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Let $X$ be a metric space. Suppose that the product $X\times\mathbb R$ admits a bi-Lipschitz embedding into $\mathbb R^{n}$. Does it follow that $X$ admits a bi-Lipschitz embedding into $\mathbb R^{n-1}$?

Definitions and comments:

  1. A map $F\colon (Y,\rho)\to\mathbb R^n$ is a bi-Lipschitz embedding if there exists a constant $L$ such that $L^{-1}\rho(a,b)\le |F(a)-F(b)|\le L\rho(a,b)$ for all $a,b\in Y$.
  2. The choice of a product metric on $X\times \mathbb R$ makes no difference. For definiteness, let the distance between $(x,t)$ and $(x',t')$ be $d_X(x,x')+|t-t'|$.
  3. The cancellation fails if bi-Lipschitz is replaced by topological or smooth embedding. For example, $S^1\times \mathbb R$ embeds into $\mathbb R^2$ smoothly, but $S^1$ does not embed into $\mathbb R$ in any sense.
  4. The cancellation is possible when $n=1$, according to $\mathbb R^0=\{0\}$. :) I'm pretty sure it's also possible when $n=2$, but don't have a proof. For $n\ge 3$ I don't have a clue.
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    I think your example in 3 should work in the Lipschitz case: You can extend a smooth embedding $S^1\times (0,1)\to \mathbb{R}^2$ to a $C^1$ diffeomorphism of open sets in $\mathbb{R}^3$, where this sets have Lipschitz boundary, then this diff. is bi-Lipschitz, and then so is the restriction.2012-06-13
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    @Jose27 You cannot replace $\mathbb R$ with $(0,1)$ -- these are different animals in the bi-Lipschitz zoo. $S^1\times \mathbb R$ does not have a bi-Lipschitz embedding to a plane, since the images of some half-lines $(x,t)$ would be trapped inside the image of $S^1$, forcing them to be bounded.2012-06-13
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    @Thursday, do you got any progress on this problem?2014-09-13

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