TITLE: Fundamental group of the plane minus a Cantor set
QUESTION [10 upvotes]: If $C⊆ℝ$ is the Cantor set, what is the rank of the (necessarily free) fundamental group $π_1(ℝ^2 - C×\{0\})$?
Since the complement of the Cantor set is open, and an open set in $ℝ$ is always a union of at most countably many disjoint open intervals, the number of "gaps" between the points of the Cantor set (through which we may braid loops in the plane) is countable. This leads me to believe that there is a countable set of generators for the fundamental group in question, if indeed every loop may be deformed to such a well-behaved finite braiding. However, I don't know how to proceed.

REPLY [3 votes]: 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.