Examples of a quotient map not closed and quotient space not Hausdorff

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?

Solution 1:

As to the first example:

Let $X$ be a Hausdorff 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.

Solution 2:

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 $p\in C.$

Solution 3:

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 $\bigcup f[A]=(0,\infty)$.