My question is whether all manifolds that can be embedded in $\mathbb R^n$ are homeomorphic to a smooth manifold?


I know that every smooth manifold can be triangulated which I think is a result of Whitehead and I think every manifold in $\mathbb R^n$ can be triangulated so this lends plausibility I think. (If the dimension 4 E8 manifold can be embedded in $\mathbb R^n$ then we have a counterexample but I'm not sure if it can be. All manifolds of dimension up to 3 can be triangulated).


I appreciate if this is a basic question but it doesn't seem to be spelt out explicitly anywhere. Most textbooks define a manifold and then a smooth manifold but don't say in which cases these concepts may be equivalent. 



Are all manifolds in $\mathbb R^n$ homeomorphic to a smooth manifold?
The answer is negative. As you observe yourself, if we allow abstract manifolds, then the E8 manifold provides a counterexample. However, it turns out that *any* abstract manifold can be embedded into $\mathbb R^n$ for a suitable $n$ (for instance, twice the dimension plus one, see e.g. [here](https://mathoverflow.net/q/34658/30186)), so in particular the E8 manifold is a submanifold of $\mathbb R^n$ (for $n=9$) which is not homeomorphic to a smooth manifold.


