Cardinality of separating families on an infinite cardinal $\kappa$

Question

Let $\kappa$ be an infinite cardinal. We say ${\cal S}\subseteq {\cal P}(\kappa)$ is separating if whenever $a\neq b\in \kappa$ there is $T\in {\cal S}$ such that $|T\cap\{a,b\}| = 1$. Let $\newcommand{\s}{\text{sep}}\s(\kappa)$ be the minimum cardinality that a separating family on ${\cal S}\subseteq {\cal P}(\kappa)$ can have.

We have $\s(\aleph_0) = \aleph_0 = \s(2^{\aleph_0})$. (For the 2nd equality, identify $2^{\aleph_0}$ with $\mathbb{R}$, and consider ${\cal S} = \{(-\infty,q):q \in \mathbb{Q}\}$.)

Question. Do we have $\s(\kappa) < \kappa$ for all cardinals $\kappa > \aleph_0$?

Answer 1

If $\kappa$ is a strong limit then $\operatorname{sep}(\kappa)=\kappa$, and otherwise $\operatorname{sep}(\kappa)<\kappa$.

Let us first note that the proof that $\operatorname{sep}(2^{\aleph_0})=\aleph_0$ does extend to showing that $\operatorname{sep}(2^\mu)\leq\mu$, for all $\mu$, replacing $\mathbb{Q}$ by $\{f\in2^\mu\mid(\exists\alpha)(\forall\beta>\alpha)f(\beta)=0\}$ in the usual linear order on $2^\mu$.

For each $\lambda<\kappa$, let $\mathcal{S}_\lambda\subseteq\mathcal{P}(2^\lambda)$ be separating, with $|\mathcal{S}_\lambda|\leq\lambda$, and let $\mathcal{S}=\bigcup_{\lambda<\kappa}\mathcal{S}_\lambda$. Then $\mathcal{S}$ is separating and $|\mathcal{S}|\leq\kappa$.

If $\mathcal{S}$ is such that $|\mathcal{S}|=\lambda<\kappa$ then for each $\alpha<\kappa$ let $B_\alpha=\{A\in\mathcal{S}\mid\alpha\in A\}$. Since $|\mathcal{S}|<\kappa$ and $\kappa$ is a strong limit, $|2^{\mathcal{S}}|<\kappa$, so there are $\alpha<\beta<\kappa$ such that $A_\alpha=A_\beta$. Hence $\mathcal{S}$ does not separate $\alpha$ and $\beta$.