Urysohn proved that any regular, Hausdorff, second-countable space $X$ is metrizable, i.e. there exists a metric space whose underlying topological space is $X$. But what if we ask the same question for *Lawvere metric spaces*?


**Definition:** Let $(X,d)$ be a Lawvere metric space. For any $\epsilon>0$ and point $x\in X$, define the *$\epsilon$-ball emanating from $x$*, denoted $B(x,\epsilon)$, to be the set
$$B(x,\epsilon):=\{x'\in X\mid d(x,x')<\epsilon\}.$$
Define the *induced topology on $X$* to be the set of those subsets $U\subseteq X$ with the property that for all $x\in U$ there exists $\epsilon>0$ with $B(x,\epsilon)\subseteq X$.


**Example:** The Sierpinski space is not Hausdorff, so it is not metrizable. But it is Lawvere metrizable. Indeed, let $S=\{o,c\}$, let $d(o,c)=2$ and $d(c,o)=0$. Then taking $\epsilon=1$ we have $B(o,1)=\{o\}$, so the singleton set $\{o\}$ is open. But for all $\epsilon>0$, we have $B(c,\epsilon)=\{o,c\}$, so $\{c\}$ is not open.


**Question:** Do you know of a characterization of those topological spaces that are Lawvere metrizable?



Analogue of Urysohn metrization for Lawvere metric spaces?
According to [this SE-post](https://math.stackexchange.com/questions/1861611/on-the-separation-axiom-in-a-lawvere-or-generalized-metric-space), a *Lawvere metric* on a set $X$ is a function $d:X\times X\to[0,+\infty)$ satisfying two axioms:


1) $d(x,x)=0$ and


2) $d(x,z)\le d(x,y)+d(y,z)$


for all $x,y,z\in X$.


Then the following theorem can be considered as a counterpart of the Urysohn metrization theorem (I strongly suspect that this theorem was known to Lawvere).



> 
> **Theorem.** Each topological space $X$ with countable base is metrizable by a Lawvere metric.
> 
> 
> 


*Proof.* Fix a countable base $\{U\_n\}\_{n\in\omega}$ of the topology of $X$. For every $n\in\omega$ consider the Lawvere metric $f\_n:X\times X\to \{0,1\}$ defined by $$f(x,y)=\begin{cases}1&\mbox{if $x\in U\_n$ and $y\notin U\_n$};\\
0&\mbox{otherwise}.
\end{cases}
$$ 
The Lawvere metrics $f\_n$, $n\in\omega$ compose another Lawvere metric $$f=\max\_{n\in\omega}\frac1{2^n}f\_n$$
which generates the topology of $X$. 


Indeed, for any open set $U\subset X$ and any $x\in U$ we can find $n\in\omega$ with $x\in U\_n\subset U$ and conclude that $B(x,\frac1{2^n})\subset U\_n\subset U$.


On the other hand, let us show that a set $U\subset X$ is open if for any $x\in U$ there exists $\varepsilon>0$ with $B(x,\varepsilon)\subset U$. Let $\Omega\_x:=\{n\in\omega:x\in U\_n\}$. Choose $m\in\omega$ with $\frac1{2^m}<\varepsilon$ and consider the open neighborhood $V:=\bigcup\{U\_n:n\in\Omega\_x,\;n\le m\}$ of $x$.
Observe that for every $y\in V$ and every $n\le m$ we have $f\_n(x,y)=0$.
So, $d(x,y)<\frac1{2^m}<\varepsilon$ and $x\in V\subset B(x,\varepsilon)\subset U$, which means that $x$ is an interior point of $U$ and $U$ is open.


**Remark.** The proof essentially uses the Lawvere metrizability of the Sierpinski two-point space and the fact that each $T\_0$-space with countable base embeds into the countable product of the Sierpinski two-point spaces.


