# “Nicely” strong measure zero sets

This question is essentially an expanded version of the unanswered half of Two strengthenings of "strong measure zero".

A set $X$ of reals is strong measure zero if, for any $f: \omega\rightarrow\omega$, there is a sequence of rational open intervals $I_n$ such that

• $X\subseteq \bigcup_{n\in\omega}I_n$, and

• $m(I_n)=2^{-f(n)}$.

(This is not the usual formulation, but equivalent; and it makes the precise phrasing below nicer.)

My question is about what happens when we try to continue dissecting this class of sets - are there strong measure zero sets which are more "easily" strong measure zero?

(Note: in this question, "easily" is interpreted topologically - the question Antirandom reals addresses what happens if we interpret it computability-theoretically .)

Specifically, to each strong-measure-zero set $X$ we associate a relation $N_X\subseteq\omega^\omega\times\omega^\omega$, where $(f, g)\in N_X$ if $g$ is a cover of $X$ by rational open intervals such that the $n$th element of $g$ has width $2^{-f(n)}$. (Here we fix some nice bijection between $\{$rational open intervals$\}$ and $\omega$.) Intuitively, we'll say that $X$ is nicely strong measure zero if the relation $N_X$ has a topologically nice selector - that is, a map $F: \omega^\omega\rightarrow\omega^\omega$ such that for all $f$ we have $(f, F(f))\in N_X$. (Here, we use the standard topology on $\omega^\omega$.) Let's call such an $F$ a modulus for $X$.

For instance, if $X$ is countable, then $X$ has a continuous modulus. Note that this means that, consistently, we don't get any new information this way: "Every strong measure zero set is countable" (=Borel's conjecture) is consistent with $ZFC$. My question is whether - consistently! - there might be some interesting structure we can tease out by looking at the moduli.

One instance of this:

• Is it consistent with ZFC that there are strong measure zero sets $X$ which (a) are uncountable and (b) have continuous moduli?

For that matter, I don't know the answer to the converse question:

• Is it consistent with ZFC that there are strong measure zero sets $X$ which do not have continuous moduli?

The continuous/discontinuous divide is the most natural one to me, but there are lots of others - e.g., Borel/non-Borel (or level-by-level in the Borel hierarchy), etc. And, of course, if we drop choice, there's the question of whether moduli exist at all.

• Is it consistent with ZF (or ZF+DC, ambitiously) that there is a strong measure zero set with no modulus?

(Note that models of determinacy won't help us here - AD implies that every uncountable set contains a perfect subset, and ZF proves that perfect sets don't have strong measure zero.)

This is, of course, a lot of questions; I'm most interested in the first bulleted question above (does continuous smz imply countability?), but I would love any information at all.

• Could you supply some motivation for this question? The reason I see a need for one is that continuity deals with the head of the sequence, whereas covering depends on the tail, so I see no clear connection. The question makes sense, but I seek for motivation. – Boaz Tsaban Dec 14 '15 at 17:09
• Please do not replace the "descriptive set theory" tag (which is appropriate, and well-used) with "selection principles" (which is appropriate, but only has 8 questions attached to it currently). – Noah Schweber Jan 11 '16 at 4:43
• @BoazTsaban Sorry for the late reply, I only just saw your comment. This question is motivated partly by the linked questions (especially Andreas' answer to "Two strengthenings"), as well as general descriptive set theoretic instincts: saying "$A$ is strong measure zero" asserts the existence of a functional with certain properties, and it's natural to ask how the complexity of that functional will affect the set $A$. In particular, "simple" witnessing functionals ought to imply that $A$ is countable (assuming reasonable set-theoretic hypotheses). That's what I'm interested in here. (cont'd) – Noah Schweber Jan 11 '16 at 4:45
• "Continuity" is just the lowest level of the hierarchy, so that's where I start. – Noah Schweber Jan 11 '16 at 4:46
• It seems to me that the construction that I sketch in mathoverflow.net/questions/63497/… will give a continuous function . (I have to check the details, because that construction uses sequences of names, not sequences of natural numbers.) – Goldstern Dec 13 '16 at 17:27