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?