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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.

Zed Tuller
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1 Answers1

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This is a very slight modification of 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

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

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 $i\in\{-1,1\}$, and $U$ is an open nbhd of $y_i$ in $Y$. 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.

Brian M. Scott
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  • 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
  • @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) – PatrickR Oct 22 '21 at 02:28
  • 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. – PatrickR Oct 22 '21 at 02:31
  • @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.) – Brian M. Scott Oct 22 '21 at 22:40
  • I don't understand the argument profered here that the quotient space is not $T_2$. What is the purpose of starting with an open neighborhood $U$ of both the points $y_{-1}$ and $y_1$. Should one not want to start with two neighborhoods $U_{-1}$ and $U_1$ of $y_{-1}$ and $y_1$, respectively, and show they must intersect (by going back to $X$). – murray Dec 09 '22 at 23:42
  • @murray: We don’t start with an open nbhd of both points: the argument shows that if $U$ is an open nbhd of either $y_{-1}$ or $y_1$, then $X_0\setminus f^{-1}[U]$ is countable. Thus, if $U$ and $V$ are disjoint open nbhds of $y_{-1}$ and $y_1$, respectively, then the countable set $(X_0\setminus f^{-1}[U])\cup(X_0\setminus f^{-1}[V])$ contains the uncountable set $X_0$, which is absurd. We conclude that $y_{-1}$ and $y_1$ do not have disjoint open nbhds. – Brian M. Scott Dec 10 '22 at 03:16
  • @BrianM.Scott: Of course that's what should have been said! But "$U$ is an open nbhd of $y_i$ for $i \in {-1, 1}$" surely means "$U$ is an open nbhd of $y_i$ for each $i \in {-1, 1}$." I suggest a minor rephrasing. – murray Dec 10 '22 at 16:15
  • @murray: It could certainly have been worded better, and I’ll change it after I post this comment, but the intended meaning really should be pretty clear by the time one gets to the end of the paragraph. – Brian M. Scott Dec 10 '22 at 19:21
  • @BrianM.Scott: I don't see why $f$ is open. Take any $q = \langle x, 1\rangle \in X_1$ and $\epsilon > 0$. Let $U = B(q, \epsilon, \emptyset)$, an open neighborhood of $q$ in $X$. If $f(U)$ is open in $Y$, then the saturation $S$ of $U$ under $\sim$ must be open in $X$. Let $y \in \mathbb{R}$ with $x - y$ irrational. If $S$ is open in $X$, it would also contain an open neighborhood $W = B(r, \delta, C)$ of the point $r = \langle x, 1\rangle \in S$. Yet such $W$ cannot be contained in the saturation $S$, since it contains irrational translates of points in $U$. – murray Feb 04 '23 at 02:27
  • @murray: Since $(x,x+\epsilon)\times{0}$ contains a representative of every $\sim$-class in $X_0$, $f[B(q,\epsilon,\varnothing)]=Y\setminus{y_{-1}}$, which is open in $Y$. – Brian M. Scott Feb 04 '23 at 03:13
  • @BrianM.Scott: And $Y \setminus {y_{-1}}$ is open in $Y$ because $f^{-1}(Y \setminus {y_{-1}}) = X_0 \cup X_1$ is open in $X$. – murray Feb 04 '23 at 21:11
  • @murray: And in fact $Y$ is $T_1$. It’s actually homeomorphic to the space obtained by topologizing $[0,1]$ so that points of $(0,1)$ are isolated, and for $x\in{0,1}$ the basic open nbhds of $x$ are the sets ${x}\cup\big((0,1)\setminus C\big)$ such that $C\subseteq(0,1)$ is countable. – Brian M. Scott Feb 04 '23 at 21:49
  • @BrianM.Scott: In the case $F \cap x_0$ uncountable, when proving that the quotient map is closed, why must such a point $x_0 \in A$ exist? – murray Apr 15 '23 at 00:51
  • @murray: It exists by the result, proved at the link two paragraphs earlier in the answer, that every uncountable subset of $\Bbb R$ contains such a point. – Brian M. Scott Apr 15 '23 at 03:41
  • @BrianM.Scott: Pardon my tunnel vision. – murray Apr 16 '23 at 00:52
  • Can you cite a source for Alster's original version of this example? If it's buried in one of his papers about products with Lindelöf spaces, I haven't been able to find it there. – murray Jun 18 '23 at 19:37
  • @murray: I’m afraid not. At this point, in fact, I’m not at all sure where I got the idea that it was a modification of an example by Alster, and I’m simply going to remove that first assertion. – Brian M. Scott Jun 19 '23 at 17:42
  • @BrianM.Scott: Chaber's example 3.2 is slightly different in that $X_i = \mathbb{Q} \times {i}$ for $i = -1, 1$. – murray Jun 21 '23 at 14:22
  • @BrianM.Scott: In the last paragaraph, for the proof that [] is closed when ∩0 is uncountable: Why does ∖[]⊂[0] imply that ∖[] is open? Is it simply that then ∖[] is the image of a subset of 0, that subset is open in because the topology on 0 is discrete, and so []=∖[] is open in because, as already established, is an open map? – murray Jun 21 '23 at 14:23
  • @murray: Yes, it’s a slight modification; I meant to mention that when I removed the Alster non-reference and forgot to do so. Fixed now. \ Yes. Or more directly, because the points of $f[X_0]$ are isolated in $Y$. – Brian M. Scott Jun 21 '23 at 19:32