Motivation: Consider for example a metric space that is a disjoint union of a point with $\mathbb R$ (with the usual metric on $\mathbb R$). Intuitively, it feels like there is some space "missing" from it. Topologically we can fix that by embedding it into $\mathbb R^n$ for the smallest $n$ possible: $2$. Intuitively it feels like $\mathbb R^2$ is some sort of "empty space completion" of the original space, in the smallest possible dimention. One natural question to ask is whether we can extend this topological embedding into an isometric embedding, by choosing some suitable metric in $\mathbb R^2$. I have no idea how to even begin attacking this question.
This can clearly be generalized much more. It would be natural to ask:
Question: If $M$ is a metric space and $N$ a metrizable space such that $M$ topologically embeds into $N$, can we choose a metric in $N$ that generates the topology of $N$ such that $M$ isometrically embeds into $N$ (perhaps even with the same embedding as the given one?)?
Is the answer to this question known, and/or what are some references studying questions similar to this one?
Best Answer
First, a counter-example. Take ${\mathbb R}^n$ with the standard metric and take a topological embedding $f: {\mathbb R}^n\to S^n$, say, given by the inverse stereographic projection. Then this map (or any other embedding) cannot extend to an isometric embedding to $S^n$ with respect to any metric on $S^n$, since $S^n$ is compact and ${\mathbb R}^n$ is unbounded.
On the other hand, Hausdorff proved in
Hausdorff, F., Erweiterung einer Homöomorphie., Fundam. Math. 16, 353-360 (1930). ZBL56.0508.03.
that if $N$ is any metrizable space and $M\subset N$ is any closed subset equipped with a compatible metric $d$, then $d$ can be extended to a metric on $N$. I found this reference in this answer by Alex Ravsky. (His answer should have been accepted long time ago, but, alas...)
A bit more modern reference is
Arens, Richard, Extension of functions on fully normal spaces, Pac. J. Math. 2, 11-22 (1952). ZBL0046.11801.
who works with pseudometrics. I suspect that in the metric case his extension is a metric, but one would have to check.