# Not Quite Metrization

Let's say I have a space $X$ with a function $d\colon X \times X \to \mathbb R$ that has the following 2 properties:

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

$d(x,x) = 0$ for all $x$, $y \in X$.

The $d$-ball of radius $r$ centered at $x$ is defined as $B_r(x) = \{y \in X : d(x,y)< r\}$. We'll say the topological space $(X,T)$ is $d$-metrizable if there is a function $d$ such that $U$ is in $T$ if and only if for all $x$ in $U$ there exists a $d$-ball centered at $x$ which is contained in $U$.

I want to show by example that a space can be "$d$-metrizable", meaning it has the properties above, and separable but not second-countable.

Note: I realize that a regular metric space which is separable is second-countable. This one, however, is not and I need an example to show that.

I also need to show that a spaces which are d-metrizable are sequential. It's a long question, I realize. Thank you for your help.

Define

$$d:\Bbb R\times\Bbb R\to\Bbb R:\langle x,y\rangle\mapsto\begin{cases} x-y,&\text{if }x\ge y\\ 1,&\text{if }x<y\;. \end{cases}$$

For $\epsilon\le 1$ we have

$$\{y\in\Bbb R:d(x,y)<\epsilon\}=[x,x+\epsilon)\;,$$

so this generates the topology of the Sorgenfrey line, also known as the lower-limit topology. This space is separable, since the rationals are dense, but it’s not second countable. (By the way, $d$ even satisfies the triangle inequality.)

• So my "example space" which is separable and "d-metrizable" is R with the lower limit topology? – user191141 Nov 10 '14 at 19:10
• @Jimmy: It certainly works. I’m sure that there are many others, but I knew that one, and it has the virtue of being a familiar space. – Brian M. Scott Nov 10 '14 at 19:12
• I see, so it's d-metrizable because we constructed it from the d-metric above, separable because Q is dense in the Sorgenfrey line, but not second countable because R with the lower limit topology has no countable basis. How can I show that all spaces which are "d-metrizable" are sequential? – user191141 Nov 10 '14 at 19:21
• Thank you for this by the way, that really helped – user191141 Nov 10 '14 at 19:21
• @Jimmy: You’re welcome. If such a space isn’t sequential, there’s a sequentially closed set $F$ that isn’t closed. Let $x\in(\operatorname{cl}F)\setminus F$, and use the $d$-balls to construct a sequence in $F$ that converges to $x$, thereby getting a contradiction. (You may need a little more than your hypotheses currently give you: you may need the $d$-balls to be open, and I’m not sure that you can prove that just from what you have.) – Brian M. Scott Nov 10 '14 at 19:29