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I'm working through Hatcher book and done $\pi_1(\mathbb{R}^2 - \mathbb{Q}^2)$ is uncountable. It's easy to see that it's true as you can imagine only non trivial maps contract in the space.

But, was wondering has anyone worked out $\pi_2(\mathbb{R}^2 - \mathbb{Q}^2)$. $\pi_2(\mathbb{R}^3 - \mathbb{Q}^3)$ I imagine this isn't that hard, shouldn't it be uncountable aswell.

Just doing a project on higher homotopy theory, but fundamental group stuff seems hard already. Like $\pi_2$ is that just a sphere doing a weird thing and looping back on itself. I know the definition but can't really see it. Plus the calculation aren't that easy.

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    @Mariano: really? It seems to me easier to see the group structure using the definition in terms of homotopies.2019-02-26

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I asked my colleague Jurek Dydak about this, and he pointed out the following paper

Topology and its Applications Volume 120, Issues 1–2, 15 May 2002, Pages 23–45 One-dimensional sets and planar sets are aspherical J. W. Cannon , G. R. Conner, and Andreas Zastrow

which shows that every subset of the plane has trivial $\pi_k, k\geq 2$. So in particular $\pi_2(\mathbb R^2\setminus \mathbb Q^2)=0$.

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    Amazing result!2012-02-17