# Strong measure zero sets and selection principles

A set of reals $X$ is strong measure zero if for any sequence of positive real numbers $( \epsilon_n ) _{n \in \omega }$ there is a sequence of open intervals $( a_n ) _{n \in \omega }$ which covers $X$ and such that each $a_i$ has length less than $\epsilon _i$.

Does anyone know where can I find a proof that the selection principle $\mathcal S_1(\mathcal O,\mathcal O)$ implies Borel property of strong measure zero? I have seen this in The Combinatorics of open covers, but the reference there is to an article which is not in English. Thank you,

• What is the Borel property of strong measure zero? What is the reference that is not in English? Jan 16, 2014 at 22:47
• The proof is very short, actually: given the sequence $\epsilon_i$, let $\mathcal{U}_i$ be the cover of $X$ by open intervals of length $\epsilon_i$. Then $\mathcal{S}_1(\mathcal{O},\mathcal{O})$ ensures that there is a cover of $X$ consisting of (at most) one member of each $\mathcal{U}_i$. A more general statement is proved in Theorem 9 in "Finite powers of strong measure zero sets" by M. Scheepers. Jan 17, 2014 at 13:05
• Baillif's answer is correct and complete. Is there a reason why this question appears in MO as unanswered? Apr 28, 2014 at 1:00
• Thank you all for your comments. The answer is very clear. but, I don't know how to mark it as answered. If anyone could please mark it. Thank you! Apr 30, 2014 at 12:18
• @Student: Checkmark the answer below. Aug 17, 2015 at 21:54

The proof is very short, actually: given the sequence $\epsilon_i$, let $U_i$ be the cover of $X$ by open intervals of length $\epsilon_i$. Then $S_1(O,O)$ ensures that there is a cover of $X$ consisting of (at most) one member of each $U_i$. A more general statement is proved in Theorem 9 in "Finite powers of strong measure zero sets" by M. Scheepers.