# Dose closed Alexandrov space admit a bi-Lipschitz embedding into $\mathbb R^N$?

as the title says. Let $A^n$ be an $n$-dimensional closed Alexandrov space. Does it admit a bi-Lipschitz embedding into the Euclidiean space $\mathbb R^N$ for sufficiently large $N$?

I know there are some spaces does not admit such embedding, for example a theorem by Pansu says that:

The Heisenberg group equipped with the Carnot-Caratheodory distance does not biLipschitz embed into $\mathbb R^n$, for any $n$.

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The Heisenberg group doesn't embed isometrically with respect to Euclidean distance, but Enrico Le Donne proved that there are Nash-type embeddings that preserve the length of every curve! See arxiv.org/abs/1005.1623 . –  Robert Young Apr 12 '13 at 16:04