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More specifically, if we only know that a complete Boolean algebra, $\mathbf{B}$, is $\kappa$-c.c., can we give a (reasonably tight) upper bound to the size of $\mathbf{B}$ in terms of $\kappa$?

Thanks in advance.

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The Cohen algebras of regular open subsets of $2^\lambda$ are ccc, and of arbitrarily large cardinality. –  Emil Jeřábek Jan 4 '13 at 12:06
    
Thanks again! Back to the drawing board I guess. –  Zoorado Jan 4 '13 at 12:44
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Emil, why not post your comment as an answer? (Unanswered questions periodically get auto-reposted to the main page.) –  Joel David Hamkins Jan 4 '13 at 15:09
    
Alright. I was worried the question was not quite research-level, if even someone as ignorant about forcing as me knows this example. –  Emil Jeřábek Jan 7 '13 at 15:53

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The $\kappa$-cc condition by itself does not put any bound on the cardinality of the algebra. For example, for each $\lambda$, the Cohen algebra of regular open subsets of $2^\lambda$ is ccc, but it has cardinality $\lambda^\omega$. However, one can bound the size of $B$ using additional cardinal characteristics: for a simple bound, if $B$ is $\kappa$-cc and has a dense subset $P$ of cardinality $\lambda$, then $|B|\le\lambda^{<\kappa}$, as every element of $B$ can be written as the join of an antichain in $P$.

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This is just like the forcing notion $\text{Add}(\omega,\lambda)$ to add $\lambda$ many Cohen reals, which is c.c.c. but has size $\lambda^{\lt\omega}$. –  Joel David Hamkins Jan 7 '13 at 16:01
    
Yes, this should be the same thing. When I was learning set theory (from not particularly up-to-date sources), I got the impression that each author has his own totally incompatible notation for basic forcing notions, and therefore I didn’t bother to use any in the answer. Is $\mathrm{Add}(\omega,\lambda)$ the standard notation nowadays? And does it denote the complete Boolean algebra, or some dense subset? –  Emil Jeřábek Jan 7 '13 at 16:22
    
Oh, I think there is no universal standard, but this notation is fairly common (especially here in New York). Sometimes this denotes a dense subset (the usual poset) and sometimes the complete Boolean algebra, but if it ever matters, then authors usually say so. I should have said "size $\lambda^\omega$" in my comment, since I usually think of it as referring to the Boolean completion. –  Joel David Hamkins Jan 7 '13 at 17:36

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