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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?

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@james: Isn't $\mathbb{R}^2-\mathbb{Q}^2$ a Baire space? So maybe suppose for contradiction that $\pi_{1}(\mathbb{R}-\mathbb{Q})$ is countable. – PEV Jan 10 '11 at 4:17
The proof of R^2-Q^2 being path connected is easy. Take any two points p,q in the space there are an uncountable number of lines that go through each of these points and only a countable number of points are removed. So take any two lines through p,q that aren't parallel and do not touch a point in Q^2 and you have a path from p to q. – JSchlather Jan 10 '11 at 4:39
Jacob, thank you --- this proof is SIGNIFICANTLY nicer than the ones I was looking at. – james Jan 10 '11 at 5:32
@tetrapharmakon: if a, b and c are linearly independent over the rationals, the line ax+by+c contains no rational points – Andrea Mori Feb 1 '11 at 18:07
There is even simpler proof for path-connectedness of $X=\mathbb R^2\setminus\mathbb Q^2$: each point of the type $(i,r)$ with $i$ irrational, $r$ rational can be connected to any point $(i',r')$ via the rectangular shaped path: $(i,r)\to(i,\pi)\to(i',\pi)\to(i',r')$, and even simpler for connecting $(i,r)$ to $(r',i')$. – mathreader Aug 18 '13 at 5:35
up vote 14 down vote accepted

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?


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.)

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"Now you can proceed to construct more loops passing through $p$ and $q$ that are not ... homotopic to the previously constructed loop by picking different line segments...". I agree with this completely, but I haven't been able to come up with a rigorous argument. Would you mind expanding on this a bit? – Jason DeVito Jan 10 '11 at 5:32
This is exactly what I wanted to do, but I was conflicted about its rigorousness: I'd need to prove that two loops are not equivalent, and that seemed like kind of a pain. In particular, I wanted to just prove that $R^2 - Q^2$ was hausdorff, and then go from there, but that kind of seemed like overkill. Edit: ha, apparently, Jason took the words out of my mouth. – james Jan 10 '11 at 5:33
How does this prove that $\pi_1(\mathbb R^2\setminus\mathbb Q^2)$ is uncountable and not just large? – Rasmus Jan 10 '11 at 18:36
I think it does give uncountable. The point is that uncountably many of the lines through a point $p$ miss $\mathbb{Q}^2$, and likewise through $q$. Assuming that each different choice of allowable pairs of lines gives a different element of $\pi_1$, you get your uncountable many elements. The validity of this assumption is very closely related to the question raised by james and I earlier - it seems intuitively plausible, but I'm still not sure of a rigorous proof. – Jason DeVito Jan 11 '11 at 2:02
Thread necromancy here, but for the benefit of anyone (like me) googling for this question: this argument is rigorous. Take two distinct loops as you've defined them; they are homotopic iff the composition of one with the inverse of the other is nulhomotopic, which is another such loop. So it suffices to show that the loops you've defined are not nulhomotopic. Take a rational point $q$ bounded by the loop: if the loop is nulhomotopic in $\mathbb{R}^2-\mathbb{Q}^2$, a fortiori it is nulhomotopic in $\mathbb{R}^2-\{q\}$, which we rigorously know is false. – forgetfulfunctor Jun 2 at 6:18

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.

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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.

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I'm pretty sure you get that the Hawaiian Earring is a retract from $\mathbb R^2\setminus \mathbb Q^2$. So you have an uncountable subgroup. – Gabriel Furstenheim Feb 1 '11 at 19:12
That's another way to say it, I agree. – Grumpy Parsnip Feb 1 '11 at 21:13
Is the infinite concatenation you wrote uncountable? – spencer Sep 10 '11 at 14:52
@spencer: No, but the set of such infinite concatenations is uncountable. – Grumpy Parsnip Sep 10 '11 at 16:42

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