Engelking in his "General Topology" states that $T_2$ separation axiom is not preserved under open closed continuous surjections. In "General Topology" by Stephen Willard I have found two separate examples showing that the continuous closed and continuous open images of a Hausdorff space need not be Hausdorff.

I would be nice to have an example of open closed continuous image of $T_2$-space that is not $T_2$. Or, equivalently, an example of quotient space of $T_2$-space by open closed equivalence relation that is not $T_2$.

Since $T_3$ separation axiom is preserved under open closed continuous surjections, and $T_1$ separation axiom is preserved under closed surjections, the task is to build such irregular Hausdorff space and open closed equivalence relation on it that the corresponding quotient space is non-Hausdorff $T_1$-space.


1 Answer 1


This is a slight modification of an example due to K. Alster; it is Example $\mathbf{3.2}$ of J. Chaber, Remarks on open-closed mappings, Fundamenta Mathematicae ($1972$), Vol. $74$, Nr. $3$, $197$-$208$.

Let $X=\Bbb R\times\{-1,0,1\}$, and for convenience set $X_i=\Bbb R\times\{i\}$ for $i\in\{-1,0,1\}$. Points of $X_0$ are isolated. For $p=\langle x,-1\rangle\in X_{-1}$, $q=\langle x,1\rangle\in X_1$, $\epsilon>0$, and countable $C\subseteq\Bbb R$ let

$$B(p,\epsilon,C)=\{p\}\cup\Big(\big((x-\epsilon,x)\setminus C\big)\times\{0\}\Big)$$


$$B(q,\epsilon,C)=\{q\}\cup\Big(\big((x,x+\epsilon)\setminus C\big)\times\{0\}\Big)\;;$$

the sets $B(p,\epsilon,C)$ and $B(q,\epsilon,C)$ for $\epsilon>0$ and countable $C\subseteq\Bbb R$ are local bases at $p$ and $q$. $X$ with this topology is Hausdorff.

Now define an equivalence relation $\sim$ on $X$ by $\langle x,i\rangle\sim\langle y,j\rangle$ iff $i=j$, and either $i\ne 0$, or $x-y\in\Bbb Q$. Let $Y=X/\!\!\sim$, and let $f:X\to Y$ be the quotient map; $f$ is of course continuous.

Let $y_{-1}$ and $y_1$ be the points of Y corresponding to $X_{-1}$ and $X_1$, respectively. Suppose that $U$ is an open nbhd of $y_i$ in $Y$ for $i\in\{-1,1\}$. Clearly $X_i\subseteq f^{-1}[U]$, so for each $p\in\Bbb R$ there are $\epsilon_p>0$ and countable $C_p\subseteq\Bbb R$ such that

$$B(\langle p,i\rangle,\epsilon_p,C_p)\subseteq f^{-1}[U]\;.$$

Let $A=\{p\in\Bbb R:\langle p,0\rangle\notin f^{-1}[U]\}$, and suppose that $A$ is uncountable; then there is $p\in A$ such that $(p-\epsilon_p,p)\cap A$ and $(p,p+\epsilon_p)\cap A$ are uncountable. But then $C_p$ must be uncountable, which is impossible. Thus, $A$ is countable, and it follows that $X_0\setminus f^{-1}[U]$ is countable. Clearly, then, $y_{-1}$ and $y_1$ do not have disjoint open nbhds in $Y$, which is therefore not Hausdorff. Indeed, since the $\sim$-equivalence classes of $X_0$ are countable, this shows that open nbhds of $y_{-1}$ and $y_1$ are co-countable in $Y$.

Conversely, if $U$ is a co-countable subset of $Y$ containing $y_i$ for some $i\in\{-1,1\}$, then $X_0\setminus f^{-1}[U]$ is countable, so $f^{-1}[U]$ and therefore $U$ are open. Thus, the open nbhds of $y_i$ are precisely the co-countable subsets of $Y$ containing $y_i$, and it follows that $f[B(\langle p,i\rangle,\epsilon,C)]$ is open in $Y$ for each $\langle p,i\rangle\in X\setminus X_0$, $\epsilon>0$, and countable $C\subseteq\Bbb R$. It’s clear that $f[X_0]$ is a discrete open subset of $Y$, so $f$ is an open map.

Now suppose that $F\subseteq X$ is closed. If $F\cap X_0$ is countable, then $f[F]$ is countable and hence closed in $Y$. If $F\cap X_0$ is uncountable, let $A=\{x\in\Bbb R:\langle x,0\rangle\in F\}$. Then there is an $x_0\in A$ such that $(u,x_0)\cap A$ and $(x_0,v)\cap A$ are uncountable whenever $u<x_0<v$, and therefore $\langle x_0,-1\rangle,\langle x_0,1\rangle\in\operatorname{cl}(F\cap X_0)\subseteq F$. Thus, $y_{-1},y_1\in f[F]$, so $Y\setminus f[F]\subseteq f[X_0]$; thus, $Y\setminus f[F]$ is open, and $f[F]$ is closed. This shows that $f$ is a closed map.

  • $\begingroup$ many thanks) It seems $W$ should be $B$, $U$ and $A$ should be $U_i$ and $A_i$ respectively. Besides, I can't understand the last paragraph. Since the point $(0,0)$ is open in $X$, its complement $F=X\backslash\{(0,0)\}$ is closed in $X$, but $F\cap X_0$ is uncountable. $\endgroup$
    – Zed Tuller
    Aug 29, 2015 at 15:16
  • $\begingroup$ @Zed: You’re right about $W$, and the $U_i$ should have been $U$; I changed the writeup at one point and evidently missed a few spots. The last paragraph was a case of my mind getting ahead of my fingers. Thanks for catching these; they should be fixed now. $\endgroup$ Aug 29, 2015 at 17:29
  • $\begingroup$ @BrianM.Scott Hi Brian, I think there may still be a problem in the last paragraph to show $f$ is a closed map, in the case of $F\cap X_0$ uncountable. "every point of $X\setminus X_0$ is a limit point of $F\cap X_0$": that's not necessarily true. For example if $F=[0,1]\times\{0\}$, points of $X_{-1}$ outside of $[0,1]\times\{-1\}$ will not be limit points of $F$. How about something like the following instead: there is an $x\in\mathbb{R}$ such that $\langle x,0\rangle$ is a double sided condensation point of $F\cap X_0$, then $p=\langle x,-1\rangle$ is a limit point of $F$, ... (ct'd) $\endgroup$
    – PatrickR
    Oct 22, 2021 at 2:28
  • $\begingroup$ so $p\in F$ and $f[F]$ contains $f(p)=y_{-1}$, and similarly for $y_1$, hence the complement of $f[F]$ is in $f[X_0]$, hence open, etc. $\endgroup$
    – PatrickR
    Oct 22, 2021 at 2:31
  • $\begingroup$ @PatrickR: You’re right. I may have had something else in mind at the time and simply wrote it up incorrectly, but if so, I’ve no idea what it was, and your approach certainly works just fine, so I’ve adopted it; thanks! (I also fixed a typo that no one caught.) $\endgroup$ Oct 22, 2021 at 22:40

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