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.

  • $\begingroup$ You will probably be interested in F. Galvin's, Indeterminacy of point-open games $\endgroup$
    – user35370
    Sep 27, 2015 at 3:11
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    $\begingroup$ I don't think you want $Eps$ to be topologized as usual. It would be more interesting to use Baire space instead of $Eps$ and let $f \in \omega^\omega$ correspond to the sequence $\epsilon_n = 2^{-f(n)}$, for example. The issue is that $Eps = (0,\infty)^\omega$ is connected. If $x \in X$ and $X$ is csmz via $F$, we can define $H_x:Eps \to \omega$ by $H_x(f) = \mu n[x \in F(f)(n)]$. Since $\omega$ is discrete and $H_x$ is continuous, $H_x$ must be constant. If $x \neq y$, then $H_x$ and $H_y$ can't have the same constant value and it follows immediately that $X$ is countable. $\endgroup$ Sep 27, 2015 at 13:35
  • $\begingroup$ @FrançoisG.Dorais That's a good point - this is what I get for not paying attention to the details! $\endgroup$ Oct 12, 2015 at 3:47
  • $\begingroup$ @bof I'm not familiar with that game, but let me check ... $\endgroup$ May 30, 2020 at 3:11
  • $\begingroup$ Your game seems to be equivalent to the Mycielski–Solovay game $G(X)$ in the paper "Strong measure zero and infinite games", Arch. Math. Logic 56 (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, 2020 at 1:54

1 Answer 1


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.

  • $\begingroup$ This is very nice! It doesn't obviously extend to the continuously strong measure zero case, though, does it? The natural game to play in that case would be: player I plays epsilons, player II plays intervals, but is allowed to pass. But in that game, many reals can be killed at a given stage, so the analysis seems to break down. Or am I missing something? $\endgroup$ Sep 26, 2015 at 23:36
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    $\begingroup$ Something I missed at first glance, which I'm mentioning here because it might be interesting to others: this argument is essentially the same as the proof that if Empty has a winning strategy in the perfect set game for $X$, then $X$ is countable. $\endgroup$ Sep 27, 2015 at 1:34
  • $\begingroup$ Thank you very much for this answer! I've accepted, and asked a new question expanding on the rest of this question: mathoverflow.net/questions/220687/…. $\endgroup$ Oct 12, 2015 at 4:18

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