TITLE: Is there a Whitney Embedding Theorem for non-smooth manifolds?
QUESTION [52 upvotes]: For smooth $n$-manifolds, we know that they can always be embedded in $\mathbb R^{2n}$ via a differentiable map.  However, is there any corresponding theorem for the topological category?  (i.e. Can every topological manifold embed continuously into some $\mathbb R^N$, and do we get the same bound for $N$?)

REPLY [4 votes]: My contribution here is very limited:
Every topological 4-manifold admits an embedding in $R^8$.
Proof. It suffices to consider the case of connected manifolds. If a connected 4-manifold is noncompact then it admits a smooth structure (equivalently, a triangulation). A proof can be found in
Quinn, Frank, Ends of maps. III: Dimensions 4 and 5, J. Differ. Geom. 17, 503-521 (1982). ZBL0533.57009.
Hence, such manifold admits an embedding in $R^8$ by Whitney's theorem.
It is a general result that every compact $n$-dimensional manifold admits an embedding in $R^{2n}$. This observation is due to Sergey Melikhov (in comments to Andy Putman's answer). All the available references prove a stronger result, namely, an embedding theorem for compact generalized manifolds:
Bryant, J. L.; Mio, W., Embeddings in generalized manifolds, Trans. Am. Math. Soc. 352, No. 3, 1131-1147 (2000). ZBL0934.57023.
and for certain classes of compact ANRs (which include all compact manifolds):
S. Melikhov, E.Shchepin,
The telescope approach to embeddability of compacta, unpublished, 2006.
Combining the two, one gets the embedding theorem for 4-manifolds. qed

My sense is that similar arguments to the ones by Bryant-Mio and Melikhov-Shchepin will go through in the noncompact case (I looked only at the paper by Melikhov and Shchepin) and can be used to prove an embedding theorem for noncompact manifolds but somebody would have to check the details...