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If A is connected, is $\bar{A}$ connected? Here $\bar{A}$ is the closure of $A$.

Here's my attempt at trying to prove this:

Suppose that $\bar{A}$ is disconnected. Then, there exists open, disjoint, non empty subsets $U, V$ such that $U \cup V = \bar{A}$ (This is the definition of disconnectedness that I've learned)

Then we can write A as $A = (U \cap A) \cup (V \cap A)$

$U \cap A$ and $V\cap A$ are open in A. Also, they are disjoint since $U$ and $V$ are disjoint. Now if I show that $U \cap A$ and $V \cap A$ are nonempty then I get a contradiction and proof is complete. However, I am not quite sure how to do this? Or is it not true that $\bar{A}$ is connected?

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  • $\begingroup$ Usually when we start these problems, we explain the setting. For example, I'm assuming you want to try to prove this for any topological space, so you should start the question by saying "Let $(X, \mathcal{T})$ be any topological space. I want to prove if $A \subseteq X$ is connected, then so is $\overline{A}$, the closure of $A$." $\endgroup$ – layman Feb 1 '15 at 21:33
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You are on the right track with your problem. Now argue by contradiction. Suppose one of the two sets $U \cap A$ or $V \cap A$ is empty (choose one without loss of generality, maybe $U \cap A$). Then this implies $A \subseteq V \cap A$, so $A \subseteq V$. But since $U$ and $V$ are nonempty and their union equals $\overline{A}$, it must be that $U$ contains some point in $\overline{A}$.

Now apply Matt Samuel's hint that since $U$ is open and it contains a point in $\overline{A}$, it must contain a point in $A$ (you should prove this hint, too). That means $U \cap A \neq \emptyset$, which contradicts our assumption that it was empty in the first place.

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Hint: Any open set containing some point in the closure of $A$ also contains some point in $A$.

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Any set containing a dense connected subset is connected. Indeed, say $X$ is your set and $C$ is dense and connected. Show that any morphism $X\to 2$ is constant by considering the restriction to $C$ and using it is connected. Note that $2$ is Hausdorff.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – Johanna Feb 1 '15 at 22:25
  • $\begingroup$ @Johanna Why do you think it doesn't? $\endgroup$ – Pedro Tamaroff Feb 1 '15 at 22:47

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