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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.

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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.

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  • $\begingroup$ 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 '15 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. $\endgroup$ – Brian M. Scott Aug 29 '15 at 17:29

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