# An example of open closed continuous image of $T_2$-space that is not $T_2$

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.

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

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

and

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

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, then every point of $$X\setminus X_0$$ is a limit point of $$F\cap X_0$$, so $$F\supseteq X_{-1}\cup X_1$$, and $$Y\setminus f[F]\subseteq f[X_0]$$; thus, $$Y\setminus f[F]$$ is open, and $$f[F]$$ is closed. This shows that $$f$$ is a closed map.

• 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. – Zed Tuller Aug 29 '15 at 15:16
• @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. – Brian M. Scott Aug 29 '15 at 17:29