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Given a complete boolean algebra B, and two partitions W and T of B, why is it true that W induces a partition on every element of T? (And is this true more generally - does W induce a partition on every non zero element of B?)

This is motivated by trying to understand the solution given here : If $B$ is an infinite complete Boolean algebra, then its saturation is a regular uncountable cardinal to theorem 7.15 in Jech's Set Theory.

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  • $\begingroup$ If $t\in T$, $\{w\land t:w\in W\text{ and }w\land t\ne 0_B\}$ is the partition of $t$ induced by $W$. $\endgroup$ – Brian M. Scott Jun 14 '16 at 3:44
  • $\begingroup$ That was my first thought, but why is that set a subset of W? For instance the product of w and w∧t is non-zero? $\endgroup$ – qwert4321 Jun 14 '16 at 4:01
  • $\begingroup$ It almost certainly isn’t a subset of $W$: the only time that it’s a subset of $W$ is when $t$ happens to be the join of some members of $W$. The claim isn’t that each member of $T$ can be partitioned into members of $W$; it’s simply that there each $t\in T$ has a partition naturally generated by $W$, and the partition that’s meant is the one that I described. $\endgroup$ – Brian M. Scott Jun 14 '16 at 4:04
  • $\begingroup$ Got it, thank you very much. So this works for any non-zero element of B. Is there a way I can upvote your comment? $\endgroup$ – qwert4321 Jun 14 '16 at 4:09
  • $\begingroup$ Yes, it works fine for any non-zero element of $B$. Don’t worry about upvoting; I’m happy enough that you got it sorted out! $\endgroup$ – Brian M. Scott Jun 14 '16 at 4:12

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