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Problem. Let $\lambda$ be the standard product measure on the Hilbert cube $[-\frac12,\frac12]^\omega$ and $A,B$ be two $\lambda$-positive Borel subsets of $[-\frac12,\frac12]^\omega$. Is it true that the sum-set $A+B=\{a+b:a\in A,\;b\in B\}\subset [-1,1]^\omega$ contains a homothetic copy $s+\varepsilon\cdot [-1,1]^\omega$ of the Hilbert cube for some $s\in[-1,1]^\omega$ and $\varepsilon>0$?

Remark. It can be shown that for any sequence $(\varepsilon_n)_{n\in\omega}\in\ell_1$ the sum-set $A+B$ contains a cube $s+\prod_{n\in\omega}[-\delta_n,\delta_n]$ with $\delta_n=|\varepsilon_n|$ for all sufficiently large $n$.

But this is rather far from the required $\varepsilon_n=\varepsilon$ for all $n\in\omega$.

By the way, is the restriction $(\varepsilon_n)\in\ell_1$ the best possible, or it can be improved to, say, $(\varepsilon_n^2)\in\ell_2$?

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  • $\begingroup$ Is there an example of a subset with positive measure that does not contain such a tiny Hilbert cube ? $\endgroup$
    – HenrikRüping
    Commented May 19, 2018 at 13:55
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    $\begingroup$ @HenrikRüping Yes: each set of positive measure contains a zero-dimensional compact set positive measure. This zero-dimensional set cannot contain a copy of the Hilbert cube. The existence of a zero-dimensional set of positive measure follows from the observation that the space contains many spheres of zero measures (by the $\sigma$-additivity of the measure). $\endgroup$
    – Taras Banakh
    Commented May 19, 2018 at 14:08
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    $\begingroup$ Obviously not. Take any square summable but not summable sequence $a_k$ and consider the set $A$ (of full measure) on which $\sum_k a_k(x_k-\frac 12)$ converges. No matter how many times you add it to itself, there won't be any $\varepsilon$-cube in it. $\endgroup$
    – fedja
    Commented May 19, 2018 at 17:09
  • $\begingroup$ @fedja Thank you for the comment. Could you explain why your set $A$ has full measure? $\endgroup$
    – Taras Banakh
    Commented May 19, 2018 at 20:36
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    $\begingroup$ Erm... The sum of independent symmetric random variables with convergent sum of variances converges almost surely, doesn't it? $\endgroup$
    – fedja
    Commented May 20, 2018 at 0:05

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