Proving that a fundamental group is uncountable

Question

Given the space $\mathbb{R}^2 - \bigl(\{0\}\times\mathbb{Q}\bigr)$, I need to show that the fundamental group of this space is uncountable. I thought of taking two points $A=(x_0,y_0)$ in the area where $x_0 < 0$ and $B (x_1,y_1)$ where $x_1>0$, now i can find a path between them since I removed the rationals form the $y$ axis , I still can find an irrational point such that my path pass by this point, now I consider a loop from $A$ passing throughout $B$, I can find another loop also, my claim is that these two loops can't be homotopic but i can't really prove this last thing. Am I thinking in the right way? if anyone could correct me and help me finish the proof, I would really appreciate it.

Answer 1

HINTS let $C_n$ be the circle of radius $\sqrt2/n$ center at $(0,\sqrt2/n)$... consider a homomorphism $R: \pi_1(X) \rightarrow \prod_\infty\mathbb{Z}$ to the direct product (not the direct sum)of infinitely many copies of $\mathbb{Z}$ s.t for every sequence of integers {$x_n$} we can construct a loop f:I$\rightarrow$X that wraps $x_n$ times the loop $C_n$ in the time interval [1-1/n,1-1/n+1]...and it is continuous since every neighbourhood of the base point $(0,0)$ in X contains all but finitely many circles...so $R$ is surjective... so $\pi_1(X)$ is uncountable.

Answer 2

Your idea is basically correct. What you need to complete the proof is the concept of winding number. Given a closed path $p : [0,1] \to \mathbb{R}^2$ based at $A$ and a point $C \in \mathbb{R}^2$ such that $C \not\in \text{image}(p)$, the winding number of $p$ around $C$ is an integer which I will denote $w(p;C)$. There are two propeties of winding number that you need:

1. For any two closed paths $p_0, p_1 :[0,1] \to \mathbb{R}^2$ based at $A$ whose images do not contain $C$, if there is a path homotopy $h :[0,1] \times [0,1] \to \mathbb{R}^2$ from $p_0$ to $p_1$ such that $C \not\in \text{image}(h)$ then $w(p_0;C) = w(p_1;C)$.
2. Suppose $p :[0,1] \to \mathbb{R}^2$ is a simple closed curve, and so by the Jordan curve theorem there are two components of $\mathbb{R}^2 - \text{image}(p)$, a bounded component called its inside and an unbounded component called its outside. If $C$ is on the inside of $p$ then $w(p;C) = \pm 1$, whereas if $C$ is on the outside of $p$ then $w(p;C)=0$. (The difference between $+1$ and $-1$ depends intuitively on whether $p$ goes around $C$ in the counterclockwise or the clockwise direction, but that's not important for this question).

Now what you write can be completed as follows. Consider an interval $[u,v]$ with irrational endpoints $u,v$. Define a simple closed curve $p$ which starts at $A$, crosses the $y$-axis from left to right at the point $(0,u)$, and crosses back from right to left at the point $(0,v)$, before returning to $A$. It follows that for any rational number $c$, we have $$w(p;(0,c)) = \begin{cases} \pm 1 & \text{if $u<c<v$} \\ 0 & \text{otherwise} \end{cases} $$ Also, since the image of each path homotopy misses the rational points on the $y$-axis, the number $w(p;(0,c))$ is an invariant of the path homotopy class of $p$ for each rational number $c$. Since there are uncountably many choices for the interval $[u,v]$, this proves that the fundamental group is uncountable.