Suppose that $\kappa$ has the property that for every family $A\subseteq\omega^{\omega}$, if $|A|<\kappa$, then there exists some $g\in\omega^{\omega}$ such that for any $f\in A, \exists^{\infty}n\;f(n)\neq g(n)$. Does it then follow that for any family $A\subseteq\omega^{\omega}$ of size $<\kappa$, there exists a $g\in\omega^{\omega}$ such that for any $f\in A, \forall^{\infty}n\;f(n)\neq g(n)$?

## 1 Answer

It does not follow. In fact, it is not true in the Cohen model. Let $X$ be a set of $\aleph_2$ mutually generic Cohen reals over $V$. (Recall: ``mutually generic'' means that if $x \in X$ and $Y \subseteq X$, then $x \in V[Y]$ if and only if $x \in Y$, and otherwise $x$ is Cohen-generic over $V[Y]$.)

In $V[X]$, if $A \subseteq \omega^\omega$ with $|A| < \aleph_2$, then (because $A$ is an object that is hereditarily of size $\leq\!\aleph_1$) there is some $X_0 \subseteq X$ with $|X_0| = \aleph_1$ such that $A \subseteq V[X_0]$. If $g \in X \setminus X_0$, then $g$ is Cohen-generic over $V[X_0]$, and hence over any $f \in A$. In particular, for any $f \in A$, $\exists^\infty n f(n) \neq g(n)$.

On the other hand, let $A \subseteq X$ with $|A| = \aleph_1$. If $g \in \omega^\omega$, then (because $g$ is a hereditarily countable object) there is some countable $A_0 \subseteq A$ such that $g \in V[A_0]$. If $f \in A \setminus A_0$, then it is Cohen-generic over $V[A_0]$, and hence $\exists^\infty n f(n) = g(n)$.