This question is about comparing the relative sizes of null sets by switching from open covers to open covering sequences (a la strong measure zero sets or microscopic sets). The main question is whether the most obvious preorder is in fact linear, and the secondary question is whether this preorder coincides with a related "game-theoretic" preorder I know is linear.

Below we either work in $\mathsf{ZF}+\mathsf{DC}+\mathsf{AD_\mathbb{R}}$ or restrict attention to appropriately tame sets of reals.

Main question

For $X\subseteq\mathbb{R}$ of measure zero, say that the efficiency of $X$ is the set $\mathsf{Eff}(X)$ of all sequences of positive reals $\alpha=(a_i)_{i\in\mathbb{N}}$ such that there is some sequence of nonempty rational open intervals $(U_i)_{i\in\mathbb{N}}$ with $m(U_i)\le a_i$ for each $i$ and $\bigcup_{i\in\mathbb{N}}U_i\supseteq X$.

We get a natural preorder from this notion, $\trianglelefteq$, via $$X\trianglelefteq Y\iff \mathsf{Eff}(X)\supseteq\mathsf{Eff}(Y)$$ (that's not a typo - the idea is that $X\trianglelefteq Y$ means that $X$ is smaller than $Y$, which is to say that it's more efficient). I would like to understand this preorder better, and in particular:

Is $\trianglelefteq$ a linear preorder?

Secondary question

There is a different relation, of similar "flavor" to $\trianglelefteq$, which I can prove is a linear preorder. Namely, given null $X,Y$ consider the game $E(X,Y)$ defined as follows:

  • Players $1$ and $2$ alternately play individual nonempty rational open intervals building a sequence $U_0,V_0,U_1,V_1,...$, with $m(U_i)\ge m(V_i)\ge m(U_{i+1})$.

  • Player $1$ wins a given play iff

    • for each $n$ we have $\bigcup_{i>n}U_i\supseteq Y$, but

    • there is some $n$ such that $\bigcup_{i>n}V_i\not\supseteq X$.

(The win condition can be rephrased as a kind of redundancy property: "$(U_i)_{i\in\mathbb{N}}$ covers $X$ with infinite repetition but $(V_i)_{i\in\mathbb{N}}$ does not cover $Y$ with infinite repetition.")

Consider the relation $\sqsubseteq$ given by $$X\sqsubseteq Y\iff \mbox{player $2$ has a winning strategy in $E(X,Y)$}.$$

Like $\trianglelefteq$ this is clearly reflexive and transitive, and linearity follows from determinacy. If $X\not\sqsubseteq Y$ then player $1$ has a winning strategy $\Sigma$ in $E(X,Y)$, and we can turn that into a winning strategy $\hat{\Sigma}$ for player $2$ in $E(Y,X)$. This is where the redundancy part of the win condition comes in: $\hat{\Sigma}$'s first move will be essentially useless, but this won't affect the ultimate outcome.

So one natural way to get a positive answer to the main question would be to prove $\sqsubseteq$ is the same as $\trianglelefteq$. However, I don't see how to do this:

Do $\trianglelefteq$ and $\sqsubseteq$ coincide?

I suspect that the answer is no, and in fact that $\trianglelefteq$ is not linear.

(As an aside, the "non-redundant-covering" version of $\sqsubseteq$ is more similar to $\trianglelefteq$ so would make the secondary question more interesting, but I don't know that that relation is linear either so it's not obviously useful to the main question.)

  • $\begingroup$ Incidentally, I've asked a couple questions at MO which generated discussion that might be relevant (although at the moment I don't see it): 1, 2. $\endgroup$ – Noah Schweber May 28 '20 at 22:02
  • $\begingroup$ Besides the Cantor set, do you please ( is there any ) say a set which its measure is zero but not strongly? $\endgroup$ – Maryam Ajorlou May 29 '20 at 6:39
  • 1
    $\begingroup$ @MaryamAjorlou It's consistent with $\mathsf{ZFC}$ that no uncountable measure-zero set is strong measure zero. On the other hand, the Cantor set is paradigmatic in the context of this question since every "tame" uncountable set has a perfect subset. $\endgroup$ – Noah Schweber May 29 '20 at 6:55
  • $\begingroup$ Noah, does AD (or its variants) prove that every strong measure zero set is countable? $\endgroup$ – Asaf Karagila May 29 '20 at 7:57
  • $\begingroup$ @AsafKaragila I think perfect sets don't have strong measure zero, so yes? $\endgroup$ – Noah Schweber May 30 '20 at 0:13

Here are two closed measure zero sets $A$ and $B$ which are incomparable under $\trianglelefteq$.

Let $A$ be the Cantor set.

Let $B=C\cup D$ where $C$ and $D$ are nonempty perfect measure zero sets with $\operatorname{diam}(C)\lt\frac14$, $\operatorname{diam}(D)\lt\frac1{10}$, and $d(C,D)\gt2$.

Let $\varepsilon=(\varepsilon_1,\varepsilon_2,\varepsilon_3,\dots)$ be a sequence of positive rational numbers such that $\varepsilon\notin\mathsf{Eff}(C)\cup\mathsf{Eff}(D)$.

Then $$(2,\varepsilon_1,\varepsilon_2,\varepsilon_3,\dots)\in\mathsf{Eff}(A)\setminus\mathsf{Eff}(B)$$ while $$\left(\frac14,\frac1{10},\frac1{28},\cdots,\frac1{3^n+1},\cdots\right)\in\mathsf{Eff}(B)\setminus\mathsf{Eff}(A).$$

Regarding the game $E(X,Y)$ it should be noted that the existence of a winning strategy for player $1$ implies that $X$ is "strategically strong measure zero" and therefore countable as shown in Andreas Blass's answer to the question Two strengthenings of "strong measure zero".


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