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What are some simple examples of metric spaces that cannot be subspaces of $\mathbb{R}^n$? I've heard there is an example with $4$ points, where two points lie between the other two, but I cannot figure out the details.

YCor
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Jake B.
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    Any metric space $(S, d_S)$ where $S$ has cardinality larger than $c$ is a simple example. – Nick S Apr 17 '18 at 19:06
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    AB equals to 2, other distances between ABCD are equal to 1. Then both points C, D should be midpoints of AB, but they do not coincide. – Fedor Petrov Apr 17 '18 at 19:07
  • @NickS, that was cute! There is nothing like a huge hammer. – Wlod AA Apr 17 '18 at 20:15
  • (1) It's useful to say "that are not isometric to", since metric spaces are usually considered in many categories (topological, lipschitz, isometric, large-scale lipschitz, etc). (2) It's also useful to say which norm is considered on $\mathbb{R}^n$ since it can matter. (3) The question is ambiguous since it's not clear if it means "not embeddable into $\mathbb{R}^n$ for some given $n$ (for each $n$ such a question makes sense, and the title rather suggests this interpretation), or not embeddable in $\mathbb{R}^n$ for any $n$. – YCor Apr 17 '18 at 20:46
  • Slightly less huge hammer: an infinite-dimensional Hilbert space. – YCor Apr 17 '18 at 20:47
  • Another version would be: a metric space that is not homeomorphic to a subset of $\mathbb R^n$ – Gerald Edgar Apr 17 '18 at 21:21
  • @FedorPetrov: is this obvious? That there is no (unusual) metric on $\mathbb{R}^n$ that this could hold? – Jake B. Apr 20 '18 at 08:56
  • what do you mean by unusual metric? How should it be related to the algebraic and topological structures of $\mathbb{R}^n$? I meant the Euclidean metric. – Fedor Petrov Apr 20 '18 at 12:06
  • I guess I'm confused. So we are comparing 4 points and the metric you provided and $\mathbb{R}^n$ with Euclidean metric, so I guess that makes sense. So does statements like "In $\mathbb{R}^n$ every Cauchy sequence is convergent" include the fact that $\mathbb{R}^n$ is equipped with the Euclidean metric or does it hold for any metric in $\mathbb{R}^n$? There can be different metrics defined for $\mathbb{R}^n$, right? – Jake B. Apr 20 '18 at 16:51

1 Answers1

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Although this questions should be closed since it is not a at the research level, let me answer it since the answer is of independent interest.

There is a complete characterization, due to Schoenberg, of finite metric spaces that admit an isometric embedding into a Euclidean space: A reference to a characterization of metric spaces admitting an isometric embedding into a Hilbert space. Any metric space that violates Schoenberg's condition cannot be realized as a subset of $\mathbb{R}^n$. A metric space given in a comment of Fedor Petrov is an example.

You can also find a lot of interesting references in fantastic comments to this answer.

Piotr Hajlasz
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    There is a nice paper by Morgan with an intuitive geometric interpretation of Schoenberg's criterion. However, Morgan seems to have been aware of Schoenberg's 1935 work. – Tobias Fritz Apr 17 '18 at 19:41
  • @TobiasFritz Thank you for the reference I was not aware of Morgan's paper. – Piotr Hajlasz Apr 17 '18 at 19:50
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    Morgan (1974) quotes a 1928 paper by Menger with the same result. This is very interesting, since it seems that Menger's criterion is the same as Schoenberg's. Menger, Karl: “Untersuchungen über allgemeine Metrik”, II Math. Ann 100 (1928), p. 113-141. – YCor Apr 17 '18 at 21:06
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    There's yet more discussion of the work of Menger and Schoenberg in this expository paper on distance geometry https://arxiv.org/abs/1502.02816 – j.c. Apr 17 '18 at 22:28
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    As well as Blumenthal's 1970 book "Theory and applications of distance geometry". – YCor Apr 18 '18 at 07:13