Connectedness of sets in the plane with rational coordinates and at least one irrational

Can someone please let me know if my solution is correct:


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.

Solutions Collecting From Web of "Connectedness of sets in the plane with rational coordinates and at least one irrational"

Removing this from the unanswered list. As the comments note the OPs reasoning is correct.

To run with the comment of a proof using the definition of connectedness.

Suppose that $\mathbb{R} \backslash A$ can be written as $U \cup V$ where $U$ and $V$ are both open and disjoint. We know that there are open sets $U’$ and $V’$ in $\mathbb{R}$ such that $U = U’ \cap \mathbb{R} \backslash A$ and $V = V’ \cap \mathbb{R} \backslash A$. The fact that $\mathbb{R}$ is connected tells us that $U’ \cap V’$ is not empty. It is clear that $U’ \cap V’ \subset A$ (otherwise $U \cap V$ isn’t empty) so we can choose some point $x=(x_1 , x_2)$ of $A$ in $U’ \cap V’$. Now $U’ \cap V’$ is open in $\mathbb{R}$ so there exists some $\epsilon$ such that $B_\epsilon(x) \subset U’ \cap V’$. There exists a rational $q$ such that $\vert q – x_1 \vert < \epsilon $ and then there exists an irrational number $s$ such that $s \in (q , x_1)$ and therefore $(s,x_2) \in B_\epsilon(x)$ so that $B_\epsilon(x) \cap \mathbb{R}\backslash A \not = \emptyset$ but this would mean that $U \cap V$ is not empty which is a contradiction so $\mathbb{R} \backslash A$ must be connected.