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Here I call a topological space "normal" to mean only that any two disjoint closed sets can be separated by disjoint neighborhoods — with no additional assumption of being $T_1$ or $T_0$.

What are some examples of a normal, preferably normal $T_1$-, space $X$ having a quotient space that is not normal. (So failure of the quotient to be $T_1$ in itself does not constitute such a counterexample.)

I'm looking for some "concrete", even "geometric" examples of such $X$ and not the sort of abstruse, Axiom-of-Choice-using example such as what is constructed in https://math.stackexchange.com/a/1682677/32337.

murray
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    At the end of Eric Wofsey's answer that you link to, Eric points out that the construction can be easily modified to avoid the use of the axiom of choice (by using convergent filters containing the cofinite filter instead of ultrafilters). Whether it's "abstruse" is, I suppose, a matter of taste (the construction is definitely very abstract, but I agree with Eric that it is still "conceptually simple") - but it seems to me that anyone interested in general topology can benefit a lot from getting comfortable with ultrafilters. – Alex Kruckman Apr 19 '21 at 18:17

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Here’s a fairly trivial example. Let $X=[0,1]\times\{0,1\}$, where $[0,1]$ has its usual topology, and $\{0,1\}$ is discrete; $X$ is certainly $T_4$ (normal and $T_1$). Define an equivalence relation $\sim$ on $X$ by setting $\langle x,i\rangle\sim\langle y,j\rangle$ iff $x=y$ and either $i=j$, or $x=0$, and let $Y=X/\!\sim$. (In other words, we’re simply identifying $\langle x,0\rangle$ and $\langle x,1\rangle$ for each $x\in(0,1]$.) Then $Y$ is $T_1$, but $\{\langle 0,0\rangle\}$ and $\{\langle 0,1\rangle\}$ are disjoint closed sets that cannot be separated by disjoint open sets.

Added: And here is a Hausdorff example. Let $\Bbb P=\Bbb R\setminus\Bbb Q$, and let

$$X=\big(\Bbb Q\times(\Bbb R\setminus\{0\}\big)\cup\big(\Bbb P\times\{0\}\big)$$

with the usual topology $\tau_X$; clearly $X$ is metrizable and hence $T_4$. Let $Y=\Bbb R$ and

$$\pi:X\to Y:\langle x,y\rangle\mapsto x\,,$$

and let $\tau_Y$ be the topology on $Y$ that makes $\pi$ a quotient map:

$$\tau_Y=\left\{U\subseteq Y:\pi^{-1}[U]\in\tau_X\right\}\,.$$

Let $\tau$ be the usual topology on $\Bbb R$; clearly $\tau\subseteq\tau_Y$, so $\tau_Y$ is Hausdorff. Moreover, if $U\in\tau$, then

$$\pi^{-1}[U\cap\Bbb Q]=\big(U\times(\Bbb R\setminus\{0\})\big)\cap X\in\tau_X\,,$$

so $\{U\cap\Bbb Q:U\in\tau\}\subseteq\tau_Y$, and it’s not hard to see that in fact

$$\tau_Y=\tau\cup\{U\cap\Bbb Q:U\in\tau\}\,.$$

Clearly $\Bbb Q$ is open in $Y$, so $\Bbb P$ is closed in $Y$. Let $y\in\Bbb Q$, and let $U$ be any open nbhd of $y$ in $Y$; there are $u,v\in Y$ such that $y\in(u,v)\cap\Bbb Q\subseteq U$. Let $x\in(u,v)\cap\Bbb P$; then every open nbhd of $x$ in $Y$ meets $U$, so $\Bbb P\cap\operatorname{cl}_YU\ne\varnothing$, and $Y$ is not regular.

Brian M. Scott
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