What is 1st homology group of $X = \text{plane} - \bigcup_{n}\{(1/n, 0)\}\cup\{(0,0)\}$? What is 1st homology group of $X = \text{plane} - \bigcup_{n}\{(1/n, 0)\}\cup\{(0,0)\}$?
My guess is since it is $\operatorname{Ab}(\Pi_1(X))$.
It is a subgroup of $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}\dotsb$ countably many times, generated by all vectors which are $0$ except at finitely many terms and the diagonal group ie. all vectors of form $[n , n , n , n ,\dotsc]$ where $n$ is any integer. As any loop either loops about finitely many of pts $(1/n ,0 )$ or it loops origin in which case it loops around all but finitely many of them.
Am I right? If yes what is a rigorous proof, what is a good reference?
 A: Think of the plane as $\mathbb C$ and apply the homeomorphism $z\mapsto 1/z$. Then your space is homeomorphic to $\mathbb C\setminus\mathbb N$. Now it's very easy to see what's happening since there the accumulation point has been pushed out to infinity. The result is then homotopy equivalent to a wedge of countably many circles. (Countably many circles identified at a single point.) The first homology of a countable wedge of circles is similar to what you described in your post. It is a direct sum of countably many copies of $\mathbb Z$: in notation, $\bigoplus_{n=1}^\infty \mathbb Z$. This is different from the direct product in that all but finitely many coordinates are required to be $0$. You might ask, what about the diagonal elements $(n,n,...)$ that you seemed to be seeing? The idea is that if a loop passes around $0$ in the original picture, you instead think of it as passing around $\infty$, in which case it only loops around finitely many points (including $\infty$), so you don't see the diagonal terms you were talking about.
