Boolean algebra of a certain size containing filter In an attempt to adapt a proof that I found in a book to I different statement, I am wondering about the following question.
Suppose $\kappa$ is strongly inaccessible. Then $ G= \lbrace \lbrace x \in \mathcal{P}_{\kappa} \kappa \mid y \subseteq x \rbrace \mid y \in \mathcal{P}_{\kappa} \kappa \rbrace$ can clearly be extended to a $\kappa$-complete filter on $\mathcal{P}_{\kappa} \kappa$ (by which we mean all subsets of $\kappa$ of size strictly less than $\kappa$. Keep in mind that since $\kappa$ is inaccessible this set itself has size $\kappa$) Now, my question is: can we find a $\kappa$-complete boolean algebra $B$ of size $\kappa$ that contains $G$ and such that the filter generated by $G$ in $B$ is still $\kappa$-complete?
 A: Very generally, suppose $X$ is an algebraic structure with at most $\kappa$ operations, each of which has arity less than $\kappa$.  Then if $S\subseteq X$, the subalgebra of $X$ generated by $S$ can be constructed by transfinite recursion, starting with $S_0=S$, taking $S_{\alpha+1}$ to be the union of $S_\alpha$ and all the elements of $X$ that can be obtained by applying an operation to elements of $S_\alpha$, and taking unions at limit steps.  This process will terminate once you reach an ordinal of cofinality greater than the arity of each operation, so in this case it will terminate by the time you reach $S_\kappa$.  If $|S|\leq\kappa$ then it is easy to see by induction that $|S_\alpha|\leq\kappa$ for all $\alpha\leq\kappa$ (each time you apply an operation of arity $\lambda<\kappa$ it only adds at most $\kappa^\lambda=\kappa$ new elements).  So, the subalgebra of $X$ generated by $S$ still has cardinality at most $\kappa$.
In particular, taking $X$ to be the power set of $\mathcal{P}_\kappa\kappa$ considered as a $\kappa$-complete Boolean algebra and $S$ to be your $G$, the $\kappa$-complete subalgebra $B\subseteq X$ generated by $G$ will have cardinality $\kappa$.  Moreover, the filter in $B$ generated by $G$ will still be $\kappa$-complete, since given any family of less than $\kappa$ elements of $G$ there is another element of $G$ contained in their intersection.
(In fact, in this particular case the transfinite iteration is not even necessary: the $\kappa$-complete subalgebra generated by any subset $S$ of a power set will be just $S_3$, since you can use the distributive law to write any element in the form $\bigcup_{i\in I}\bigcap_{j\in J_i}s_{ij}$ where each $s_{ij}$ is either an element of $S$ or the complement of an element of $S$.)
