My Math Forum fundamental group, free group

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 March 21st, 2009, 02:35 PM #1 Newbie   Joined: Dec 2008 Posts: 4 Thanks: 0 fundamental group, free group Let $Y$ be the complement of the following subset of the plane $\mathbb{R}^2:$ $\{(x,0) \in \mathbb{R}^2: x \in \mathbb{Z} \}$ Prove that $\pi_1(Y)$ is a free group on a countable set of generators. I don't know how to start this problem. I just need a few helpful hints. Thanks in advance.

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