A set $X\subseteq\mathbb{R}$ is *strong measure zero* if, for every sequence $(\epsilon\_i)\_{i\in\mathbb{N}}$ of positive reals, there is a sequence $(I\_i)\_{i\in\mathbb{N}}$ of open intervals covering $X$ such that $\mu(I\_n)<\epsilon\_n$.


Looking at various computability theoretic questions around strong measure zero (see e.g. <https://math.stackexchange.com/questions/1446602/anti-random-reals>), I've run into two possible strengthenings; I'm curious what is known about them.


For the first, we demand that the cover depend on the sequence one bit by bit:


* $X\subseteq\mathbb{R}$ is *strategically strong measure zero* if player II has a winning strategy in the following game: on move $n$, player I plays a positive real $\epsilon\_n$, and player II plays an interval $I\_n$ with $\mu(I\_n)<\epsilon\_n$; and player II wins if the $I\_n$ form a cover of $X$.


For the second, we merely ask that the cover depend on the sequence continuously. Let $Eps$ be the set of infinite sequences of positive reals, and let $Int$ be the set of infinite sequences of open intervals in $\mathbb{R}$, each topologized as usual.


* $X\subseteq\mathbb{R}$ is *continuously strong measure zero* if there is a continuous $F$ from $Eps$ to $Int$ such that, for every $f\in Eps$, $F(f)$ is a cover of $X$ and $\mu(F(f)(n))<f(n)$.


Clearly strategically strong measure zero implies continuously strong measure zero implies strong measure zero; consistently all strong measure zero sets are countable (this is *Borel's conjecture*), so no nonimplications can be proved over ZFC. My question is whether we can say anything else; in particular,



> 
> Are any other implications provable over ZFC+A, where A is some reasonable axiom which does not imply Borel's conjecture?
> 
> 
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Note that ZFC proves that there are continuum many continuously strong measure zero sets $S\_r$ such that every continuously strong measure zero set is contained in one of the $S\_r$. This seems likely to not be the case in general for strong measure zero sets in ZFC, so I suspect that the situation is not entirely trivial.


EDIT: To clarify, this informal argument merely suggests that not every strong measure zero set is continuously strong measure zero; as Andreas' answer below shows, it's quite likely that every continuously strong measure zero set is *countable*, which would be triviality in the other direction.


FURTHER EDIT: See ["Nicely" strong measure zero sets](https://mathoverflow.net/questions/220687/nicely-strong-measure-zero-sets) for a continuation of this question.



Two strengthenings of "strong measure zero"
Strategically strong measure zero is equivalent to countable. To prove the nontrivial direction, suppose $X$ is strategically strong measure zero and $s$ is a winning strategy for player II. Consider the tree of finite plays in which player I plays only rational numbers and player II follows $s$. For any partial play $p$ in which player I is to move next (i.e., the length of $p$ is even), say that an element $x\in X$ is killed at $p$ if, no matter what (rational) move player I makes next, player II's response (using $s$) is an interval that covers $x$. 


I claim first that, at any $p$, at most one $x$ is killed. The reason is that, if $x\neq x'$, then player I could play an $\epsilon<| x-x'|$ and then player II's response would be an interval too short to cover both $x$ and $x'$.


Since the tree of partial plays is countable (that's why I restricted player I to rational $\epsilon$'s), only countably many $x$'s are killed anywhere in the tree.


To finish the proof, I claim that every element of $X$ is killed somewhere in the tree. Suppose, toward a contradiction, that $x$ is a counterexample. Then, at the initial position, player I can move so that II, using $s$, doesn't immediately cover $x$ (because $x$ isn't killed at the initial position). Then, player I can again move so that II doesn't immediately cover $x$ (because $x$ isn't killed at the position after II's first move). continuing in this way, player I can ensure that II never covers $x$. Since $x\in X$, this contradicts the assumption that $s$ is a winning strategy for II.


