Is the topology generated by this weaker notion of a metric necessarily metrisable?

Question

The triangle inequality seems much stronger than necessary for a lot of analysis. So I will define a "loose metric" on a set $X$ to be a function $d \colon X \times X \to [0,\infty)$ with the following properties:

• $d(x,y)=d(y,x)\ $ for all $x,y \in X;$
• every $\, x,y \in X\ $ has $\ d(x,y)=0\ $ if and only if $\ x=y$;
• there exists a function $\rho \colon [0,\infty) \to [0,\infty]$, with $\rho(t) \to 0$ as $t \to 0$, such that for all $x,y,z \in X$, $$ | d(x,z) - d(y,z) | \,\leq\, \rho(\, d(x,y) \, )\text{.} $$

Is the topology generated by a loose metric necessarily metrisable?

Obviously, by "the topology generated by a loose metric $d$ on $X$", I mean the collection of all $d$-open subsets of $X$, where a set $U \subset X$ is called $d$-open if for all $x \in U$ there exists $\delta>0$ such that $\{y \in X : d(x,y) < \delta\} \subset U$. I believe it is easy to show that this topology is a regular Hausdorff topology, and so the Nagata-Smirnov metrisation theorem would give that it is sufficient to show that the topology has a $\sigma$-locally finite basis. However, I suspect that there will be a more direct way of constructing a classical metric on $X$ that is topologically equivalent to a given loose metric $d$.

Answer 1

For a loose metric $d$ as above, we can consider the function $$d_1(x,y):=\sup\{|d(x,z)-d(y,z)|;z\in X\}.$$

It is easy to verify that $d_1$ is a metric, and $d(x,y)\leq d_1(x,y)\leq\rho(d(x,y))$ for all $x,y$, so $d_1$ and $d$ generate the same topology.

Edit: As mentioned in the comments, we can let $d_2=\min(d_1,1)$ if we prefer metrics which don't take the value $\infty$.