Show that \(\pi_1(\mathbb{R}^2 - \mathbb{Q}^2)\) is uncountable.
Solution
1. \(\mathbb{R}^2 - \mathbb{Q}^2\) is the plane with the countable dense set \(\mathbb{Q}^2\) removed.
\(\mathbb{Q}^2\) is countable.
2. For each irrational \(\alpha\), the vertical line \(L_\alpha = \{(\alpha, y) : y \in \mathbb{R}\}\) is contained in \(\mathbb{R}^2 - \mathbb{Q}^2\) (since \(\alpha \notin \mathbb{Q}\)).
a point \((\alpha, y)\) is in \(\mathbb{Q}^2\) only if \(\alpha \in \mathbb{Q}\).
3. For each irrational \(\alpha\), choose a basepoint \(p_\alpha = (\alpha, 0)\) and a loop \(\gamma_\alpha\) that goes around the point \((\alpha, 0)\) once (a small circle).
construct a loop.
4. The loops \(\gamma_\alpha\) for distinct irrationals \(\alpha\) represent distinct elements of \(\pi_1(\mathbb{R}^2 - \mathbb{Q}^2)\).
the winding number of \(\gamma_\alpha\) around a point \((\beta, 0)\) is \(1\) if \(\beta = \alpha\) and \(0\) otherwise; since winding number is a homotopy invariant, distinct \(\alpha\) give non-homotopic loops.
5. There are uncountably many irrationals \(\alpha\).
\(\mathbb{R} \setminus \mathbb{Q}\) is uncountable.
6. Hence \(\pi_1(\mathbb{R}^2 - \mathbb{Q}^2)\) contains uncountably many distinct elements, so it is uncountable.
4 and 5.
7. Q.E.D.
6.