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.
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5$\begingroup$ Any metric space $(S, d_S)$ where $S$ has cardinality larger than $c$ is a simple example. $\endgroup$– Nick SCommented Apr 17, 2018 at 19:06
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12$\begingroup$ 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. $\endgroup$– Fedor PetrovCommented Apr 17, 2018 at 19:07
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$\begingroup$ @NickS, that was cute! There is nothing like a huge hammer. $\endgroup$– Wlod AACommented Apr 17, 2018 at 20:15
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$\begingroup$ (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 $n$. $\endgroup$– YCorCommented Apr 17, 2018 at 20:46
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$\begingroup$ Slightly less huge hammer: an infinite-dimensional Hilbert space. $\endgroup$– YCorCommented Apr 17, 2018 at 20:47
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1 Answer
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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.
answered Apr 17, 2018 at 19:37
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3$\begingroup$ 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. $\endgroup$ Commented Apr 17, 2018 at 19:41
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$\begingroup$ @TobiasFritz Thank you for the reference I was not aware of Morgan's paper. $\endgroup$ Commented Apr 17, 2018 at 19:50
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2$\begingroup$ 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. $\endgroup$– YCorCommented Apr 17, 2018 at 21:06
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1$\begingroup$ There's yet more discussion of the work of Menger and Schoenberg in this expository paper on distance geometry arxiv.org/abs/1502.02816 $\endgroup$– j.c.Commented Apr 17, 2018 at 22:28
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1$\begingroup$ As well as Blumenthal's 1970 book "Theory and applications of distance geometry". $\endgroup$– YCorCommented Apr 18, 2018 at 7:13