0
$\begingroup$

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 ${\frak S}$?

Improve this question
$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ Commented Jan 7, 2020 at 10:42
  • 2
    $\begingroup$ $\liminf a_n=\limsup a_n=1/2$ can be abbreviated as $\lim a_n=1/2$. $\endgroup$ Commented Jan 7, 2020 at 10:45
  • $\begingroup$ Oh right :-) Will simplify the post accordingly, thanks @GHfromMO $\endgroup$ Commented Jan 7, 2020 at 10:46
  • 5
    $\begingroup$ 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 $1$. $\endgroup$ Commented Jan 7, 2020 at 10:56
  • $\begingroup$ Isn't this sometimes called relative density (of the set $A$ with respect to the set $S$)? $\endgroup$ Commented Jan 7, 2020 at 10:57

1 Answer 1

Reset to default
6
$\begingroup$

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 $|\mathfrak S|\lt\kappa$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.