# Is $\mathbb{R}$ the disjoint union of finitely many congruent dense sets?

Both the set of rational numbers $\mathbb{Q}$ and its complement are dense in $\mathbb{R}$, but the relationship between them is very asymmetric. For instance, the rationals are countable and have Lebesgue measure 0, whereas the irrationals are uncountable and have infinite Lebesgue measure. Is it possibly to decompose the real numbers into dense subsets in a more symmetric way, so that $\mathbb{R}$ can be written as a union of finitely many disjoint sets which can be mapped into each other by translation or reflection (i.e. are congruent)?

• Presumably, you want these sets to be disjoint? – Alex Wertheim Jan 25 '17 at 1:47
• Yes. I'll edit the question accordingly. – Ross Jennings Jan 25 '17 at 1:49
• @AlexWertheim: None of the sets can be open because the complement of a nonempty open set is not dense. – Jonas Meyer Jan 25 '17 at 2:13
• @Neal: the Banach-Tarski paradox doesn't work in $\mathbb R$ or $\mathbb R^2$. – TonyK Jan 25 '17 at 2:14
• Technically $\{\mathbb R\}$ is a partition of $\mathbb R$ into a finite number of congruent sets. To be more precise you should say "more than one". – bof Jan 25 '17 at 2:39

## 2 Answers

$$A=\bigcup_{n\in\mathbb Z}[2n,2n+1)$$ $$B=\bigcup_{n\in\mathbb Z}[2n+1,2n+2)$$ $$\mathbb R= [(A\cap\mathbb Q)\cup(B\setminus\mathbb Q)] \cup [(B\cap\mathbb Q)\cup(A\setminus\mathbb Q)]$$ The translation $x\mapsto x+1$ maps $[(A\cap\mathbb Q)\cup(B\setminus\mathbb Q)]$ onto $[(B\cap\mathbb Q)\cup(A\setminus\mathbb Q)].$

• Nice use of $\mathbb{Q}$ to get around the density issue. Also, +1 because this example works without choice! – Noah Schweber Jan 25 '17 at 2:19

Yes, this is doable, via a construction like that of the Vitali set but for integers. Indeed, with Choice we can get an uncountably dense example! (Note that bof's answer solves the problem as stated, without using Choice at all.)

For $x, y\in\mathbb{R}$, let $x\sim y$ if $x-y\in\mathbb{Z}$. Now via Choice we can get an uncountably dense transversal $T$ for $\sim$ - that is, $T$ is uncountably dense and contains exactly one real from each $\sim$-class. (Note that $[0, 1)$ is a non-dense transversal - the existence of a transversal, full stop, does not require choice.)

Now let $$A=\{t+2k: t\in T, k\in\mathbb{Z}\},\quad B=\{t+2k+1: t\in T, k\in\mathbb{Z}\}.$$ It's not hard to see that $B$ is gotten by shifting $A$ one unit (in either direction!), and that $A$ and $B$ are disjoint and cover $\mathbb{R}$.

• Awesome! Now I wonder if the answer depends on the axiom of choice... – Ross Jennings Jan 25 '17 at 2:20
• @RossJennings See bof's answer! – Noah Schweber Jan 25 '17 at 2:20
• Do we need Choice? $[0, 1)$ is a transversal. – user4894 Jan 25 '17 at 2:21
• But I guess choice is needed if you want the sets to be uncountably dense? – bof Jan 25 '17 at 2:22
• @RossJennings Between any two distinct points, there are uncountably many points from the set. For example, $\mathbb{Q}$ is dense but not uncountably dense, while $\mathbb{R}\setminus\mathbb{Q}$ is uncountably dense. – Noah Schweber Jan 25 '17 at 2:37