The  concept of continuously symmetrizable Recently，I am reading a paper, in which there is a space called continuously symmetrizable. What's "continuously symmetrizable"? Could someone give me an explanation with examples?
Thanks ahead:)
 A: A space $X$ is symmetrizable iff there is a function $d:X\times X\to\Bbb R$ such that:


*

*$d(x,y)\ge 0$ for all $x,y\in X$;  

*$d(x,y)=0$ iff $x=y$;  

*$d(x,y)=d(y,x)$ for all $x,y\in X$; and  

*for each $V\subseteq X$, $V$ is open iff for each $x\in V$ there is an $\epsilon>0$ such that $B(x,\epsilon)\subseteq V$, where $B(x,\epsilon)=\{y\in X:d(x,y)<\epsilon\}$. Note that the $\epsilon$-balls $B(x,\epsilon)$ need not be open sets in $X$.


Such a function $d$ is called a symmetric on $X$. $X$ is continuously symmetrizable if the function $d$ is continuous as a function from $X\times X$ with the product topology to $\Bbb R$ with the usual Euclidean topology.
Added: As you can see, a symmetric satisfies all of the conditions imposed on a metric generating the topology of $X$ except the triangle inequality. In particular, every metrizable space is continuously symmetrizable: any metric generating the topology of $X$ is a continuous symmetric. Here’s an example of a symmetrizable space that fails badly to be metrizable: it isn’t even first countable.
Example: Let $X$ be the quotient space of $\Bbb R$ obtained by identifying $n$ with $1/n$ for each $n\in\Bbb Z^+$, and let $q:\Bbb R\to X$ be the quotient map. For $n\in\Bbb Z^+$ let $p_n=q(n)=q(1/n)\in X$. For each $n\in\Bbb Z^+$ the sets 
$$q\left[\left(\frac1n-\epsilon,\frac1n+\epsilon\right)\cup\left(n-\epsilon,n+\epsilon\right)\right]\tag{1}$$ with $\epsilon>0$ form a nbhd base at $p_n$. For $x\in \Bbb R\setminus\left(\{0\}\cup\Bbb Z^+\right)$, $q(x)$ has basic open nbhds of the form $q\left[(x-\epsilon,x+\epsilon)\right]$, where $\epsilon>0$ is small enough so that $(x-\epsilon,x+\epsilon)\cap\Bbb Z^+=\varnothing$. Finally, $p_0=q(0)$ has a local base consisting of sets of the form $$q\left[(-\epsilon,\epsilon)\cup\bigcup_{n>1/\epsilon}(n-\epsilon_n,n+\epsilon_n)\right]\;,$$ where $\epsilon>0$ and $\langle\epsilon_n:n>1/\epsilon\rangle$ is any sequence of positive real numbers. The fact that $\epsilon>0$ and $\langle\epsilon_n:n>1/\epsilon\rangle$ can be any sequence of positive real numbers means that $X$ is not first countable at $p_0$ and hence cannot be metrizable. However, the function $$d:X\times X\to\Bbb R:\langle x,y\rangle\mapsto\rho\left(q^{-1}[\{x\}],q^{-1}[\{y\}]\right)$$ is a symmetric on $X$, where $\rho$ is the usual Euclidean metric on $\Bbb R$.
Thus, if $x,y\in\Bbb R\setminus\Bbb Z^+$, $d\big(q(x),q(y)\big)=|x-y|$; for $m,n\in\Bbb Z^+$, $$d(p_m,p_n)=\rho\left(\left\{m,\frac1m\right\},\left\{n,\frac1n\right\}\right)=\left|\frac1m-\frac1n\right|\;;$$ and if $x\in\Bbb R\setminus\Bbb Z^+$ and $m\in\Bbb Z^+$, $$d\big(q(x),p_m\big)=\rho\left(\{x\},\left\{m,\frac1m\right\}\right)=\min\left\{|x-m|,\left|x-\frac1m\right|\right\}\;.$$ Clearly $d$ satisfies (1)-(3) of the definition of a symmetric on $X$. To check that it also satisfies (4), note that for $n\in\Bbb Z^+$, the sets $B(p_n,\epsilon)$ are the basic open nbhds of $p_n$ described by $(1)$, and that for $x\in\Bbb R\setminus\left(\{0\}\cup\Bbb Z^+\right)$, $B(x,\epsilon)=q\left[(x-\epsilon,x+\epsilon)\right]$ for sufficiently small $\epsilon$, while $B(p_0,\epsilon)=q\left[(-\epsilon,\epsilon)\right]$. The $d$-balls at points of $X\setminus\{p_0\}$ actually form local bases at those points. The sets $B(p_0,\epsilon)$ are not nbhds of $p_0$, but a set $V\subseteq X$ containing $p_0$ is open only if it contains $B(p_0,\epsilon)$ for some $\epsilon>0$, so (4) is satisfied as well.
The symmetric $d$ is not continuous, however. To see this, let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $(1,2)$ converging to $2$, and let $y=3/4$. Then $$\left\langle d\big(q(x_n),q(y)\big):n\in\Bbb N\right\rangle\to\frac54\;,$$ but $$d\big(q(2),q(y)\big)=d\big(p_2,q(y)\big)=\min\left\{\left|2-\frac34\right|,\left|\frac12-\frac34\right|\right\}=\frac14\;.$$
