Fundamental Group of complement of Integers Let $X=\mathbb{R}\times (-1,1)-\mathbb{Z}\times \{0\}$.  I want to compute the fundamental group of that space.
I would be tempted to do some division like the following to apply Van Kampen, but to do it all the sets must intersect in the basepoint of $\pi(X,x)$, so I can't really do it. 

If I incorrectly use Van Kampen, I will get to the correct result that it's free in a countable number of generators, though.
Is there any other way to do that, I mean, with the basic theory of Hatcher's of Massey's book (It's an exercise from Massey's book)?
 A: I suggest to proceed as follows. Define $Y = \bigcup_{n \in \mathbb{Z}} S(n)$, where $S(n) = \{ x \in \mathbb{R}^2 \mid \lVert x - n \rVert =  \frac{1}{2}) \}$. This is an infinite chain of circles. It is easy to see that $Y$ is a strong deformation retract of $X$. Now $Y$ has the structure of a CW-complex, $0$-cells being the sets $\{n - \frac{1}{2} \}$ and $1$-cells being the upper and lowers half circles of the $S(n)$. Then $Y_0 = Y \cap \mathbb{R} \times (1,0]$ is subcomplex of $Y$ which is homeomorphic to $\mathbb{R}$. In particular, $Y_0$ is contractible. It is known that then $p : Y \to Y/Y_0$ is a homotopy equivalence. But $Y/Y_0$ is nothing else than a countably infinite wedge of circles and your arguments apply.
A: Use Van Kampen inductively to prove that for all $n$ the set $X_n = B_{-n} \cup \cdots \cup B_n$ is free of rank $2n+1$, and that the inclusion $X_n \to X_{n+1}$ takes the rank $2n+1$ free group $\pi_1(X_n)$ to a free factor of the rank $2n+3$ free group $\pi_1(X_{n+1})$. Then use a direct limit argument to conclude that $\pi_1(\cup_{n=1}^\infty X_n)$ is free of infinite rank.
A: I want to say a rose of countably infinitely many petals;  but I'm not sure it's correct.  At least it seems like it might deformation retract onto it...  
