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:*

- 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 $a,b\in Y$.
- 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 $d_X(x,x')+|t-t'|$.
- 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.
- 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.