a problem about quotient spaces Suppose $X$ is a topological space and $A\subset X$ is closed. Consider an equivalence relation on $X$ such that its classes are $A$ and $\{x\}$ for $x\in {X-A}$. Now we denote the induced quotient space by $X/A$. Prove that $X/A$ is hausdorff if $X$ is regular and it is normal if $X$ is normal.
I think we can say every pair of elements in $X/A$ could be separated by disjoint nbhds if both of them belongs to ${X-A}$ , because $X$ is regular and if one of them was $A$ then the statement is again true because $A\subset X$ is closed & $X$ is regular.
Am I right?
Is it really as easy as I said?
What can be done for the next part of the statement?
Hints are welcomed.
Thanks
 A: It is indeed that easy, but you have to be careful that your open neighborhoods are really open in $X/A$. So if $U$ and $V$ are disjoint open neighborhoods around points in $X-A$, we can assume that both are subsets of $X-A$ since this set is open. Then $q[U]$ and $q[V]$ have preimages $U$ and $V$, respectively, hence they are open as $U$ and $V$ are open.
The second part is similar: If two closed disjoint sets in $X/A$ are represented by closed sets in $X-A$, the same reasoning as for points in $X-A$ gives the disjoint neighborhoods. If one of these sets contains $A$, then its preimage has $A$ as a subset, but they are still closed and disjoint, hence we can find disjoint open sets and these are open preimages.
Another approach is more general: We can show that the quotient map $q:X\to X/A$ is a closed (open) map if $A$ is closed (open). Now every closed continuous surjection preserves normality. Here normal means disjoint closed sets have disjoint neighborhoods, not necessarily Hausdorff. I'm used to the term $T_4$ for a space that is normal and Hausdorff.
