TITLE: Uniform Embedding into Euclidean Space
QUESTION [9 upvotes]: Given a locally compact, separable, metric space $X$.

When does $X$ uniformly embed into some Euclidean space? 

This means, when does there exist some integer $n$ and a closed subset $Y\subset\mathbb{R}^n$ such that $X$ and $Y$ are uniformly equivalent, i.e., there exist a one-to-one map $f:X\to Y$ such that $f$ and the inverse $f^{-1}$ are uniformly continuous?
Background/Motivation
If we just ask for a topological embedding (i.e. $f$ and $f^{-1}$ are continuous), then such an embedding exists if and only if the covering dimension of $X$ is finite. Neccessity is clear, and sufficiency is shown in Corollary 2.6 of Luukkainen: "Embeddings of n-dimensional locally compact metric spaces to 2n-manifolds".
A neccessary condition for uniform embedding of $X$ is that the uniform covering dimension (as defined by Isbell) of $X$ is finite. 
Recall that an open cover of $X$ is called uniform if there exists some $\epsilon>0$ such that for every $x\in X$ the open ball of radius $\epsilon$ around $x$ is contained in some element of the cover.
Recall also that the order of a cover is at most $k$ if the intersection of any $k+1$ different elements of the cover is empty.
The uniform covering dimension of $X$ is at most $k$ if every uniform open cover of $X$ can be refined by a uniform open cover that has order at most $k+1$.
Is this condition also sufficient, i.e., does $X$ uniformly embed into some Euclidean space if and only if it has finite uniform covering dimension?

REPLY [2 votes]: Assume $X$ is uniformly embedded in $\mathbb E^d$ 
and $\varepsilon>\delta>0$.
Then there is $N$ such that any $\varepsilon$-ball in $X$ 
can be covered by $N$ balls of radius $\delta$.
It is easy to construct a rotationally symmetric Riemannian metric on $\mathbb R^2$
which does not have this property for some fixed $\varepsilon$ and $\delta$. Its  uniform covering dimension is still  2...