, $X$ is connected.

Now,we see for what you have asked...

$A$ is dense in $\bar{A}$ and $A$ is connected, thus by previous result we see that $\bar{A}$ has to be connected.

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  • $\begingroup$ Note that $A$ being dense means $U\cap A$ and $V\cap A$ must be nonempty if $U,V$ are nonempty. But this gives a disconnection of $A$ by means of $U'=A\cap U$, $V'=A\cap V$. Your proof is not very clear, and you have taken unions when I guess you meant to take intersections. $\endgroup$ – Pedro Tamaroff Oct 4 '13 at 5:58
  • $\begingroup$ Its not clear for me what is not clear for you in what i have written... $\endgroup$ – user87543 Oct 4 '13 at 6:01
  • $\begingroup$ You say "Without loss of generality we assume...". Can you explain why? And again, some intersections turned into unions. $\endgroup$ – Pedro Tamaroff Oct 4 '13 at 6:22
  • $\begingroup$ as $A$ is connected and $U\cap A, V\cap A$ are open in $A$ and $A=(U\cap A)\cup (V\cap A)$, one of $(U\cap A)$, $(V\cap A)$ has to be empty... $\endgroup$ – user87543 Oct 4 '13 at 6:26
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    $\begingroup$ I give up. Are you even reading what I write? ${}{}{}$ $\endgroup$ – Pedro Tamaroff Oct 4 '13 at 6:26

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