If A is connected, is $\bar{A}$ connected? 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?
 A: Hint: Any open set containing some point in the closure of $A$ also contains some point in $A$.
A: 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.
A: 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. 
