the Whitney embedding theorem
/ WIT-nee /
Abstract manifolds are defined intrinsically, with no surrounding space — they are charts glued by transition maps, floating free. A natural worry is whether such an abstraction is more general than honest surfaces sitting in Euclidean space. Whitney's theorem dissolves the worry: every smooth manifold, however abstractly defined, can be realized as a genuine embedded submanifold of some R^N. Intrinsic and extrinsic descriptions are equivalent.
The strong form states that every smooth n-manifold (second-countable, Hausdorff) embeds in R^{2n}, and immerses in R^{2n-1}. The easy form, which is all most courses prove, gives an embedding in R^{2n+1} and is reachable by elementary means: cover the manifold by finitely many charts using compactness (or proper exhaustion), patch the local coordinate maps together with a partition of unity to get a smooth injective immersion into a high-dimensional R^M, then use Sard's theorem to project down repeatedly, each generic projection lowering the ambient dimension by one without destroying the embedding, until you reach 2n+1.
The dimension 2n is sharp in general — the real projective plane RP^2 does not embed in R^3, only in R^4 — and getting from 2n+1 down to 2n is the hard part, requiring Whitney's trick to remove double points. Two honesty notes: first, embeddability is about smooth manifolds; it does not say a given topological manifold is smooth, and indeed some are not. Second, the theorem grants an embedding but says nothing about geometry — it is purely about the smooth structure, so the embedding generally distorts any metric. That a manifold CAN be embedded isometrically is a different, much harder story (Nash), not part of this result.
The Klein bottle is a closed 2-manifold that cannot embed in R^3 (its R^3 pictures always self-intersect), but by Whitney it embeds in R^4. The model self-intersection in three dimensions is an immersion, not an embedding; the extra dimension lets the surface pass over itself.
The Klein bottle immerses in R^3 but only embeds in R^4.
The theorem is about smooth structure, not metric: it never claims an isometric embedding, which is Nash's far harder theorem. And 2n is optimal in general, so do not expect every n-manifold in R^{n+1}.