Is there any example of a closed relation $\sim$ on a Hausdorff space $X$ such that $X/\sim$ is not Hausdorff?

Also, is there any example of a closed relation ~ on a Hausdorff space $X$ such that a quotient map $f:X→X/ \sim $ is not closed? Here a relation is called closed when $R = \{(x,y) \in X \times X : x \sim y \}$ is closed in $X \times X$ in the product topology.

While thinking about this problem, I thought about whether there are any non closed subset C of the set of real numbers such that C×C is closed. Will there be such a set?

  • 1
    $\begingroup$ What does it mean for $f$ to be Hausdorff? Also, the answer to your second question is no. $\endgroup$ – Forever Mozart Apr 16 '16 at 2:44
  • $\begingroup$ I meant X/~ the quotient space to be a Hausdorff. $\endgroup$ – Emily Apr 16 '16 at 2:50
  • $\begingroup$ So you just want $X / \sim$ to be non-Hausdorff? $\endgroup$ – Forever Mozart Apr 16 '16 at 2:51
  • $\begingroup$ Can you also define "closed relation" for me? $\endgroup$ – Forever Mozart Apr 16 '16 at 2:52
  • $\begingroup$ Yes I mean just X/~ to be non Hausdorff. I meant a relation ~ to be closed if {(x,y):x~y} is closed in X×X. $\endgroup$ – Emily Apr 16 '16 at 2:55

As to the first example:

Let $X$ be a Hausorff but non-normal space. Let $A, B$ be two closed sets that cannot be separated by disjoint open sets. Define $R$, the equivalence relation as a subset of $X \times X$ by $\Delta \cup (A \times A) \cup (B \times B)$, which is closed, where $\Delta = \{(x,x): x \in X\}$ is the diagonal. So we identify all points in $A$ to a new point $[A]$ and all points in $B$ to a new point $[B]$ in the closure. Let $q$ be the quotient map $X \rightarrow X/R$.

Then if $U \subseteq X/R$ and $V \subseteq X/R$ would be disjoint open neighbourhoods of $[A]$ and $[B]$, then $q^{-1}[A]$ and $q^{-1}[B]$ would be disjoint open sets in $X$, and this cannot be. So $X/R$ is not Hausdorff.

For $X$ we can take the Sorgenfrey plane, or Tychonoff's plank, or $\mathbb{R}^I$ where $I$ is uncountable, to name some concrete spaces that are all Tychonoff (so Hausdorff) but not normal.

I proved directly that if is $X/R$ is Hausdorff, then $R \subseteq X \times X$ is closed in this answer.

A slightly slicker argument: let $\Delta'$ be the diagonal of $(X/R) \times (X/R)$ which is closed as $X/R$ is Hausdorff, and note that $q \times q: X \times X \rightarrow (X/R) \times (X/R)$ is continuous and $R = (q \times q)^{-1}[\Delta']$ is thus closed.

So closedness of the relation is necessary but not sufficient for Hausdorff $X$.

As to the second question: suppose that $f: X \rightarrow Y$ is a quotient map onto the Hausdorff space $Y$. Define $R_f = \{(x,x') \in X^2 : f(x) = f(x')\}$, the equivalence relation induced by $f$. Standard facts on quotient spaces show that $X/R_f$ is homeomorphic to $Y$ using the map incuced by $f$. Let $q: X \rightarrow X/R_f$ be the quotient map then it's easy to see that $q^{-1}[q[A]] = f^{-1}[f[A]]$ for all $A \subseteq X$. It follows that $q$ is closed iff $f$ is closed. And as $R_f = (f \times f)^{-1}[\Delta_Y]$, where $\Delta_Y$ is the (closed as $Y$ is Hausdorff) diagonal of $Y$, $R_f$ is a closed relation.

So e.g. if $f$ is open, continuous (so quotient), but not closed, like the projection from $\mathbb{R}^2$ onto one of its factors, then $R_f$ is an example of a closed relation with Hausdorff (even metric) quotient, such that $q$ is not a closed map.

As to the final wondering: if $C \subset X$ and $x$ would be in $\overline{C}\setminus C$, $(x,x)$ would be in $\overline{C \times C}\setminus (C \times C)$. So a non-closed set has a non-closed square.

  • $\begingroup$ I think you mean the Niemitsky plane. +1. $\endgroup$ – DanielWainfleet Apr 16 '16 at 7:15
  • $\begingroup$ @user254665 No, the Sorgenfrey plane is the square of the Sorgenfrey line (reals in the lower limit topology). But the Niemitzky plane also works, indeed (IIRC this is also called the Bing plane by more US-centred mathemeticians). $\endgroup$ – Henno Brandsma Apr 16 '16 at 7:37
  • $\begingroup$ Thank you. I had not heard of it called by the name Sorgenfrey plane before. $\endgroup$ – DanielWainfleet Apr 16 '16 at 8:54

Q2: NO. If $C\subset R$ and $C^2$ is closed, take any $p\in \bar C.$ Consider any nbhd $W$ of $(p,p).$ We have $W\supset U\times V$ for some nbhds $U,V$ of $p.$ Let $T=U\cap V.$ Then $T$ is a nbhd of p, so $C\cap T\ne \phi.$ Let $p'\in C\cap T.$ Then $T^2\subset U\times V\subset W$, so $$(p',p')\in (C\cap T)^2\subset C^2\cap T^2\subset C^2\cap W.$$ So every nbhd $W$ of $(p,p)$ intersects the closed set $C^2,$ so $(p,p)\in C^2$, so $p\in C.$

  • $\begingroup$ Actually, my question was the opposite way. I was wondering if I can find a nonclosed $C \subseteq R$ such that $C×C$ is closed. $\endgroup$ – Emily Apr 16 '16 at 6:35
  • 1
    $\begingroup$ i re-wrote it completely. $\endgroup$ – DanielWainfleet Apr 16 '16 at 7:10

A third answer to the second question: let $X=\mathbb{R}$ and $x\sim y$ iff $x=y\lor (x>0\land y=1/x$). Then $R=\{(x,y)\in\mathbb{R}^2\!:x\sim y\}$ is the union of the diagonal and one branch of a hyperbola, which is clearly closed. The corresponding quotient map $f\colon\mathbb{R}\to\mathbb{R}/\mathord{\sim}$ maps a closed set $A=[1,\infty)$ to $f[A]=\{\{x,1/x\}\colon x>0\}$ which is not closed because $\bigcup f[A]=(0,\infty)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.