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.

  • $\begingroup$ "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}$. $\endgroup$
    – cch
    Jan 14 '11 at 6:49
  • $\begingroup$ 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. $\endgroup$
    – user2959
    Jan 14 '11 at 6:50
  • $\begingroup$ @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? $\endgroup$
    – undergrad
    Jan 14 '11 at 6:55
  • 6
    $\begingroup$ 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. $\endgroup$
    – user2959
    Jan 14 '11 at 6:56

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.

  • $\begingroup$ Part of why I added in the proof is that part of an exam I was doing was find an interesting Topology question here and answer it for the exam (and if it didn't have an answer post it here as well). Also I think it may be helpful for someone who comes across this question in the future and wonders about james's first comment on the question. $\endgroup$
    – user171177
    Dec 13 '14 at 3:20

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