How is $\mathbb R^2\setminus \mathbb Q^2$ path connected?

Question

Prove $(\mathbb R \times \mathbb R)-(\mathbb Q \times \mathbb Q)$ is path connected.

I know I need to let $(x_0, y_0), (x_1, y_1) \in (\mathbb R \times \mathbb R)-(\mathbb Q \times \mathbb Q)$ and then consider each of the cases where $x_0 \in \mathbb Q, x_1 \notin \mathbb Q$ and $x_0 \in \mathbb Q, x_1 \in \mathbb Q$ but I don't know where to go from there.

Answer 1

Hint: for every irrational number $\alpha\in \mathbb{R}\setminus\mathbb{Q}$, the lines

$$ \{\alpha\}\times\mathbb{R}, \qquad \mathbb{R}\times\{\alpha\} $$

belong to $\mathbb{R}^2 \setminus \mathbb{Q}^2$.

Answer 2

This is just the set of all points in the plane with at least one irrational coordinate. This space can be viewed as a union of horizontal lines and vertical lines which have an irrational intersection with the axes. We can move from one point in the space to another by moving along these lines, which implies it is path connected. We had this question in our topology class.