A recent issue of American Math. Monthly has a paper that partitions $\mathbb{R}$ into an arbitrary finite number of uncountable sets such that every real number is a condensation point of all the sets in the partition. In fact, we can find uncountably many uncountable subsets $S_\alpha$ of $\mathbb{R}$ that are (1) pairwise disjoint, (2) have union $\mathbb{R}$, and (3) every real number is a condensation point of all the sets. One way to do this is the following. For any infinite binary sequence $\alpha=(a_1, a_2, \dots)$, let $S_\alpha$ be the set of all real numbers whose fractional part has the binary expansion $0.b_1b_2\cdots$ such that the sequence $(b_2,b_4,b_6,\dots)$ differs from $\alpha$ in only finitely many terms. When $\alpha=(1,1,1,\dots)$ we must remove from $S_\alpha$ all numbers whose binary expansion ends in infinitely many 1's, due to the nonuniqueness of the binary expansion. The set of distinct $S_\alpha$'s then has the desired properties. This partition of $\mathbb{R}$ also has the property that any two of the sets $S_\alpha$ and $S_\beta$ are translates of each other, except for $S_{1,1,1,\dots}$. This suggests the question: does there exist a partition of $\mathbb{R}$ satisfying (1), (2), and (3) such that every pair of sets in the partition are translates?
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2(Not very constructive) $\mathbb{R}$, as a vector space over $\mathbb{Q}$, may be decomposed in direct sum of $\mathbb{Q}$-vector subspaces, $S\oplus T$, both of continuum dimension. Then we can take $S_\alpha:=S+\alpha$ for $\alpha\in T$. – Pietro Majer May 21 '18 at 22:13
1 Answers
I think the following works:
Fix an uncountable subgroup $G$ of $(\mathbb{R}, +)$ - note that any such is dense in $\mathbb{R}$, and in fact intersects each nonempty open interval uncountably often - of uncountable index, and consider the set of cosets of $G$.
(Such a group can be produced by transfinite recursion using a well-ordering of $\mathbb{R}$: alternate between throwing reals into the group, and barring reals from entering the group.)
Note that the argument above relies on the axiom of choice. However, this isn't necessary: provably in ZF, there is a $\mathbb{Q}$-linearly independent set $S$ of reals of size continuum. We can split this into two disjoint sets $S_1, S_2$ of size continuum (pick some appropriate real $a$ and consider the points to the left of $a$ versus the right of $a$). Now consider the $\mathbb{Q}$-vector space generated by $S_1$.
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1It might be worth crediting @PietroMajer's comment for the idea in your edit. – Greg Martin May 21 '18 at 22:34
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3@GregMartin My answer actually preceded that comment - note the timestamps. – Noah Schweber May 21 '18 at 22:37
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Your answer did, but the second section was added after the comment. – Greg Martin May 22 '18 at 03:43
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1@GregMartin The content of the second section was the provability of the claim without choice, which was not present in the comment. My answer was not dependent on or inspired by the (quite correct) comment in question. – Noah Schweber May 22 '18 at 07:21
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(And in fact I didn't see Pietro's comment until you pointed it out to me.) – Noah Schweber May 22 '18 at 07:27