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In ZFC, does there exist an uncountable set of reals $A$ such that for every closed measure zero set of reals $B$, we have that $ A + B = \{a+b : a \in A, b \in B\} \neq \mathbb{R}$?

This question is motivated from the results in the following paper.

Note that under CH, there are such sets (Lusin/Sierpinski sets) and a model where every such set is countable must satisfy the Borel and the dual-Borel conjecture.

The fact that Sierpinski sets are strongly meager is due to Pawlikowsi. A more general result appears in Pawlikowski, Strongly meager sets and subsets of plane, Fundamenta Math., 156, 1998. Lusin sets are strongly null because for any sequence $\langle \epsilon_n : n \geq 1 \rangle$, the set $\bigcup \{(r_n - \epsilon_{2n}, r_n + \epsilon_{2n}) : n \geq 1\}$ , where $r_n$'s run over all rationals, covers all but countably many points in the Lusin set. Finally, if there is no uncountable set $A$ as above, then there cannot be any uncountable strongly meager/null set because if $A$ is uncountable strongly meager/null set then $A + B \neq \mathbb{R}$ for any $B$ which is both meager and null.

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    $\begingroup$ To complement the question, the recent paper by Goldstern, Kellner, Shelah, Wohofsky establishing the relative consistency of BC+dBC is here and on the arxiv. Also, Goldstern's overview of the proof is on the arxiv. $\endgroup$ – François G. Dorais Apr 15 '14 at 11:19
  • $\begingroup$ Could you add proofs to your notes? $\endgroup$ – Boaz Tsaban May 14 '14 at 22:33
  • $\begingroup$ I added some explanations and one reference. $\endgroup$ – Ashutosh May 15 '14 at 3:23
  • $\begingroup$ Could you point out some more implications among your property and more well-studied ones (e.g., selection principles)? $\endgroup$ – Boaz Tsaban Jan 17 '16 at 10:42
  • $\begingroup$ Did you consider a union of a $\mathfrak{b}$-scale (a ZFC object) and the rationals? That's the candidate I would try first. It is SMZ if, e.g., $\mathfrak{b}\le \text{cov}(\mathcal{M})$. It is perfectly meager, though I guess it is not necessarily strongly meager. But you seek for weaker properties, don't you? $\endgroup$ – Boaz Tsaban Jan 17 '16 at 10:45

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