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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?


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?

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Why do you require $Y$ to be closed? $X=(0,1)\subseteq\mathbb R$ meets your conditions, but AFAICS it is not uniformly equivalent to a closed subset $Y\subseteq\mathbb R^n$ ($X$ is bounded, hence so is $Y$, hence $Y$ is compact, hence $X$ is, but it actually isn’t). –  Emil Jeřábek Jul 5 '12 at 16:17
Emil, I agree. Every closed subset of $\mathbb{R}^n$ is complete, and uniform equivalence of metric spaces preserves completeness. Thus, when asking for spaces that are uniformly equivalent to closed subsets of Euclidean space, one has to add completeness to the list of neccessary conditions. But it is maybe more natural to drop the assumption that $Y\subset\mathbb{R}^n$ be closed. –  Hannes Thiel Jul 6 '12 at 14:44
Note that even if you drop the assumption that $Y$ be closed, this set cannot get too wild: the other assumptions imply $Y$ is locally compact, which means it can be written as a difference of two closed sets. –  Emil Jeřábek Jul 7 '12 at 18:48
Hannes, your question is discussed at length in Isbell's book "Uniform spaces", and in some of his preceding papers. He gives some necessary conditions which make it clear that the question is not likely to have a simple answer. –  Sergey Melikhov Jul 9 '12 at 14:56
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1 Answer

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...

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Thank you Anton, I agree that the ($\epsilon$,$\delta$)-covering property that you mention is preserved by uniform equivalence. I just remark that it reminds me of the box-counting dimension of metric spaces, and this dimension is preserved by Lipschitz equivalence, but I think not by uniform equivalence (of course, you didn't claim that). Do you think that the ($\epsilon$,$\delta$)-covering property (together with finite uniform covering dimenion) will be sufficient for uniform embedding into Euclidean space? Is this covering property totally independent of the uniform covering dimension? –  Hannes Thiel Jul 6 '12 at 15:03
@Hannes, I do not think so. It might be no nice "iff"-description. –  Anton Petrunin Jul 6 '12 at 17:08
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