# A complete Boolean algebra $B$ satisfies the $\kappa$-chain condition if and only if $B$ is $\kappa$-saturated

Let $B$ be a Boolean algebra. Then we say $B$ is $\kappa$-saturated if there is no partition $W$ of $B$ such that $|W| = \kappa$. We say that $B$ satisfies the $\kappa$-chain condition if there is no descending $\kappa$-sequence $u_0 > u_1 > \ldots > u_{\alpha} > \ldots$ $(\alpha < \kappa)$ of elements of $B$.

Let $B$ be complete. I am trying to prove the statement "$B$ satisfies the $\kappa$-chain condition if and only if $B$ is $\kappa$-saturated" (in Jech's book, Chapter 7). I think one direction is easy. If we have a decreasing sequence, set $W'= \{ \{u_{\alpha} \} \mid \alpha < \kappa \}$, and define a partition $W =\left( B \setminus W' \right) \cup W'$. This is a partition of size $\kappa + 1 = \kappa$, so $B$ is not $\kappa$-saturated.

Does anyone have any ideas of how to prove the converse? I supposed that $B$ is not $\kappa$- saturated, and tried to construct a decreasing sequence from the resulting partition of size $\kappa$, but to no avail. Any help would be appreciated.

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I find that an extremely useful heuristic when calculating in a Boolean algebra is to pretend you're working in an algebra of sets. We tend to have more intuition about what's going on in this case. For instance, via this heuristic, a $\kappa$-partition is just $\kappa$ many disjoint sets, which cover a bigger set. From this point of view, the answer (provided by Henno) becomes more or less obvious. Additionally, this heuristic is very much valid, due to Brikhoff's theorem. – Miha Habič Feb 7 '12 at 19:20
That is good to know. I did use that heuristic a couple of times to get a feel for what's going on, but it's good to know it is valid! – Paul Slevin Feb 7 '12 at 19:48

Let $\{ u_\alpha \mid \alpha < \kappa \}$ be a partition of $B$ (of non-$0$ elements). Then by completeness $v_{\alpha} = \lor \{ u_\beta \mid \beta \ge \alpha \}$ is well defined and forms a decreasing $\kappa$-sequence of elements of $B$.