TITLE: If A is connected, is $\bar{A}$ connected?
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: 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.