TITLE: Are homeomorphic open subsets of $\mathbb{R}^n$ also diffeomorphic?
QUESTION [23 upvotes]: Let $U_1, U_2$ be open subsets of $\mathbb{R}^n$.  Both are naturally differentiable submanifold, getting the differentiable structure from $\mathbb{R}^n$.  Further, both are natural topological manifolds, as submanifolds of $\mathbb{R}^n$.
Question:
If $U_1$ and $U_2$ are homeomorphic, are they also diffeomorphic?
Of course two general topological manifolds which are homoemorphic do not need to be diffeomorphic.  But here the differentiable structure is a very special one.
The answer might depend on the dimension $n$.  For $n=1,2,3$ it is yes, as there each topological manifold has a unique differentiable structure.  For $n \geq 5$ and $U_1$ an open ball the answer is yes by the uniqueness of differentiable structures on $\mathbb{R}^n$ for $n \geq 5$.
Some special cases are:

What happens if $U_1$ (and hence $U_2$) is contractible?
What happens if $U_1$ is a ball and $n=4$?  Is there an exotic $\mathbb{R}^4$ which can be realized as an open subset of the standard $\mathbb{R}^4$?

(The question came up because I encountered a sloppy definition of a manifold.  One can view the above manifolds as being defined by only one chart.  [That of course depends on your definition of chart, if you require it to start from a ball or not.]  So the questions basically asks: How do manifolds with only one chart look like?)

REPLY [8 votes]: Adding to Andy's answer: there are lots of contractible open subset of $\mathbb R^n$ that
are not homeomorphic to $\mathbb R^n$. For example, any compact contractible manifold of dimension $n>4$ embeds into $\mathbb R^n$: the double of any compact contractible manifold is simply-connected and hence a homotopy sphere, which after removing a point becomes $\mathbb R^n$. For constructions of compact contractible manifolds, see Kervaire's paper 
Smooth homology spheres and their fundamental groups.
You mention a confusion about a "sloppy definition of a manifold" and ask which manifolds
have one chart. By a chart you seem to mean any open subset of a manifold together with a homeomorphism onto an open subset of $\mathbb R^n$, which I think is a valid definition.
Any atlas of charts whose transition functions are smooth defines a smooth structure on
your manifold, which by definition is the set of all atlases compatible the given one. 
If there is only one chart, then the (only) transition function is the identity,
which is smooth. However, this merely implies that your manifold with one chart it diffeomorphic to an open subset of $\mathbb R^n$.