Let $n \in \mathbb{N}$. Is there a standard example of a subset of $\mathbb{R}^{n+1}$ that is contained in the image of a Lipschitz map $\mathbb{R}^n \to \mathbb{R}^{n+1}$ (or, more generally, that is $n$-rectifiable) but which is not locally bi-Lipschitz equivalent to a subset of $\mathbb{R}^n$? The example should be topologically simple, and as "sparse" as possible.
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2$\begingroup$ A circle $t \mapsto (cos t, sint)$ is an example for $n=1$. Perhaps you wanted to rephrase the question? $\endgroup$– Yuval PeresCommented Sep 9, 2019 at 9:45
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$\begingroup$ Thanks, I overlooked that example. So I'm thinking more of local bi-Lipschitz equivalence rather than global equivalence. Of course, you can take a shrinking "Hawaiian earing" sequence of circles as a variation on the standard circle. But now the failure is more for "topological" reasons than "metric" reasons. $\endgroup$– mdrCommented Sep 9, 2019 at 14:21
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After posting the question, I thought of the example of a line with a cusp point on it, such as the graph of $|x|^{1/2}$, denoted by $C$. You can project the line $y=|x|$ horizontally onto $C$, which is locally Lipschitz. The projection from $C$ onto the $x$-axis is also Lipschitz.
On the other hand, $C$ is not linearly locally connected (there is no $\lambda \geq 1$ such that any two points in an arbitrary ball $B(x,r) \cap C$ can be connected in $C$ without leaving the ball $B(x,\lambda r)$), so $C$ is not bi-Lipschitz equivalent to a line.
I'd still be interested if there are other meaningful examples out there -- if there are essentially different ways to rule out bi-Lipschitz equivalence.