TITLE: Homotopy type of the plane minus a sequence with no limit points
QUESTION [10 upvotes]: It is well known that the plane minus $n$ points is homotopy equivalent to a wedge of circles and hence its fundamental group is free on $n$ letters.
Question: Is the plane minus an infinite sequence of points having no limit point 
homotopy equivalent to an infinite wedge circles?
I'm pretty sure that this could fail if the sequence had a limit point, since then the space in question might be something more like the Hawaiian earing. But is the sequence having a limit point the only thing that could break down?

REPLY [2 votes]: Expanding on John's answer a bit: If $X = \{x_1,x_2,\ldots\} \subset R^2$ is a countable discrete set, then there is a diffeomorphism $f:R^2 \to R^2$ with $f(x_i)=i$. Proof: by Sards theorem, we can assume that $X \cap N =0$ (translate a bit if necessary). 
Claim 1: we can find paths $p_n: [0,1] \to R^2$ which are embeddings, $p_n (0)=x_n$, $p_n (1)= n$ and such that the images of all the $p_n$ are all disjoint. 
Proof by induction on $n$. Suppose that $p_1,\ldots ,p_{n-1}$ have been constructed. By homological duality theory, the complememt $R^2 \setminus \cup_{i=1}^{n-1} p_i ([0,1])$ is path-connected. We can then find a path from $x-n$ to $n$, avoiding $\cup_{i=1}^{n-1} p_i ([0,1])$ and also $X \cup N$.
If we have found these paths, we can find neighborhoods $U_n$ of the images, $U_i \cap U_j=\emptyset$ for $i \neq j$. There exist diffeomorphisms $f_n:U_n \to U_n$ with compact support and $f_i(x_i)=i$. These glue together to a global diffeomorphism $f$ with the desired property.