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Can someone point me to (or provide) a standard (fairly elementary) proof that every (abstract, e.g. as defined in Tu, An Introduction to Manifolds) smooth manifold embeds in Euclidean space for some $n$. I know this is more or less what Whitney's embedding theorem tells us, but Whitney's theorem is hard and from what I can tell, most of the difficult comes from putting the nice numerical bound on the dimension of the Euclidean space. Also I have seen the proof for the compact case, but I would like the general case. Ideally the proof should be more or less from first principles.

A little context is that I am trying to demonstrate the equivalence of definitions of smooth manifolds (and other associated machinery) given by different authors (namely Tu vs. Guillemin and Pollack). I have spelled out every detail, but my proof needs to cite some embedding theorem to go from abstract smooth manifolds to embedded submanifolds of Euclidean space.

Note: I am talking about manifolds without boundary.

Physical Mathematics
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    As far as I know in the non-compact case there is not any argument that is easy comparable to how the compact case is. You either need to use something like the methods used to prove the Whitney embedding theorem or some nontrivial theorems from topological dimension theory to guarantee you have a finite cover by charts and so can imitate the proof in the compact case. – Eric Wofsey Jan 20 '20 at 01:11
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    See https://math.stackexchange.com/questions/3236140/existence-of-an-embedding-of-a-smooth-manifold for essentially a duplicate of your question which contains some discussion of this and has not received an answer. – Eric Wofsey Jan 20 '20 at 01:12
  • To be clear, when I say "the methods used to prove the Whitney embedding theorem" I'm referring to the weak version that gives an embedding in $\mathbb{R}^{2n+1}$ rather than $\mathbb{R}^{2n}$. For that version you don't really need to do any hard work to get the specific numerical bound--proving it for $2n+1$ is just as easy as proving it for any higher dimension. Going from $2n+1$ down to $2n$ is where it gets trickier. – Eric Wofsey Jan 20 '20 at 01:41
  • @EricWofsey That's fine that it can't be done analogously to the compact case. Basically I'm looking for the "most elementary" proof, i.e. trying to avoid highly specific subfields of math, but if the simplest proof is still difficult, that's how it is. Do you know of a good, somewhat self-contained reference for the Whitney's theorem ($2n+1$) case? Now that I know that I'm effectively just looking a proof of Whitney's theorem, I'll look around for that. – Physical Mathematics Jan 20 '20 at 03:37

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There is a proof in Bredon's differential topology book. It takes maybe 2 pages more than the proof for compact manifolds.

Esteban Perez
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