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Show that $\pi_1(\Bbb R^2 - \Bbb Q^2)$ is uncountable.

Since $\Bbb Q^2$ is countable we can enumerate the elements as $q_1, \dots, q_k$ and now $\Bbb R^2 - \{q_1, \dots, q_k\}$ is essentially the plane with countable many holes. Now $$\Bbb R^2 - \{q_1, \dots, q_k\} \simeq \bigvee_{i=1}^k S^{1}$$ so by Van-Kampen's theorem $$\pi_1(\Bbb R^2 - \Bbb Q^2) \cong \pi_1\left(\bigvee_{i=1}^k S^{1}\right) \cong \Bbb Z \ast \dots \ast \Bbb Z \text{ (k times)}$$

so this would be the same thing as showing that the free product of the integers $k$ times is an uncountable set? However I think this is a false statement as the countable free product of countable sets is countable so what is going on here?

Arctic Char
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Emil Grönberg
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    Just because $\mathbb{Q}$ is countable does not mean $\mathbb{Q}$ is finite. There is no enumeration of $\mathbb{Q}$ that consists of only $k$ elements. – Tob Ernack Jul 21 '22 at 12:27

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$\Bbb Q^2$ is countably infinite.

So you actually get the so-called Hawaiian earring for the homotopy type of $\Bbb R^2-Q^2$.

Its fundamental group is uncountable.

The group is called the Hawaiian earring group. $\Bbb Z^\Bbb N$ can be embedded in it, which proves it's uncountable.

suckling pig
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