# proving it is Hausdorff

Question:

Let $X$ be a topological space and let $A ⊂ X$. Define an equivalence relation $∼$ on $X$ such that the equivalence classes are:

• $A$ itself, and, • all singletons $\{x\}$ such that $x \notin A$.

Then define $X/A$ to be the quotient space $X/{∼}$.

Prove that if $X$ is regular and $A$ is closed then $X/A$ is Hausdorff.

The quotient mapping $q(x)$ is $$q(x)=\begin{cases} x,&\text{if }x∉A\\ A,&\text{if }x∈A\;. \end{cases}$$
Pick two distinct points $x,y∈X/A$. If neither point is $A$, then $x,y∈$$X \backslash A. With the given information in the problem, particularly that A⊂X is closed, we may deduce that X \backslash A is open in X. Then there exist open neighborhoods U_A of x and V_A of y such that U_A and V_A are disjoint from the closed set A. Given that X is regular, we can deduce that X is Hausdorff. x and y are distinct and thus we can find open sets U' and V' disjoint from each other such that x∈U' and y∈V'. Now define U=U_A∩U′ and V=V_A∩V′. Then U and V are open, disjoint neighborhoods such that x∈U and y∈V, and U, V are also disjoint from A. If one of the points, say y, is A, then x, the other distinct point, is ∈ X \backslash A. Because X is regular, every non-empty closed subset of X,including A ⊂ X, and every point in X contained in a closed subset, admit non-overlapping open neighborhoods. Then the point x ∈ X \backslash A has an open neighborhood and the closed set A has a non-overlapping open neighborhood. If y is A, then x and y are contained in disjoint open neighborhoods. Therefore any two distinct points x,y∈X/A, whether or not they are A, have open neighborhoods U of x and V of y such that U and V are disjoint. X/A is Hausdorff. □ Is my proof even correct? Please help me make my proof more rigorous and accurate. I need everything to be absolutely clear and rigorous. Thank you. • In the case where x, y \in X - A, you said "we may deduce that X - A is open in X". Where did you use this afterward? – layman Dec 18 '14 at 1:49 • Oh, I see. Here is what I think: the U and V that you just mentioned in the comment are disjoint from A, but they are not necessarily disjoint from each other. Let's actually denote these U_{A}, V_{A}, since they are disjoint from A. Now, since x and y are distinct, and X is regular (which means it is Hausdorff, as you said), then we know we can find open U', V' disjoint from each other such that x \in U', y \in V'. But these U' and V' are not necessarily disjoint from A. But if U = U_{A} \cap U' and V = V_{A} \cap V', then U and V are open, disjoint – layman Dec 18 '14 at 1:56 • (continued) from each other, and x \in U and y \in V, and U, V are also disjoint from A. Does this make sense? – layman Dec 18 '14 at 1:57 • Please stop deleting your questions; you've deleted the content of your questions multiple times. This is harming the site, and is rendering useless the answers that people have donated their time writing. – Milo Brandt Jan 5 '15 at 3:01 • Please, stop deleting, editing to new questions and or vandalizing posts. Regards, – Pedro Tamaroff Jan 5 '15 at 3:55 ## 1 Answer Pick two distinct points x,y∈X/A. If neither point is A, then x,y∈$$X$ $\backslash$ $A$. With the given information in the problem, particularly that $A⊂X$ is closed, we may deduce that $X$ $\backslash$ $A$ is open in $X$. Then there exist open neighborhoods $U_A$ of $x$ and $V_A$ of $y$ such that $U_A$ and $V_A$ are disjoint from the closed set A. Given that $X$ is regular, we can deduce that $X$ is Hausdorff. $x$ and $y$ are distinct and thus we can find open sets $U'$ and $V'$ disjoint from each other such that $x∈U'$ and $y∈V'$. Now define $U=U_A∩U′$ and $V=V_A∩V′$. Then $U$ and $V$ are open, disjoint neighborhoods such that $x∈U$ and $y∈V$, and $U$, $V$ are also disjoint from $A$.
So you take disjoint open neighborhoods $U'\ni x$ and $V'\ni y$, and you want them to also be disjoint from $A$. (Why is that sufficient in order to have disjoint neighborhoods of $\{x\}$ and $\{y\}$ in $X/A$? Here you could explain a bit more.) What you do is correct, but one could shorten it a bit by simply intersecting $U'$ and $V'$ with the open set $X/A$.
If one of the points, say $y$, is $A$, then $x$, the other distinct point, is $∈$ $X$ $\backslash$ $A$. Because $X$ is regular, every non-empty closed subset of $X$, including $A ⊂ X$, and every point in $X$ contained in a closed subset, admit non-overlapping open neighborhoods. Then the point $x$ $∈$ $X$ $\backslash$ $A$ has an open neighborhood and the closed set $A$ has a non-overlapping open neighborhood. If $y$ is $A$, then $x$ and $y$ are contained in disjoint open neighborhoods.
What do you mean by "every point in $X$ contained in a closed subset"? The disjoint neighborhoods exist for every closed set $A$ and every point outside of $A$. Again, why does this show that there are disjoint open neighborhoods around $\{x\}$ and $A$ in $X/A$?
One more thing: It is bad style to write something like "then $x$, the other point, is $\in X/A$", as one should not mix text and symbols. You could write "then $x$, the other point, is in $X/A$" or "then $x\in X/A$"
Therefore any two distinct points $x,y∈X/A$, whether or not they are $A$, have open neighborhoods $U$ of $x$ and $V$ of $y$ such that $U$ and $V$ are disjoint. $X/A$ is Hausdorff. $□$