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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?

Carla_
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  • In your example you neither explicited your metric on $\Bbb R\sqcup{\bullet}$ nor your embedding of it into $\Bbb R^2$ but e.g. if you identify $\Bbb R\sqcup{\bullet}$ with $(\Bbb R\times{0})\cup{(0,1)},$ then the usual metric on $\Bbb R^2$ answers the question. – Anne Bauval Jan 31 '23 at 14:01
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    @AnneBauval That is precisely the interest of the question. Some metrics on that space won't make it isometrically embedabble into $\mathbb R^2$ with the usual metric. – Carla_ Jan 31 '23 at 14:04
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    @AnneBauval Consider $\mathbb R^2$ with the max distance and the subspace $(\mathbb R \times {0}) \cup {(0,1)}$. That subspace cannot be isometrically embedded into the euclidean norm of $\mathbb R^2$ because it contains a line and a point with infinitely many points in the line at the same distance of that point. – Carla_ Jan 31 '23 at 15:05
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    You have to assume that the image of the embedding is closed, otherwise there are counter-examples. Under the "closed" assumption, the answer is positive, see here. – Moishe Kohan Jan 31 '23 at 15:28
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    @AnneBauval: I will add an example. I trust Alex Ravsky, but (in the link) his reference is to Hausdorff's paper which I did not read (I could, but it would take me a long time, since my German is too basic). – Moishe Kohan Jan 31 '23 at 15:49

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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.

Moishe Kohan
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