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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.

My Answer:

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.

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  • $\begingroup$ 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? $\endgroup$ – layman Dec 18 '14 at 1:49
  • $\begingroup$ 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 $\endgroup$ – layman Dec 18 '14 at 1:56
  • $\begingroup$ (continued) from each other, and $x \in U$ and $y \in V$, and $U, V$ are also disjoint from $A$. Does this make sense? $\endgroup$ – layman Dec 18 '14 at 1:57
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    $\begingroup$ 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. $\endgroup$ – Milo Brandt Jan 5 '15 at 3:01
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    $\begingroup$ Please, stop deleting, editing to new questions and or vandalizing posts. Regards, $\endgroup$ – Pedro Tamaroff Jan 5 '15 at 3:55
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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. $□$

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