Timeline for On Lipschitz embeddability of certain compact metric spaces into $\mathbb{R}^n$

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Mar 26, 2013 at 12:08	comment	added	Sergei Ivanov		I think it is true for $n=1$ and false for $n=2$. For $n=2$ you can do a similar construction using circles and points inside them (but the resulting metric is no longer intrinsic). For $n=2$ and an intrinsic metric on a disc, it is an open problem although not a very popular one. For $n=1$, it seems that one can can construct a Lipschitz embedding using the linear order on the real line: divide the interval containing the set in half, then in 4 pieces, etc, and choose images of the division points carefully.
Mar 26, 2013 at 10:35	vote	accept	Pedro Kaufmann
Mar 26, 2013 at 10:34	comment	added	Pedro Kaufmann		Thank you for your answer. The question remains open for n=1,2. It's good to keep in mind that when $K$ is a compact and countable metric space, there's always a Lipschitz injection from $K$ into $\mathbb{R}$. Indeed, for each distinct $x,y\in K$, $H_{x,y} = \{f\in Lip(K):f(x)=f(y)\}$ is a hyperplane of the Banach space $Lip(K)$ of real-valued Lipschitz functions on $K$. Baire's Theorem implies that $Lip(K)\neq \cup_{x\neq y} H_{x,y}$, thus there exists $f\in Lip(K) \setminus \cup_{x\neq y}H_{x,y}$ -- which has to be one-to-one. (this result is from a personal communication from G. Godefroy)
Mar 22, 2013 at 16:49	history	answered	Sergei Ivanov	CC BY-SA 3.0