Analogue of Urysohn metrization for Lawvere metric spaces?

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?

• Please add to your question the definition of a Lawvere metric. Are they close to quasimetrics? If yes, then the metrizability by a Lawvere metric is the same as quasiuniformizability by a quasiuniformity with a countable base. – Taras Banakh Sep 1 '18 at 6:19
• From what I can find online: a Lawvere metric on $X$ is a function $d: X \times X \to [0,+\infty]$ (where the latter set has the "obvious" order and addition function) that obeys the axioms $d(x,x) = 0$ for all $x \in X$ and $d(x,z) \le d(x,y) + d(y,z)$ for all $x,y,z \in X$. No symmetry or only $0$ values for equal points. Is that indeed the right notion? – Henno Brandsma Sep 1 '18 at 6:31
• The slick definition of a Lawvere metric space should be mentioned: Let $V$ be the symmetric monoidal poset $([0,\infty],\geq,+)$. A Lawvere metric space is precisely a category enriched in $V$. For some discussion and references, see here. It's maybe worth mentioning that there's a notion of approach space which interpolates between Lawvere metric spaces and topological spaces. Roughly, one assigns a distance from each point to each subset of the space. – Tim Campion Sep 2 '18 at 15:48

According to this SE-post, 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.