Fundamental group of open subsets of $\mathbb{R}^n$ Suppose that $U$ is an open subset of $\mathbb{R}^n$. What can be said about its fundamental group? I'm sure that the answer should be well known, since this is rather natural question.
 A: The fundamental group of a (second-countable, Hausdorff) manifold is countable. See Lee, Smooth manifolds, proposition 1.16. 
The fundamental group of an orientable noncompact surface (in particular an open subset of $\mathbb R^2$) is free. See here. You can realize rather easily a free group on $n$ generators or $\mathbb N$ generators as the fundamental group of a subset of $\mathbb R^2$ - just delete a discrete set of that cardinality. This is a complete characterization.
The fundamental group of an open subset of $\mathbb R^5$ can be whatever you like. Here's a construction. Pick a countable presentation of your group $\langle w_1, w_2, \dots \mid r_1, r_2, \dots \rangle$. Take a 5-ball centered at the origin; attach 1-handles for each $w_i$. (This will be a manifold with boundary, but not a compact one - the handles will head out to infinity. This manifold looks like what you get when you take a circle of radius 1, glue another on to the rightmost point, glue another on to the rightmost point, etc, out to infinity, and take a tubular neighborhood. Think about what happens in $\mathbb R^3$ for inspiration on this picture.) Call this $\Sigma_0$. Inductively suppose we have the manifold $\Sigma_i \subset \mathbb R^5$. Pick a loop in the boundary of $\Sigma_i$ representing $r_i$; by a version of the Whitney embedding theorem, this bounds a disc in $\mathbb R^5 \setminus \text{int} \Sigma_i$. Call a small neighborhood of this disc $D_i$. Let $\Sigma_{i+1} = \Sigma_i \cup D_i$; this has killed $r_i$. Now take the limit of these (with care!) to get $\Sigma_\infty$, a 5-manifold with boundary with desired fundamental group in $\mathbb R^5$. Now delete the boundary.
What I've done here is a more straightforward way of picturing taking an embedding of a 2-dimensional CW complex into $\mathbb R^5$ and taking a regular neighborhood. Because you can give a 2-dimensional CW complex whatever fundamental group, you're done. PVAL assures me in the comments to the question that you can embed a 2-dimensional CW complex with planar 1-skeleton into $\mathbb R^4$, so that pushes the result down one dimension (because you can realize any countable group as the fundamental group of such a thing), but that this is a folklore result (that is, not written down anywhere). See here.
It remains to say something about $\mathbb R^3$. Here's one obstruction. If a finitely generated group is the fundamental group of a noncompact 3-manifold $N$, then there is a compact 3-manifold $M \subset N$ such that $\pi_1(M) \to \pi_1(N)$ is an isomorphism. This is known as Scott's Compact Core theorem. In particular $G = \pi_1(M)$, and thus $G$ is finitely presented. So the fundamental group cannot be finitely generated but not finitely presented. An example is given here. 
