$\pi_{1}({\mathbb R}^{2} - {\mathbb Q}^{2})$ is uncountable Question: Show that $\pi_{1}({\mathbb R}^{2} - {\mathbb Q}^{2})$ is uncountable.
Motivation: This is one of those problems that I saw in Hatcher and felt I should be able to do, but couldn't quite get there.
What I Can Do:  There are proofs of this being path connected (though, I'm not exactly in love with any of those proofs) and this tells us we can let any point be our base-point.  Now, let $p$ be some point in ${\mathbb R}^{2} - {\mathbb Q}^{2}$ and let's let this be our base-point.  We can take one path from $p$ to $q$ and a second one from $q$ to $p$, and it's not hard to show that if these paths are different then there is at least one rational on the "inside" of it.  Since there are uncountably many $q$, this would seem to imply uncountably many different elements of the fundamental group; the problem I'm having is showing that two loops like we've described are actually different!  For example, a loop starting at $p$ and passing through $q$ should be different from a loop starting at $p$ and passing through $q'$ for none of these points the same, for at least an uncountable number of elements $q'$.  Is there some construction I should be using to show these elements of the fundamental group are different?
 A: In case anyone doesn't like the arguments from uncountability, here's a more concrete argument:
Fix a base-point $(p,p)$ for an arbitrary $p\in\mathbb{R}-\mathbb{Q}$. For any $q\in\mathbb{R}-\mathbb{Q}$, $q\not=p$, consider the loop $L_q$ consisting of the following line segments: $(p,p)\rightarrow(p,q)\rightarrow(q,q)\rightarrow(q,p)\rightarrow(p,p)$
We claim that for any $q_1<q_2$, the loops $L_{q_1}$ and $L_{q_2}$ are non-homotopic. To see this, pick a rational $r$ such that $q_1<r<q_2$. Embed $\mathbb{R}^2-\mathbb{Q}^2$ into $\mathbb{R}^2-(r,r)$ (aka, the punctured plane). Clearly, $L_{q_1}\equiv 0$ while $L_{q_2}\equiv 1$, so the loops are non-homotopic, as desired.
A: The way you get uncountably many loops is similar to the way you get uncountably many for the Hawaiian Earrings. Choose a sequence of loops $\alpha_i$ whose diameters approach $0$. Then consider the infinite concatenations $\alpha_1^{\epsilon_1}*\alpha_2^{\epsilon_2}\cdots*\alpha_n^{\epsilon_n}*\cdots$ where $\epsilon_n=\pm 1$. Since the diameters go to zero, this really represents a loop in the space. However, there are clearly uncountably many such loops. The tricky part is showing that no pair of them is homotopic.
A: So now that Jacob has explained why the space is path connected let us finish the problem. In his explanation he points out that there are an uncountable number of lines passing through $p$ and an uncountable number of lines passing through $q$ and since we are removing only countably many points not all of these lines can hit a point in $\mathbb{Q}^2$. So we can construct a path from $p$ to $q$ missing $\mathbb{Q}^2$, in fact more than one path. That gives us a loop, inside this loop will be an element of $\mathbb{Q}^2$ just by looking at the coordinates of $p$ and $q$. This shows that the loop is not contractible, ie it is a nontrivial element of $\pi_1(\mathbb{R}^2-\mathbb{Q}^2)$. Now you can proceed to construct more loops passing through $p$ and $q$ that are not contractible or homotopic to the previously constructed loop by picking different line segments as above. Now just construct another path from $p$ to $q$ and notice that an element of $\mathbb{Q}^2$ is inside each of your new loops (there should be at least 2 new loops, try drawing a picture).
Does this help?
Edit:
Here is, in my mind, a clearer construction of uncountably many non-homotopic loops. Let $p \in \mathbb{R}^2-\mathbb{Q}^2$ and $L$ a line in $\mathbb{R}^2-\mathbb{Q}^2$ not containing $p$. As established, there is a continuum of lines passing through $p$, and once we remove all of the ones passing through a point in $\mathbb{Q}^2$ we still have uncountably many. All remaining lines, with one possible exception, intersect $L$ at distinct points. We are left with a triangle for each pair of distinct remaining lines passing through $p$. There is a rational point inside each of these triangles (maybe this is something that was unclear, let me know and I think I can make this part more explicit). There is no way to move "homotope" the triangle past one of the rational points so that it would be homotopic to any adjacent triangle. Suppose two "loops" created in this fashion were homotopic, and suppose further that the differ in the second line passing through $p$, then there would have to be a homotopy between two paths which enclose a point of $\mathbb{Q}^2$ which is impossible.
If you want I can draw a picture on my tablet and try and post. (I don't really know the proper way to go about posting such a picture.)
