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Are there sets $S_i\subseteq\mathbb{R}$ with $i\leq n$ such that

  1. $S_i$ are disjoint,
  2. $S_i$ have same cardinality,
  3. $S_i$ are dense in $\mathbb{R}$?
Prajwal Kansakar
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Analysis
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    You can do better: $\mathbb R $ is the union of $|\mathbb R|$ many pairwise disjoint sets, all of them meeting any interval on a set of size $|\mathbb R| $. – Andrés E. Caicedo Oct 17 '14 at 15:34

3 Answers3

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Partition the set of cosets of $\mathbb{Q}$ into $n$ sets of equal uncountable cardinality, and take the union of each partition element. The resulting partition of $\mathbb{R}$ consists of $n$ sets, each contains a coset of $\mathbb{Q}$ and so is dense, and their cardinalities are equal.

Lee Mosher
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  • How can I partition $\mathbb{R}/\mathbb{Q}$? – Analysis Oct 17 '14 at 15:20
  • @Analysis: I suppose you could well-order $\mathbb{R}/\mathbb{Q}$ and then define a function $f : \mathbb{R}/\mathbb{Q} \to \mathbb{Z}/n\mathbb{Z}$ with the property that if $f(X)=i$ then $f(X+1)=i+1$ modulo $n$. – Lee Mosher Oct 17 '14 at 15:25
  • Thanks. I have more question in advance. Is each partition Lebesgue measurable? If not, is there other construction to make them measurable? – Analysis Oct 17 '14 at 15:33
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Split $\mathbb{Q}$ into $n$ dense subsets, where $p/q$ in lowest terms goes into the subset $p\pmod n$. Let the $i$th set be the union of the irrational numbers $\{x\in\mathbb{R}\backslash\mathbb{Q}|\lfloor x\rfloor=i\pmod n\}$ with one of the rational subsets.

Empy2
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Write every number in base $n$. Then $S_1$ is the set of the numbers that have only a finite number of digits $\geq 2$. $S_2$ is the set of the numbers that are not in $S_1$ and have only a finite number of digits $\geq 3$, etc. $S_n$ is the set of the numbers that are not in any of the former.

ajotatxe
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