Special Type of Locally Metrizable Space

Question

We say a topological space is locally metrizable if for every $x\in X,$ there is an open set $U$ containing $x$ which is metrizable. That is, the subspace topology on $U$ is the topology induced by some metric $d_U.$ In my research, I've come across a space which is locally metrizable such that for some $d:X^2\rightarrow[0,\infty],$ all such metrizable neighborhoods $U$, we have $d_U$ is a metric equivalent to $d$ restricted to $U^2.$ Intuitively, the space only fails to be a metric space insofar as there exist $x,y\in X$ such that $d(x,y)=\infty.$ This notion seems much stronger as it states that each pair of local metrics coincide (up to equivalence) everywhere they are defined. Is there a name for such a space, or a related condition? Are there any obvious corollaries? (ideally that $X$ is metrizable haha)

Answer 1

Another perspective on this. Suppose your $d:X^2\to[0,\infty]$ has the usual metric properties, where for the purpose of the triangle inequality we have $\infty+x=\infty$ for all $x\in[0,\infty]$, and you require that your topology is generated by the sets $B(x,r)=\{y:d(x,y)<r\}$ defined by this extended metric.

Then let $d':X^2\to[0,1]$ be defined by $d'(x,y)=\lim_{z\to d(x,y)}\frac{z}{z+1}$. It can be shown that $d'$ satisfies the properties of a metric, and generates the same topology.

Therefore, it's okay to define an infinite distance between points to prove a space is metrizable, provided the metric requirements hold globally. But if you can only show your function locally satisfies the metric properties, then your locally metrizable space need not have any strong separation properties, as MW's answer shows.

Answer 2

Intuitively, the space only fails to be a metric space insofar as there exist $x,y\in X$ such that $d(x,y)=\infty$.

Unfortunately this intuition isn't quite right. Consider the line with two origins $X=\mathbb R\cup \{0'\}$, where $0'$ is an extra origin whose open neighborhoods are those of the form $\{0'\}\cup U\backslash \{0\}$ for some open $U\subseteq \mathbb R$.

Note that this space is locally metrizable (even locally Euclidean), and the function $d\colon X^2\to \mathbb [0,\infty)$ given by

$$ d(p,q)= \begin{cases} |p-q| & p,q\in\mathbb R\\ |p| & p\in \mathbb R, q=0'\\ |q| & q\in \mathbb R, p=0'\\ 0& p=q=0'. \end{cases} $$

will restrict to a metric coinciding with the topology on any metrizable open subset of $X$ (note that the metrizable open subsets of $X$ are precisely those which contain at most one of $0$ and $0'$).

Yet $X$ is (in some sense) not a particularly well-behaved space - it is very far from Hausdorff, as it does not even have unique sequential limits. And this holds even though $d$ is always finite, and you can even verify easily that it satisfies the triangle inequality (it is a pseudo-metric, in fact, despite the fact that $X$ is not pseudo-metrizable).

So while I cannot say for certain, it is unlikely you will find a name for this condition, as it is does not entail very strong separation properties, nor even pseudo-metrizability.

Remark.

If we require that $d$ is an actual metric, (or even an extended metric, allowed values of $\infty$ but still required to be nondegenerate, so that $d(x,y)=0$ if and only if $x=y$), then the topology on $X$ is at least as fine as the topology from $d$. That is, we must at least have that $X$ is submetrizable. In particular, $X$ must be Hausdorff in this case.

Steven Clontz suggests in the comments that in this case $X$ is likely metrizable, and I suspect similarly, though I don't see the proof immediately. I will update if I think of a proof or counter-example to this.

Update.

Even when $d$ is a metric, $X$ need not be metrizable, and the counter-example was staring us in the face in the "Related" links section to the side. This answer (and other posts on this site) discusses the $K$-topology on $\mathbb R$, where the open subsets are precisely those of the form $U\backslash A$, where $U\subseteq \mathbb R$ is Euclidean open, and $A\subseteq K$, where $K=\{\frac{1}{n}\mid n\in\mathbb N\}$.

In this topology, $\mathbb R\backslash\{0\}$ and $\mathbb R\backslash K$ are both open, and have the same topology as induced by the Euclidean metric, whereby the space is locally metrizable, and the Euclidean metric serves as the function $d$ in the given question. Yet the $K$-topology is not regular (see the above link), hence not metrizable.