We denote this property by $(*)$.

Of course, if we can isometrically embed $X\hookrightarrow\mathbb{R}^n$, then $X$ is $(*)$. Is this sufficient to characterize all $X$? I've tried to construct counterexamples, but they all seem to have the problem of open balls being too big, so to speak. Obviously, the infinite-dimension normed vector spaces from functional analysis are not $(*)$. However, I suspect that the techniques necessary to prove that $(*)$ implies the existence of an isometric embedding are way beyond me.

For example, if $X$ is discrete, then it is $(*)$ iff it is finite, iff it can be isometrically embedded in $\mathbb{R}^n$ for some $n$.

Even if there exists some $(*)$ $X$ that cannot be embedded in $\mathbb{R}^n$, then my intuition tells me that we nonetheless have some constraint on the "dimension" of $X$, suitably defined.

So, what would be an intrinsic property of a metric space that is equivalent to $(*)$?

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    $\begingroup$ Look up the necessary and sufficient conditions for compactness (of metric spaces). $\endgroup$ – avs Dec 2 '16 at 22:38
  • $\begingroup$ So, is it necessary and sufficient that a closed ball be compact in this metric space? $\endgroup$ – Monstrous Moonshine Dec 2 '16 at 22:54

Let $H$ be the Hilbert cube. For another description see this question. In either version $H$ is a compact metric space, so every subset of $H$ is totally bounded. Every separable metric space can be embedded in $H$. In particular, $\Bbb R^{n+1}$ can be embedded in $H$. If $H$ could be embedded in $\Bbb R^n$, then so could $\Bbb R^{n+1}$, but in fact it cannot: $\Bbb R^m$ embeds in $\Bbb R^n$ if and only if $m\le n$. Thus, $H$ is a counterexample to the conjecture that a metric space with $(*)$ can be embedded in some $\Bbb R^n$.

  • $\begingroup$ So, can we just replace $\mathbb{R}^n$ with $H^\infty$? Is the conjecture true in that case? $\endgroup$ – Monstrous Moonshine Dec 2 '16 at 23:19
  • $\begingroup$ @MonstrousMoonshine: You can simply use $H$ itself: a metric space satisfying $(*)$ is separable, and every separable metric space embeds in $H$. $\endgroup$ – Brian M. Scott Dec 2 '16 at 23:24

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