This question was prompted by my answer to [this question](https://math.stackexchange.com/q/3222502/39599).


An exotic $\mathbb{R}^4$ is a smooth manifold homeomorphic to $\mathbb{R}^4$ which is not diffeomorphic to $\mathbb{R}^4$ with its standard smooth structure. An exotic $\mathbb{R}^4$ is said to be *small* if it can be embedded in the standard $\mathbb{R}^4$ as an open subset. An exotic $\mathbb{R}^4$ which is not small is called *large*.



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> Does every large $\mathbb{R}^4$ embed in $\mathbb{R}^5$?
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Freedman and Taylor showed there is a maximal exotic $\mathbb{R}^4$, into which all other $\mathbb{R}^4$'s can be smoothly embedded as open subsets. So it would be enough to show that this one embeds in $\mathbb{R}^5$.


By the Whitney Embedding Theorem, every large $\mathbb{R}^4$ embeds in $\mathbb{R}^8$, but I suspect one can do better.



Does every large $\mathbb{R}^4$ embed in $\mathbb{R}^5$?
The product of any two smooth open contractible manifolds is diffeomorphic to a Euclidean space, see e.g. remark 5.3 in  [my paper](https://arxiv.org/abs/1208.5220); the result is of course not due to me but I don't know any other place where all references are collected. 


In particular, this applies to the product of $\mathbb R$ and any exotic $\mathbb R^4$, so the latter embeds into $\mathbb R^5$. In this case the point is that the product is simply-connected at infinity, and hence by a result of Stallings it is PL homeomorphic to $\mathbb R^5$, but any PL structure on $\mathbb R^5$ is induced by a unique smooth structure (as proved by Munkres). 


