Can someone please let me know if my solution is correct:
Define:
1) Let $A = \{x \in \mathbb{R}^{2}: \text{all coordinates of x are rational} \}$. Show that $\mathbb{R}^{2} \setminus A$ is connected.
My answer: just note that $A = \mathbb{Q} \times \mathbb{Q}$ so countable and there's a standard result that $\mathbb{R}^{2}$ minus a countable set is path-connected so connected, thus the result follows.
2) $B = \{x \in \mathbb{R}^{2}: \text{at least one coordinate is irrational} \}$. Show that $B$ is connected. Well isn't $B$ just $B = \mathbb{R}^{2} \setminus \mathbb{Q}^{2}$ so it is exactly the same problem as above isn't it?
This is a problem in Dugundji's book.