The stationary reaping number $\mathfrak{r}_{cl}$ Let $\kappa$ be at least inaccessible (but measurable is what I am primarily interested at the moment). Let $x,y \in [\kappa]^\kappa$ both be stationary. We say that $y$ stationary-splits $x$ iff $x \cap y$ and $x\setminus y$ are both stationary.
Define $\mathfrak{r}_{cl}:=\min \{\vert \mathcal{R}\vert \colon \,\, \forall x \in \mathcal{R} \,\, x \, \text{is stationary} \, \land \neg \exists y \in [\kappa]^\kappa \,\, \mathcal{R} \, \, \text{is stationary-split by} \,\,  y\}$.
Of course, $\mathfrak{r}_{cl}$ has to be infinite and $\mathfrak{r}_{cl}= 2^\kappa$ in the $\kappa$-Sacks model. But can ZFC prove a nontrivial lower bound ($\omega_1,\, \kappa \,\, \text{or even} \,\, \kappa^+$) for $\mathfrak{r}_{cl}$ ??
Note that this question is motivated by investigating cardinal characteristics on the 'higher' Cantor/ Baire space modulo the non-stationary (not the bounded) ideal.
 A: I think that $\omega_1$ is certainly a lower bound, and that this is the case for any regular, uncountable $\kappa$. To see this, suppose that $\langle x_n \mid n < \omega \rangle$ is a sequence of stationary subsets of $\kappa$. We'll find a $y$ that stationary-splits every $x_n$. We'll do this by constructing a sequence $\langle x_n^* \mid n < \omega \rangle$ such that $x_n^*$ is a stationary subset of $x_n$ and $x_n^* \cap x_m^* = \emptyset$ for all $n < m < \omega$. Then we can let $y_n$ be a stationary subset of $x_n^*$ such that $x_n^* \setminus y_n$ is also stationary, and let $y = \bigcup_{n < \omega} y_n$.
The construction of $\langle x_n^* \mid n < \omega \rangle$ can be done by recursion on $n$, where we maintain the following additional  recursion hypothesis: for all $n \leq m < \omega$, the set $x_{m,n} := x_m \setminus \bigcup_{k < n} x_k^*$ is stationary. Fix $n < \omega$ and suppose that $\langle x_k^* \mid k < n \rangle$ has been constructed. By hypothesis, $x_{n,n}$ is stationary; partition it into $\omega_1$-many disjoint stationary subsets, $\langle x^\alpha_{n,n} \mid \alpha < \omega_1 \rangle$. For each $m > n$, there is at most one $\alpha < \omega_1$ such that $x_{m,n} \setminus x^\alpha_{n,n}$ is nonstationary, so we can choose an $\alpha^* < \omega_1$ such that $x_{m,n} \setminus x^{\alpha^*}_{n,n}$ is stationary for all $m > n$, and then set $x_n^* = x^{\alpha^*}_{n,n}$, and continue to the next step.
It's not immediately clear to me how to achieve a better lower bound. This particular construction breaks down at limit steps, but it seems conceivable that a more clever version of the argument might yield a lower bound of $\kappa$ or even $\kappa^+$.
