Cardinality of splitting families

Question

For any set $X$, let $[X]^2 = \big\{\{a,b\}:a\neq b\in X\big\}$. If $\kappa>1$ is a cardinal, then a splitting family is a collection ${\cal S} \subseteq {\cal P}(\kappa)$ such that for every $Q \in [\kappa]^2$ there is $S\in {\cal S}$ such that $|Q \cap S| = 1$. A trivial splitting family for any cardinal $\kappa$ is given by ${\cal S} = \{\alpha\in\kappa: \alpha > 0\} = \kappa \setminus\{0\}$.

Let $\kappa$ be infinite.

Questions.

1. Is there a splitting family of cardinality $<\kappa$?
2. Let $N_\kappa=\{S\subseteq \kappa: |S|=|\kappa \setminus S| = \kappa\}$. Does $N_\kappa$ contain a splitting family of cardinality $\leq \kappa$?

Answer 1

For any set $X$ let $\operatorname{SF}(X)$ be the the set of all families $\mathcal S\subseteq\mathcal P(X)$ such that, for each pair $\{x,y\}\in[X]^2$, there is a set $A\in\mathcal S$ with $|A\cap\{x,y\}|=1$.

1. If $\kappa$ and $\lambda$ are cardinal numbers (finite or infinite) and $X$ is a set of cardinality $|X|=\kappa$, then the following statements are equivalent:
(i) there is a family $\mathcal S\in\operatorname{SF}(X)$ of cardinality $|\mathcal S|=\lambda$;
(ii) $\kappa\le2^\lambda$ and $\lambda\le2^\kappa$.

Proof. If (i) holds, then $\kappa\le2^\lambda$ because we can define an injection $f:X\to\mathcal P(\mathcal S)$ by setting $f(x)=\{A\in\mathcal S:x\in A\}$, and $\lambda\le2^\kappa$ because $\mathcal S\subseteq\mathcal P(X)$.

Suppose (ii) holds. Since $\kappa\le2^\lambda$, we may assume that $X\subseteq\{0,1\}^\lambda$. Let $\mathcal A=\{A_\alpha:\alpha\in\lambda\}$ where $A_\alpha=\{x\in X:x_\alpha=0\}$. Then $\mathcal A\in\operatorname{SF}(X)$ and $|\mathcal A|\le\lambda\le2^\kappa$, so we can extend $\mathcal A$ to a family $\mathcal S$ such that $\mathcal A\subseteq\mathcal S\subseteq\mathcal P(X)$ and $|\mathcal S|=\lambda$.

2. If $|X|=\kappa$ is an infinite cardinal, and if $\lambda\le\kappa\le2^\lambda$, then there is a family $\mathcal S\in\operatorname{SF}(X)$ such that each element of $\mathcal S$ is a moiety of $X$, i.e., $|A|=|X\setminus A|=|X|$ for each $S\in\mathcal S$. Let $\mathcal M(X)=\{A\subseteq X: |A|=|X\setminus A|=|X|\}$.

Proof. For $\alpha\in\lambda$ and $\varepsilon\in\{0,1\}$ let $Y_{\alpha,\varepsilon}=\{y\in\{0,1\}^\lambda:y_\alpha=\varepsilon\}$. Since $\{Y_{\alpha,\varepsilon}:\alpha\in\lambda,\varepsilon\in\{0,1\}\}$ is a family of $\lambda$ subsets of $\{0,1\}^\lambda$, each of cardinality $|Y_{\alpha,\varepsilon}|=2^\lambda\ge\kappa$, we can choose subsets $X_{\alpha,\varepsilon}\subseteq Y_{\alpha,\varepsilon}$ with $|X_{\alpha,\varepsilon}|=\kappa$. Since $|\bigcup_{\alpha\in\lambda,\varepsilon\in\{0,1\}}X_{\alpha,\varepsilon}|=\kappa$, we may assume that $X=\bigcup_{\alpha\in\lambda,\varepsilon\in\{0,1\}}X_{\alpha,\varepsilon}$. Let $\mathcal A=\{A_\alpha:\alpha\in\lambda\}$ where $A_\alpha=\{x\in X:A_\alpha=0\}$. Then $\mathcal A\in\operatorname{SF}(X)$ and $\mathcal A\subseteq\mathcal M(X)$. Finally, since $|\mathcal A|\le\lambda\le2^\kappa$, we can extend $\mathcal A$ to a family $\mathcal S$ such that $\mathcal A\subseteq\mathcal S\subseteq\mathcal M(X)$ and $|\mathcal S|=\lambda$.