# disjoint translates of a dense uncountable set

If {c(n)} is an arbitrary sequence of irrational numbers converging to 0 then Q + c(n), the set obtained by adding c(n) to the set of rational numbers Q, is clearly disjoint from Q for each n.

Is there an uncountable dense set of the real numbers, say D, for which a sequence {c(n)} converging to 0 exists such that D + c(n) does not intersect D for all n?

If such a set exists, and is borel, then it must have measure 0 since D - D would contain an interval if D had a positive measure.

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Does anyone know if the following problem (a special version of Erdos similarity problem) is still open: Is there a measurable set of reals with positive measure that doesn't contain a similar copy of the sequence (1/2)^n? – Ashutosh Jul 20 '10 at 19:32

Choose your favorite c(n), and define $D$ as any complementary $\mathbb{Q}$-vector space of $\mathrm{span}(\{c(n): n\in\mathbb{N}\})$ in $\mathbb{R}$ as $\mathbb{Q}$ vector space. This $D$ is not measurable, of course (countably many translates of it cover $\mathbb{R}$, so it can't be of measure zero, and on the other hand as you said it can't be of positive measure). It's a version of the construction of the Vitali set (actually the standard Vitali set already works with a sequence of rationals c(n) ).

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I am sorry, I didn't get you, could you expand a bit? – acarchau Jul 19 '10 at 21:29
I understood it Daniel's comment, thanks. – acarchau Jul 19 '10 at 21:32
expanded enough? – Pietro Majer Jul 19 '10 at 21:36
Yes. This is quite clear. – acarchau Jul 19 '10 at 21:39
note: the existence of a complementary space to $\mathbb{Q}$ requires the axiom of choice, but as François G. Dorais remarks, you just need an uncountable rational subspace with$D\cap \mathbb{Q}=(0)$, that can be exhibited explicitly. – Pietro Majer Jul 21 '10 at 7:49

Use of the Axiom of Choice is unnecessary here. One can write down an explicit $\mathbb{Q}$-linearly independent subset $\mathcal{T}$ of $\mathbb{R}$ with size $2^{\aleph_0}$ as I wrote in this answer. If it so happens that some of the $c(n)$ are in the $\mathbb{Q}$-span of this set $\mathcal{T}$, then remove the finitely or countably many elements which are involved in expressing the $c(n)$ as a $\mathbb{Q}$-linear combination of elements of the original set $\mathcal{T}$; what remains of $\mathcal{T}$ still has size $2^{\aleph_0}$. The $\mathbb{Q}$-span of this remainder has the desired property. The set in question will be a Borel set (of Lebesgue measure 0).

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Yes. We proceed using the axiom of choice. Choose an basis of $\mathbb{R}$ with cardinality $|\mathbb{R}/\mathbb{Q}|$ and a bijection between this basis and the elements of $\mathbb{R}/\mathbb{Q}$. Pick an element from each coset of $\mathbb{Q}$ contained in the corresponding basis set (we may do this as each coset is dense); this will be our set $D$. It is dense as each basis set has non-empty intersection with $D$. Now any sequence of rationals tending to zero works as $c(n)$, as if $D+c(n)$ had non-empty intersection with $D$, then two elements of $D$ would differ by a rational.

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Ah, I see Pietro's answer is identical to mine (and preceded mine chronologically); note that it also requires the axiom of choice. – Daniel Litt Jul 19 '10 at 21:29
Why should D be dense? – acarchau Jul 20 '10 at 15:15
You're right, I missed that. Should be fixed now. – Daniel Litt Jul 20 '10 at 15:47