connected in topology space? I'm having difficulty in topology...How to verify or disprove the following~
Question: 
If A is a connected subset of topology space X,
Which of the following must be connected?
I. the interior of A
II. the closure of A
III. the complement of A  
I think (I) is not correct, (II) is correct, (III) don't know....
But I just don't know how to argue it in a formal way...
Thank for your time!!
 A: HINT: You’re correct about (I) and (II). (II) is a standard theorem: if $X$ is a space, and $A$ is a connected subset of $X$, then $\operatorname{cl}_XA$ is also connected. One way to prove it is to assume that $\operatorname{cl}_XA$ is not connected and get a disconnection of $A$. (I’m pretty sure that there are also proofs of it on this site, though it may not be easy to find one.)
For (III) try taking $X$ to be the real line and your connected set $A$ to be any bounded connected subset of $\Bbb R$.
For (I) consider a set in the plane that looks like two disks connected by a line segment.
A: For $(2)$ : 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 $A$. (U,V are open in X so, $U\cap A,V\cap A$ are open in $A$)
But, $A$ is connected.... So, either $U\cap A$ or $V\cap A$ is empty...
Suppose $U\cap A$ is empty.
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.
