8
$\begingroup$

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,

$\endgroup$
5
  • 3
    $\begingroup$ What is the Borel property of strong measure zero? What is the reference that is not in English? $\endgroup$
    – Goldstern
    Jan 16, 2014 at 22:47
  • 2
    $\begingroup$ 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. $\endgroup$ Jan 17, 2014 at 13:05
  • 2
    $\begingroup$ Baillif's answer is correct and complete. Is there a reason why this question appears in MO as unanswered? $\endgroup$ Apr 28, 2014 at 1:00
  • $\begingroup$ 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! $\endgroup$
    – Student
    Apr 30, 2014 at 12:18
  • $\begingroup$ @Student: Checkmark the answer below. $\endgroup$ Aug 17, 2015 at 21:54

1 Answer 1

2
$\begingroup$

To help with @Student's last comment, I post here Baillif's answer.

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.

– Mathieu Baillif

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.