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I am looking for the reference to the following (surely known) characterization of metric spaces that embed into $\mathbb R^n$:

Theorem. Let $n$ be positive integer number. A metric space $X$ is isometric to a subspace of the Euclidean space $\mathbb R^n$ if and only if every subspace $A\subseteq X$ of cardinality $|A|\le n+3$ is isometric to a subspace of $\mathbb R^n$.

Remark. It is interesting to notice that $n+3$ in this theorem cannot be lowered to $n+2$. A counterexample in dimension 1 is the square $$S=\{(1,1),(1,-1),(-1,1),(-1,-1)\}$$ in the plane $\mathbb R^2$ endowed with the $\ell_1$-metric. Every 3-element subset of $S$ is isometric to a subspace of the real line but $S$ is not isometric to a subspace of a Euclidean space. Analogous examples can be constructed in any dimension.

Example. For every positive integer number $n$ there exists a metric space $X$ of cardinality $|X|=n+3$ such that every subspace $A\subseteq X$ of cardinality $|A|\le n+2$ is isometric to a subspace of $\mathbb R^n$ but $X$ is not isometric to a subspace of a Euclidean space.

So, I am interested in finding references to the above Theorem and Example (or at least Theorem, which should be known). The proof of Theorem uses Kuratowski--Zorn Lemma and the rigidity of Euclidean spaces. The space $X$ in Example is a regular n-dimensional simplex with duplicated center. Can such an example have larger cardinality than $n+3$?

Problem. Is it true that a metric space $X$ of cardinality $|X|> n+3$ is isometric to a subspace of the Euclidean space $\mathbb R^n$ if every subset $A\subseteq X$ of cardinality $n+2$ is isometric to a subspace of $\mathbb R^n$.

Remark. The Problem has affirmative solution for $n=1$.

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1 Answer 1

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It is in

Menger, K., Untersuchungen über allgemeine Metrik., Math. Ann. 100, 75-163 (1928). EuDML:159284. JFM 54.0622.02.

A modern exposition can be found in

Bowers, John C.; Bowers, Philip L., A Menger redux: embedding metric spaces isometrically in Euclidean space, Am. Math. Mon. 124, No. 7, 621-636 (2017). Zbl 1391.51007.

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  • $\begingroup$ Thank you very much for the answer and the link. I hope it is exactly what I needed. $\endgroup$ Commented Apr 22, 2023 at 19:23
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    $\begingroup$ Pleasure is mine :) The result is certainly what you said, but maybe proofs are different from what you indicated. $\endgroup$ Commented Apr 22, 2023 at 19:25