MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $K$ be a compact subset of a Euclidean space of very large dimension $N$. Assume that any point $x\in K$ has a neighborhood $U\subset K$ which is contained in a smooth $l$-dimensional submanifold of $\mathbb R ^N$, for some fixed not very large $l$.

Is the whole set $K$ contained in a smooth $l$-dimensional submanifold?

In the case that I have encountered, I had the additional assumption that $K$ is topologically an $l$-dimensional manifold with boundary, in which case the statement is not very difficult to prove. I was wondering if the statement is true without any additional assumptions on $K$.

share|cite|improve this question
In particular, I am wondering if $K = [0,1]\times J$ where $J = \{ k^{-1} | k \in \mathbb{N}\}$ satisfies your assumptions, and if not, why not. – Willie Wong Dec 9 '10 at 15:26
oop, I meant $J$ to also include $0$. – Willie Wong Dec 9 '10 at 15:27
Yes, one considers $K$ with the subspace topology. The subset $K$ satisfies the assumption and conclusion with $l=2$ but not with $l=1$ (at any point of $[0,1] \times 0$ – Alexander Lytchak Dec 9 '10 at 16:51
Then does the usual partition of unity + glue together argument not work? – Willie Wong Dec 9 '10 at 17:59
One simple remark: I think that if there is no condition that $l$ is much more smaller than $N$ than the statement is wrong. I think that in general situation one can construct a counterexample when $K$ is a submanifold of dimension less then $l$. Indeed, if $K$ is a submanifold in $l$-dimensional submanifold then the normal bundle to $K$ admits a non-trivial splitting into the direct sum. Stiefel-Whitney classes form an obstruction to such a splitting. So, in particular, it looks like partition of unity arguments do not work here. – Petya Dec 9 '10 at 18:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.