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I define a Boolean Algebra B to be canonical iff it is isomorphic to the powerset of some set S. (under union, intersection, and complement, of course). Any Boolean Algebra that is both atomic and complete is canonical. Can this theorem be made stronger by dropping the condition "complete"? In other words, is every atomic Boolean algebra canonical? If not, exhibit an explicit atomic Boolean algebra that is not complete.

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  • $\begingroup$ Any finite Boolean algebra is complete. What infinite Boolean algebras do you know? $\endgroup$ – universalset Dec 18 '13 at 11:18
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The class of atomic Boolean algebras is an elementary class. Since there are infinite atomic Bookean algebras, it follows by the Löwenheim-Skolem theorem that there is an atomic Boolean algebra $B$ of cardinality $\aleph_0$. A canonical Boolean algebra has cardinality $2^\kappa$ for some $\kappa$. Therefore, the atomic Boolean algebra $B$ is not canonical.

If you need an explicit example, here it is under spoiler protection:

$\{X\subseteq\mathbb N:\text{ either }X\text{ is finite or }\mathbb N\setminus X\text{ is finite}\}$ is an atomic Boolean algebra which is not complete.

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