Prove that $\mathbb{R^2}\setminus\mathbb{Q}^2$ is connected.

First of all, I know this was asked many times over in this site.

However I'm looking for a proof that does not use path-connectedness (which I know, but this exercise is from a previous lecture), nor the knowledge that $\mathbb{R}^2\backslash E$, where $E$ is countable, is connected (it's the second part of the exercise).

I tried using disjoint open sets and separated sets, and also using the fact that they should also be the intersection of sets in $\mathbb{R}^2$ and $\mathbb{R}^2\setminus\mathbb{Q}^2$, but I hit a wall there.

I have the feeling I'm forgetting something very basic and obvious, or I may be overcomplicating things.



Suppose there are two disjoint open sets $U$ and $V$ whose union is all of $\mathbb{R}^2\setminus\mathbb{Q}^2$.

Consider any line $L$ of the form $\{p\} \times \mathbb{R}$ or $\mathbb{R} \times \{p\}$ for $p \in \mathbb{R}\setminus\mathbb{Q}$.

Show that $L \subset U$ or $L \subset V$.

(Hint: $U \cap L$ and $V \cap L$ are open subsets of $L$ in the subspace topology.)

  • $\begingroup$ ah, I think you beat me to it. :) +1 $\endgroup$ – Dustan Levenstein May 29 '17 at 1:09
  • $\begingroup$ @DustanLevenstein Yes, noticed yours showed up at about the same time. Good to have slightly different takes on the idea =). +1 to you as well. $\endgroup$ – aes May 29 '17 at 1:13
  • $\begingroup$ Got it done. Thanks for both of you! $\endgroup$ – AspiringMathematician May 29 '17 at 2:24
  • $\begingroup$ @aes one of the intersection must be empty. How do we proceed next? $\endgroup$ – MathMan May 5 '20 at 23:44

Hint: Assume $T := \mathbb R^2\setminus \mathbb Q^2 = U \cup V$ is a disjoint union of nonempty open subsets.

Since $\mathbb R \times \{\pi\} \subset T$ is connected, it must be a subset of either $U$ or $V$. Say it's contained in $U$. Now show that $\{r\} \times \mathbb R \subset U$ for every irrational $r$, and that $\mathbb R \times \{r\} \subset U$ as well.


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