Embedding vs embedding as a closed subset

A version of Whitney's Theorem state that any $n-$dimensional manifold can be embedded in $\mathbb{R}^{2n+1}$ as a closed subset. Another version states that that any $n-$dimensional manifold can be embedded in $\mathbb{R}^{2n}$.

Is there an example of $n-$manifold which can be embedded in $\mathbb{R}^{2n}$ but not as closed subset? Or is it possible to extend Whitney's theorem to have an embedding as a closed subset in $\mathbb{R}^{2n}$.

The motivation for this question come from this comment I found reading Milnor's book on Characteristic classes (p. 120)

[...](Compare §4.8. According to [Whitney, 1944] every smooth $n-$manifold whose topology has a countable basis can be smoothly embedded in $\mathbb{R}^{2n}$. Presumably it can be embedded as a closed subset of $\mathbb{R}^{2n}$, although Whitney does not prove this).

Edit: In order to clarify what I'm looking for, you can consider the open Moebius strip: it can be embedded in $\mathbb{R}^{3}$ but not as a closed subset. I'm looking for a manifold for which this happen in the maximal dimension (i.e. $2n$).

• Open disk would be one. – Moishe Kohan Oct 24 '14 at 19:56
• @studiosus But if you embed the open $n-$disk as an $n-$plane in $\mathbb{R}^{2n}$ it is a closed subset... – Dario Oct 25 '14 at 18:41
• I mean, embed open unit n-disk in $R^n$ by the identity map. – Moishe Kohan Oct 25 '14 at 19:40
• @studiosus Yes, probably I wasn't very precise in my question, I'm looking for an $n-$manifold that **can** be embedded in $\mathbb{R}^{2n}$ and for which there exists no embedding in $\mathbb{R}^{2n}$ whose image is a closed subset. – Dario Oct 25 '14 at 20:48
• I think I read somewhere in Milnor-Stasheff that the authors expect one should be able to do this in general (give a closed embedding), but that they don't recall a reference. Unfortunately, I don't have my copy to verify. – user98602 Feb 23 '15 at 19:02