# Does there exist a metrizable space of which do we not know its metric?

It seems that in topology, we have very good ways to deduce whether a space is metrizabe or not.

For example, if a space $(X, \mathcal{T})$ is metrizable, then it is Hausdorff, $T_3$, $T_4$, $\ldots$, $T_6$, first countable, separable $\Leftrightarrow$ second countable, and we can do even further by applying metrizability theorems like Urysohn's theorem.

Then by negation, if we are missing some of those properties, then it is not metrizable.

But how do people come up with the actual metric on those spaces? It seems to me as far a $\mathbb{R}$ is concerned, there is sort of this intuitive/common sense feeling what the metric should be. For example, the absolute value of distance between two points. And lo and behold, it just happens to satisfy the metric axioms.

So to me it is not very clear if this way will always work for weirder topological spaces that happens to be metrizable. Are there metrizable spaces for which we do not know what metric actually generates the topology?

• $\ldots$ is a lot more to type than $T_5$ – Aweygan Aug 5 '16 at 3:55
• Furthermore, there are plenty of metric (metrizable) spaces which are not separable. Look at $\ell^\infty$ for example. – Aweygan Aug 5 '16 at 4:06

## 1 Answer

There is no such thing as "the" metric. There are lots of equivalent metrics that will generate the same topology. How you might construct one might have a lot to do with how you know the space is metrizable.

• Given one metric can we in an easy way identify all the other metrics? – Jacob Wakem Aug 5 '16 at 4:24
• @BrianM.Scott Those are not all the metrics on a discrete space. $a \le d(x,y) \le 2a$ is sufficient, but not necessary. For example, $d(x,y) = |x-y|$ on $\mathbb N$ does not satisfy it. – Robert Israel Aug 5 '16 at 15:49
• You’re right, of course. – Brian M. Scott Aug 5 '16 at 15:53
• On a discrete space with $n$ points, the metrics correspond to solutions of a system of inequalities in $n(n-1)/2$ variables ($d_{ij}$ for each unordered pair of distinct points): $d_{ij} > 0$ and the triangle inequalities $d_{ik} \le d_{ij} + d_{jk}$. These solutions form a convex cone. I suspect we can identify the extreme rays of the closure of this cone. – Robert Israel Aug 5 '16 at 16:01
• Yes, it seems to me that the extreme rays are of the following form: for some nonempty proper subset $A$ of the points and $t > 0$, $d_{ij} = t$ if one of $i$ and $j$ is in $A$ and the other is not, otherwise $d_{ij} = 0$. The metrics are the points of the convex hull of these rays that have $d_{ij} > 0$ for all $i \ne j$. – Robert Israel Aug 5 '16 at 16:53