# 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:

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.

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"At least one coordinate is irrational" means that one coordinate could be rational, i.e. $B$ isn't the same as $\mathbb{R}^{2} \setminus \mathbb{Q}^{2}$. –  cch Jan 14 '11 at 6:49
If this is homework, the professor may want you to explicitly show that fact in (1). Either way, the path-connected way is nice. It may be kind of fun to show this right from the definition of connectedness, too. –  james Jan 14 '11 at 6:50
@james: yeah, thank you to both. Actually this is not homework but practice for the exam, homework is a little bit easier. Yeah, I got confused with 2). How can I solve 2)? Can you please give a hint? –  undergrad Jan 14 '11 at 6:55
cch, if one coordinate is rational and one is irrational, that element, $(p,x)$, say, is not in ${\mathbb Q}^{2}$. Conversely, if both coordinates are rational, the element is not in $B$, so I think undergrad's characterization of $B$ is correct. –  james Jan 14 '11 at 6:56