A Boolean algebra $B$ is defined (e.g. in Jech) to be $\kappa$-saturated if there is no partition $W$ of $B$ where $|W|=\kappa$. He seems to assume that this implies $|W|<\kappa$ for any partition $W$. But why should this be the case?


For example, say that $B$ is $\aleph\_1$-saturated. Why does this imply that $B$ is $\aleph\_2$-saturated? It's clearly true in the case where $B$ is complete or even $\aleph\_3$-complete, but suppose we're not given that. How would one construct a partition of size $\kappa$ given a partition of size $\lambda>\kappa$ in the absence of completeness?



Why is a Boolean algebra being $\kappa$-saturated upward closed in $\kappa$?
Jech defines a partition of a Boolean algebra $B$ as a *maximal* antichain. Now the cardinalities of maximal antichains in $B$ and its completion can indeed differ: Take $B$ as the finite, cofinite subsets of $\omega\_1$ with the canonical Boolean algebra strucure. $B$ has maximal antichains of every nonzero finite cardinality and of size $\omega\_1$, but no countably infinite one, so $B$ is $\omega$-saturated but not $\omega\_1$-saturated according to Jech's definition.


It seems that, to make the definition work as intended for all Boolean algebras instead of just complete Boolen algebras, one should drop the requirement of maximality and define $B$ to be $\kappa$-saturated if there is no antichain in $B$ of size $\kappa$. This definition agrees with the old one for complete Boolean algebras, but now any Boolean algebra $B$ is $\kappa$-saturated iff $B$'s completion is $\kappa$-saturated. The reason is that $B$ has the same cardinalities of antichains as its completion (as you noted in your answer).


