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?