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In user642796's answer at Prove that $Y$ is Hausdorff iff $X$ is Hausdorff and $A$ is a closed subset of $X$, he writes "If $A \subseteq X$ is not closed, let $x \in \overline{A}\setminus A$. Show that the points $p(x,0),p(x,1)$ (which are distinct) cannot be separated in $Y$." But I do not see how to show that part - any assistance would be appreciated.

For context: the original problem statement. Let $X$ be a topological space and $A$ a subset of $X$. On $X\times\{0,1\}$ define the partition composed of the pairs $\{(a,0),(a,1)\}$ for $a\in A$, and of the singletons $\{(x,i)\}$ if $x\in X\setminus A$ and $i \in \{0,1\}$. Let $R$ be the equivalence relation defined by this partition, let $Y$ be the quotient space $[X \times \{0,1\}]/R$, and let $p:X \times \{0,1\} \to Y$ be the quotient map. Prove that $Y$ is Hausdorff if and only if $X$ is Hausdorff and $A$ a closed subset of $X$.

Prasiortle
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1 Answers1

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Suppose that $U_0$ is an open nbhd of $p(\langle x,0\rangle)$ in $Y$. Then by definition $p^{-1}[U_0]$ is an open nbhd of $\langle x,0\rangle$ in $X\times\{0,1\}$, so there is an open nbhd $V_0$ of $x$ in $X$ such that $V_0\times\{0\}\subseteq p^{-1}[U_0]$. Similarly, if $U_1$ is an open nbhd of $p(\langle x,1\rangle)$ in $Y$, there must be an open nbhd $V_1$ of $x$ in $X$ such that $V_1\times\{1\}\subseteq p^{-1}[U_1]$.

Now $x\in\operatorname{cl}_XA$, and $V_0\cap V_1$ is an open nbhd of $x$ in $X$, so there is a point $a\in V_0\cap V_1\cap A$. Then $\langle a,i\rangle\in V_i\times\{i\}\subseteq p^{-1}[U_i]$ for $i\in\{0,1\}$, so $p(\langle a,0\rangle)\in U_0$, and $p(\langle a,1\rangle)\in U_1$. But $a\in A$, so $\langle a,0\rangle\mathrel{R}\langle a,1\rangle$, and therefore $p(\langle a,0\rangle)=p(\langle a,1\rangle)=\{\langle a,0\rangle,\langle a,1\rangle\}\in Y$. In other words,

$$\{\langle a,0\rangle,\langle a,1\rangle\}\in U_0\cap U_1\,,$$

and $U_0$ and $U_1$ are not disjoint. Since they were arbitrary open nbhds of $p(\langle x,0\rangle)$ and $p(\langle x,1\rangle)$, respectively, those two points cannot be separated by disjoint open sets in $Y$, and $Y$ is not Hausdorff.

Brian M. Scott
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