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?

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 for a continuation of this question.

Arch. Math. Logic56 (2017), 725–732, a version of which is available for free on Arnold W. Miller's home page. They show that player II (aka "Black") has a winning strategy iff $X$ is countable (agreeing with Andreas Blass's answer) and that player I (aka "White") has a winning strategy iff $X$ does not have strong measure zero. $\endgroup$ – bof May 31 at 1:54