Fundamental group of the plane minus a Cantor set

Solution 1:

The space $X=\mathbb{R}^2\setminus (C\times\{0\})$ is a smooth manifold and thus is triangulable, with countably many cells. It follows that $\pi_1(X)$ is countable, and so must have countable rank. It is clear the rank cannot be finite (since, for instance, $\pi_1(X)$ has a surjection to a free group of rank $n$ for each $n$, by mapping $X$ to a plane with $n$ points removed), so it must be countably infinite.

For an alternate way to prove $\pi_1(X)$ is countable, note that we can write $C$ as an intersection $\bigcap C_n$ where each $C_n$ is a finite union of closed intervals. We can thus write $X$ as a union of open subsets $U_n=\mathbb{R}^2\setminus(C_n\times \{0\})$ such that $\pi_1(U_n)$ is free of finite rank for each $n$. By compactness, every loop in $X$ is contained in some $U_n$, and so $\pi_1(X)$ must be countable since each $\pi_1(U_n)$ is countable.