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Is it possible to partition $\mathbb R$ into an countable number of disjoint dense subsets with the same cardinality?

Furthermore, is it possible to partition the reals into an uncountable number of disjoint dense subsets with the same cardinality?

This is a follow up question on an old question that was answered here.

Can $\mathbb{R}$ be partitioned into $n$ dense sets with same cardinality?

There, someone was able to construct a partition of $\mathbb R$ into $n$ dense subsets of the same cardinality.

Community
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Patrick Tam
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    For the second question, you can partition the reals into singletons... – KotelKanim Aug 30 '15 at 13:42
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    Or, again for the second, given any additive countable subgroup $G$ of $\mathbb R$ which is not all of $\mathbb R$, (say $G=\mathbb Q$,) you can partition it as $r\sim s$ iff $r-s\in G$. The partitions then amount to cosets of $G$. If $|G|<|R|$ then there an an uncountable number of cosets, all of the same cardinality as $|G|$. (Kotel's case is just $G={0}$.) – Thomas Andrews Aug 30 '15 at 13:57
  • I'm sorry, I misstated my question. I of course want the subsets to be dense. – Patrick Tam Aug 30 '15 at 13:57
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    Then my second example, with $G=\mathbb Q$, works for the second part. Or any dense $G$ with $|G|<|\mathbb R|$. – Thomas Andrews Aug 30 '15 at 13:58
  • Oh, I see. Thank you very much. – Patrick Tam Aug 30 '15 at 14:00
  • Alright, I think your answer to the second question answers my first question. Basically, just take a countable subset of the uncountable cosets. Then take the complement of the union of that countable subset in $\mathbb R $. – Patrick Tam Aug 30 '15 at 14:04
  • @PatrickTam could you explain that more clearly, with symbols please? I don't follow it. – Patrick Stevens Aug 30 '15 at 14:10
  • Let $S$ be Thomas Andrews' set of cosets. We pick a subset $U \subset S$ such that $U$ is of countable cardinality. If we take the union of all cosets of $U$, we get some subset $V$ of $\mathbb R$. Take the complement of $V$ in $\mathbb R$. Call it $V^C$. $V^C$ has to be dense since it is the union of all the other elements in $S$. Then taking the cosets of $U$ and $V^C$ will give us a partition of $\mathbb R$ into countable disjoint dense subsets. – Patrick Tam Aug 30 '15 at 14:17
  • ahh, I see the error in my reasoning. $V^C$ is not the same cardinality as the other elements of $U$. – Patrick Tam Aug 30 '15 at 14:26

2 Answers2

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For the first part, find a $\mathbb Q$-basis of the real numbers. This requires the axiom of choice or something similar. Then remove one of the basis element, and you get a subgroup $G$ of $\mathbb R$ with $\mathbb R/G\cong\mathbb Q$. Show that $G$ is dense.

The partition elements are then the cosets of $\mathbb G$ as a subgroup of $\mathbb R$ - that is $x\sim y\iff x-y\in G$.

A similar technique also works for the second question, but a much easier approach there is to take the cosets of $\mathbb Q$.

Thomas Andrews
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  • $G$ is obviously dense, since it is a non-empty $\mathbb Q$-vector subspace of $\mathbb R$, so it contains $\mathbb Qz$ for some $z\neq 0$, which is dense. @NateEldredge Of course $G$ needs to be dense, because the partition is the cosets, the partitions are supposed to be dense, and $G$ is one of the partitions. – Thomas Andrews Aug 30 '15 at 14:36
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    And how does this work for the second question, which wants uncountably many sets? – Nate Eldredge Aug 30 '15 at 14:37
  • Sorry, I mixed up the basis with the subspace it generates. – Nate Eldredge Aug 30 '15 at 14:38
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    Yeah, I was wrong there. @NateEldredge You need to split the basis then into two uncountable sets, and generate $G$ from one of those. – Thomas Andrews Aug 30 '15 at 14:38
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Consider the binary expansion of a number. (Here, I'm using "finitely many" to include zero.)

  • Let $A_0$ be the set of those numbers that have finitely many $01$s in the expansion.
  • Let $A_1$ be those that have infinitely many $01$s but finitely many $011$s in the expansion.
  • Let $A_2$ be those that have infinitely many $01$s and $011$s but finitely many $0111$s.
  • etc.
  • Let $A_\omega$ be those that have infinitely many $01$s, $011$s, $0111$s, etc.

Does this work?

Akiva Weinberger
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