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Let $\omega$ denote the set of non-negative integers and let $[\omega]^\omega$ be the collection of infinite subsets of $\omega$.

If $S\in [\omega]^\omega$ and $A\subseteq \omega$ we say that $A$ is well-balanced with respect to $S$ if $$\lim_{n\to\infty}\frac{|A\cap S\cap \{1,\ldots,n\}|}{|S\cap\{1,\ldots,n\}|+1} = 1/2.$$ Intuitively speaking, this means that $A$ contains "half of the members of $S$" (which also implies that $A$ is infinite.)

Question. Given ${\frak S}\subseteq [\omega]^\omega$ with $|{\frak S}| = \aleph_0$, is there $A\in[\omega]^\omega$ such that $A$ is well-balanced with respect to every member of ${\frak S}$?

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  • $\begingroup$ If we take ${\frak S}$ to be the collection of infinite recursive sets of $\omega$, then any such $A$ would correspond to a bitstream that is computationally random. $\endgroup$
    – Dominic van der Zypen
    Commented Jan 7, 2020 at 10:42
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    $\begingroup$ $\liminf a_n=\limsup a_n=1/2$ can be abbreviated as $\lim a_n=1/2$. $\endgroup$
    – GH from MO
    Commented Jan 7, 2020 at 10:45
  • $\begingroup$ Oh right :-) Will simplify the post accordingly, thanks @GHfromMO $\endgroup$
    – Dominic van der Zypen
    Commented Jan 7, 2020 at 10:46
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    $\begingroup$ A random subset of $\omega$ (meaning that $n\in A$ iff the $n^{\text{th}}$ toss in an infinite sequence of fair coin tosses comes of heads) will do the job with probability $1$. $\endgroup$
    – bof
    Commented Jan 7, 2020 at 10:56
  • $\begingroup$ Isn't this sometimes called relative density (of the set $A$ with respect to the set $S$)? $\endgroup$
    – Martin Sleziak
    Commented Jan 7, 2020 at 10:57

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By the strong law of large numbers, if $S$ is an infinite subset of $\omega$, a random subset of $\omega$ will be well-balanced with respect to $S$ with probability one.

By the countable additivity of Lebesgue measure, if $\mathfrak S$ is a countable collection of infinite subsets of $\omega$, a random subset of $\omega$ will be well-balanced with respect to every member of $\mathfrak S$ with probability one.

If Lebesgue measure is $\kappa$-additive (the union of fewer than $\kappa$ measure zero sets has measure zero) then the same holds for a collection $\mathfrak S$ of infinite subsets of $\omega$ with $|\mathfrak S|\lt\kappa$.

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