I am trying to prove that $\mathfrak b \leq \mathfrak r$ where $\mathfrak r$ is the minimal cardinality of a reaping family of sets.

A family $\mathcal R \subseteq [\omega]^\omega$ of sets $\mathcal R$, is called a reaping family if there is no set $U \in [\omega]^\omega$ that splits every set in $\mathcal R$

We say that a set $U$ splits a set $V$ if the sets $U \cap V$, $U\setminus V$ are both infinite.

Here is my proof:

Claim: $\mathfrak b \leq \mathfrak r$

Proof: We have to show that every reaping family $\mathcal R$ of sets admits an unbounded family of sequences. Note that we can obtain from every infinite set an infinite sequence, by applying the natural order of $\mathbb N$ on $\mathcal R$. So we have $\mathcal R'$ which is a family of infinite sequences from $\mathbb N$, which is in one to one correspondence with $\mathcal R$ and is identical to $\mathcal R$ when ignorring the order on it's elements. Now, suppose that $|\mathcal R'| < \mathfrak b$. Then, $|\mathcal R'|$ is an (almost everywhere) bounded family. So, Take a function $g \in \omega^\omega$ such that $g \leq^*f$ for every $f \in \mathcal R'$. Then, viewing $f$ as a set (i.e ignoring the order of it's elements), $f$ is a splitting set of every sequence in $\mathcal R'$ (viewed as a set), since for every $g \in \mathcal R'$, $g\setminus f$ is finite.

What do you think, is this correct?

Thank you! Shir

  • $\begingroup$ (I have removed the general-topology tag from these questions.) $\endgroup$ – Andrés E. Caicedo Jan 15 '14 at 16:43
  • 1
    $\begingroup$ Looking at this question makes me go "Huh boy. I've got a lot more to learn." $\endgroup$ – Conor O'Brien Dec 9 '14 at 1:34
  • $\begingroup$ Edit: In the def'n of "U splits V" the last word should be "infinite", not "finite". $\endgroup$ – DanielWainfleet Jun 16 '16 at 4:32

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