This question is about exercise 7.16 from Jech's Set Theory (3rd Edition):
Let $\kappa$ be a regular uncountable cardinal, and let $A$ be a set of size at least $\kappa$. With $S \triangleq P_\kappa(A)$ (i.e., the subsets of $A$ of size at most $\kappa$), let $F$ be the set of all $X \subseteq S$ such that $\hat{P} \subseteq X$ for some $P \in S$, where $\hat{P} \triangleq \{Q \in S\colon P \subseteq Q\}$. Show that $F$ is a $\kappa$-complete filter on $S$.
I believe to have a proof, but my argument (presented below) does not make use of the fact that $\kappa$ is regular and uncountable. This leads me to believe that my proof is wrong, and I kindly ask for help in pinpointing the mistake(s):
Proof. We only want to verify closedness under intersection of $< \kappa$ elements of $F$, as the properties
- $S \in F$
- $X \in F \land X \subseteq Y \implies Y \in F$
follow fairly easily from the definition of $F$. We thus consider $Y = \cap_{\alpha < \lambda} X_\alpha$, where $\lambda < \kappa$ and $X_\alpha \in F\, \forall \alpha$. Let $P_\alpha$ and $\hat{P}_\alpha$ denote the sets $P$ and $\hat{P}$ related to $X_\alpha$ as described in the definition of $F$. Since $\cap_{\alpha < \lambda} \hat{P}_\alpha \subseteq Y$, it suffices (by 2.) to show that $\cap_{\alpha < \lambda} \hat{P}_\alpha$ is in $F$ (to see that $|Y| \leq \kappa$, note that $|Y| = |\cup_{\alpha < \lambda} X_\alpha| \leq \lambda\kappa \leq \kappa$). Indeed, we claim that $\hat{Q} \subseteq \cap_{\alpha < \lambda} \hat{P}_\alpha$, where $Q \triangleq \cup_{\alpha < \lambda} P_\alpha$. To see this, we show that if $X \in \hat{Q}$, then $X \in \hat{P}_\alpha\, \forall \alpha$. To that end first note that $|Q| = |\cup_{\alpha < \lambda} P_\alpha| \leq \lambda\kappa \leq \kappa$, i.e., $Q \in S$. Since $P_\alpha \subseteq Q\, \forall \alpha$, we observe that $Q \in \hat{P}_\alpha\, \forall \alpha$; to conclude, it suffices to apply the same argument to any superset $X$ of $S$.
Remark: I am aware that $\kappa$ being regular is in general required to prevent situations such as the one described in Asaf's answer here (or see Exercise 7.13 for a similar example with ideals), but this issue does not seem to matter here, as my argument (if correct) would also extend to intersection of up to $\kappa$ elements of $F$.
Best Answer
$F$ is the filter on $S$ generated by $G=\{\hat P: P\in S\}.$ So for $\kappa$-completeness of $F$ it suffices to show that if $0<\lambda<\kappa$ and $T=\{\hat P_a: a\in \lambda\}\subset G$ then $\cap T\in G.$ By definition of $\hat P$ we have $\cap T=\{Q\in S: \cup_{a\in \lambda}P_a\subset Q\}.$
According to the comment by Andreas Lietz, in Jech's book $P_{\kappa}(A)$ is the set of subsets of $A$ that have cardinal $less $ $ than $ $\kappa.$
Assuming that is correct, each $|P_a|<\kappa,$ and $\lambda<\kappa,$ and $\kappa$ is regular, so $|\cup_{a\in \lambda}P_a|<\kappa .$ So $\cup_{a\in \lambda}P_a\in P_{\kappa}(A)=S$ so $\cap T\in G.$
A common notation for $P_{\kappa}(A)$ is $[A]^{<\kappa},$ which I prefer (as the "$<$" is part of the notation). And also used are $[A]^k=\{B\subset A: |B|=k\}$ and $[A]^{\le k}=\{B\subset A:|B|\le k\}.$ (Including the cases where $k$ or $\kappa$ is any finite or infinite cardinal.)