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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.

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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.

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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.

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