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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)?

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    Presumably, you want these sets to be *disjoint*? – 2017-01-25
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    Yes. I'll edit the question accordingly. – 2017-01-25
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    only translation? – 2017-01-25
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    I suppose reflection would be fine too, but I want them to be congruent, and I would be fairly surprised if there was an example where the congruence could be accomplished by a reflection but not by a translation. – 2017-01-25
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    I should add I'm assuming this can't be done unless the sets are pretty messy and hard to draw, in the way of $\mathbb{Q}$ (for example, trying to decompose the real line into two disjoint, congruent rays fails because if the rays are open, you have to leave out a point, and if they are closed, you have to cover a point twice). Then again, I could be wrong about that. – 2017-01-25
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    Actually, I just realized I am wrong about that -- the complement of the union of half-open intervals $[2n,2n+1)$, for $n\in\mathbb{Z}$, is congruent to the original set. I guess I should specify that the sets should be dense. – 2017-01-25
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    You already did :-) – 2017-01-25
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    Well, not in the title. :) – 2017-01-25
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    It seems like a Baire Category Theorem argument might be effective here. In particular, if we restrict to (finitely many) sets which are translates of each other, this is the same as considering finitely many translates of a single set $U$. Since translation is an open map, if one could prove that $U$ is open, then the translates of $U$ must be open. In this case, BCT gives a negative answer: the result would be finitely many open dense sets whose intersection is empty, a contradiction. However, based on the lack of answers so far, this naive approach is probably either false or ineffective.. – 2017-01-25
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    Perhaps see if you can adapt the construction in the proof of the Banach-Tarski paradox. – 2017-01-25
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    @AlexWertheim: None of the sets can be open because the complement of a nonempty open set is not dense. – 2017-01-25
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    Well if you ever decide you can live with countably many congruent dense sets, I wrote up a great answer that I had to delete before it got more negs! – 2017-01-25
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    @Neal: the Banach-Tarski paradox doesn't work in $\mathbb R$ or $\mathbb R^2$. – 2017-01-25
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    @JonasMeyer: doh! That's why that approach wasn't working, I suppose. :) Thanks! – 2017-01-25
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    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". – 2017-01-25
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    @bof True. Although it only vacuously consists of disjoint sets... – 2017-01-25

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

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    Nice use of $\mathbb{Q}$ to get around the density issue. Also, +1 because this example works without choice! – 2017-01-25
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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 $\mathbb{R}$.

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    Awesome! Now I wonder if the answer depends on the axiom of choice... – 2017-01-25
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    @RossJennings See bof's answer! – 2017-01-25
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    Do we need Choice? $[0, 1)$ is a transversal. – 2017-01-25
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    But I guess choice is needed if you want the sets to be *uncountably* dense? – 2017-01-25
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    @user4894 Yes, but it's not a dense transversal. You can get a dense transversal without choice (similarly to what bof did); choice, though, does appear necessary for an uncountably dense transversal. – 2017-01-25
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    @bof Yes, at least it appears necessary right now . . . – 2017-01-25
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    @NoahSchweber I see, thanks. – 2017-01-25
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    What do you mean by "uncountably dense"? I'm not familiar with that term. – 2017-01-25
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    @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. – 2017-01-25
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    OK. Thanks for explaining. – 2017-01-25