Let $X$ be a regular Hausdorff space and $C$ a closed subset of $X$. Show that $X/C$ is Hausdorff.
Since $X$ is regular for every $F$ closed in $X$ and $x \in X \setminus F$ there exists neighborhoods $U_F$ and $U_x$ such that $U_F \cap U_x = \emptyset$.
Now if $X/C$ is not Hausdorff, then for $C$ and any $[x]$ in $X/C$ we have that $V_C \cap V_{[x]} \ne \emptyset$ for all neighborhoods $V_C$ and $V_{[x]}$.
Now considering the projection $p :X \to X/C$ we have that $$\emptyset \ne p^{-1}(V_C \cap V_{[x]}) = p^{-1}(V_C) \cap p^{-1}(V_{[x]})$$ and since $p^{-1}(V_C)$ is an open neighborhood of $A$ and $p^{-1}(V_{[x]})$ an open neighborhood of $x \in X \setminus C$ we have a contradiction with regularity as this is true for any $V_C$ and $V_{[x]}$.
Is my claim correct? I have the feeling that there could be some open sets that satisfy the regularity condition which aren't preimages of open sets under the projection?
Best Answer
$X$ is a regular Hausdorff topological space and $C\subset X$ is a closed subset. We want to show that $X/C$ is Hausdorff.
Let $[x_1]$, $[x_2]\in X/C$ and $[x_1]\neq[x_2]$. We have two cases.
(i) $x_1,x_2\not\in C$.
There are disjoint open nbd's $U_1,U_2$ of $x_1,x_2$ respectively, because $X$ is Hausdorff. Since $C$ is closed, $p(U_1\cap C')$ and $p(U_2\cap C')$ are disjoint open nbd's of $[x_1]$ and $[x_2]$ respectively.
(ii) $x_1$ or $x_2\in C$ but not both. Say, $x_2\in C$ and $x_1\not\in C$.
Since $X$ is regular, there are disjoint open nbd's $U_1,U_2$ of $x_1$ and $C$ respectively. Then, $p(U_1)$ and $p(U_2)$ are disjoint open nbd's of $[x_1]$ and $[x_2]$ respectively.
The greatest topologist, my supervisor, RIP http://www.turkmath.org/beta/haber.php?id_haber=184 never liked my answers. I hope you do.