Let $\kappa$ be the smallest cardinality of a family $\mathcal F$ of subsets of $\omega$ such that for any bijective function $f:A\to B$ between disjoint infinite subsets of $\omega$ there exists a set $F\in\mathcal F$ such that the set $\{x\in A\cap F:f(x)\notin F\}$ is infinite.


It can be shown that $\mathfrak s\le \kappa\le\mathfrak c$, where $\mathfrak s$ is the splitting number.



> 
> **Question.** Is it consistent that $\kappa<\mathfrak c$? Or $\kappa=\mathfrak c$ in ZFC?
> 
> 
> 


**Remark.** The affirmative answer to the first part of this question will imply a consistent negative answer to [this MO-problem](https://mathoverflow.net/questions/309467/is-beta-mathbb-n-a-unique-compactification-with-the-smallest-possible-permuta?noredirect=1&lq=1). 



A new cardinal characteristic of the continuum?
Recall that $\mathbf{non}(\mathcal{B})$ is the least cardinality of a non-meager subset of $\mathbb{R}$; the choice of presentation of $\mathbb{R}$ does not matter for this, so we take $\mathbb{R} = \mathcal{P}(\omega)$. It is well-known that consistently, $\mathbf{non}(\mathcal{B}) < \mathfrak{c}$; indeed, start from a model of CH and add $\aleph\_2$-many Cohen reals.


**Theorem.** $\kappa \leq \mathbf{non}(\mathcal{B})$. 


*Proof.* Suppose $\mathcal{F} \subseteq \mathcal{P}(\omega)$ is nonmeager. It suffices to show that $\mathcal{F}$ works for the definition of $\kappa$. Let $A, B$ be disjoint infinite subsets of $\omega$. Let $X$ be the set of all $F \subseteq \omega$ such that there are infinitely many $n \in A \cap F$ with $f(n) \not \in F$. I claim that $X$ is comeager. Indeed, for each finite $J \subseteq A$, let $X\_J$ be the set of all $F \subseteq \omega$ such that there is some $n \in A \backslash J$ with $n \in A \cap F$ and $f(n) \not \in F$. Each $X\_J$ is open dense, and $X = \bigcap\_J X\_J$.




---


Thus, consistently $\kappa < \mathfrak{c}$.


Incidentally, the following is perhaps a cleaner formulation of $\kappa$:


**Claim.** Let $\lambda$ be the least cardinality of a family $\mathcal{G}$ of subsets of $\omega$, such that for each increasing sequence $(x\_n: n < \omega)$, there is some $F \in \mathcal{G}$ such that for infinitely many even $n < \omega$, $x\_{n} \in F$ but $x\_{n+1} \not \in F$. Then $\kappa = \lambda$.


*Proof.* It is straightforward to check that $\lambda \leq \kappa$: given $\mathcal{F}$ witnessing the definition of $\kappa$ and $(x\_n: n < \omega)$, write $A = \{x\_{2n}: n < \omega\}$, $B = \{x\_{2n+1}: n < \omega\}$ and let $f: A \to B$ be $x\_{2n} \mapsto x\_{2n+1}$.


So we show $\kappa \leq \lambda$. Let $\mathcal{G}$ witness the definition of $\lambda$; let $\mathcal{F} = \mathcal{G} \cup \{\omega \backslash S: S \in \mathcal{G}\}$. We show that $\mathcal{F}$ is as in the definition of $\kappa$.


Suppose $A, B$ are infinite disjoint sets and $f: A \to B$ is a bijection. Note that since $\mathcal{F}$ is closed under complements, we are free to replace $(A, B, f)$ by $(B, A, f^{-1})$, if desired. 


Choose $A' \subseteq A$ infinite such that either for all $x \in A'$, $f(x) > x$, or else for all $x \in A'$, $f(x) < x$. After possibly interchanging $A$ and $B$, we can suppose that for all $x \in A'$, $f(x) > x$. Write $B' = f[A']$.


Define an increasing sequence $(x\_n: n < \omega)$ inductively so that for all $n$, $x\_{2n} \in A'$ and $x\_{2n+1} = f(x\_{2n}) \in B'$. Namely, let $x\_0 = \min(A')$, let $x\_1 = f(x\_0)$, and having defined $x\_{2n-1}$, let $x\_{2n}$ be the least element of $A'$ bigger than $x\_{2n-1}$, and let $x\_{2n+1} = f(x\_{2n})$. 


Choose $F \in \mathcal{F}$ such that there are infinitely many even $n < \omega$ with $x\_n \in F$ but $x\_{n+1} \not \in F$. Then for each such $n$, $x\_n \in A \cap F$ but $f(x\_n) = x\_{n+1} \not \in F$, so $F$ is as desired.


**Remark.** For each infinite $I \subseteq \omega \times \omega$, we get a variant notion $\kappa\_I$ of $\kappa$, namely $\kappa\_I$ is the least cardinality of a family $\mathcal{G}$ of subsets of $\omega$, such that for each increasing sequence $(x\_n: n < \omega)$, there is some $F \in \mathcal{G}$ such that for infinitely many $(n, m) \in I$, $x\_n \in F$ but $x\_m \not \in F$ (possibly $\kappa\_I = \infty$). 


It is easy to check that $\kappa = \kappa\_{\{(2n, 2n+1): n < \omega\}}$ and $\mathfrak{s} = \kappa\_{\{(n, n+1): n < \omega\}}$. The above proof shows that each $\kappa\_I \leq \mathbf{non}(\mathcal{B})$, provided that for each finite $J \subset \omega$, there are infinitely many $(n, m) \in I$ with $n, m \not \in J$. It is interesting to ask which of these $\kappa\_I$'s can be separated...


