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How can I construct a countably infinite Boolean algebra with $n$ atoms, for $n \in \mathbb{N}$?

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Do you know any examples of countably infinite atomless Boolean algebras? –  Arthur Fischer Feb 5 '13 at 15:45
@Arthur: You make it sound as if there are several... :-) –  Asaf Karagila Feb 5 '13 at 15:53
@Asaf: There are lots of such BAs. Sure, they all look the same if you stand far enough away. But once you get close enough to inspect their innards, you can tell them apart. ;-) –  Arthur Fischer Feb 5 '13 at 16:05
@Arthur: Well, never look a gift atomless countable Boolean algebra in the innards, that's what ma daddy used to say. (This comment is to be read in a Dixie accent) :-) –  Asaf Karagila Feb 5 '13 at 16:19
I think the periodic sequences form a (the) countable atomless Boolean algebra, but how does this help? –  natural Feb 5 '13 at 16:29

2 Answers 2

Hint: every boolean algebra is isomorphic to an algebra of sets. What happens if you add together (and generate algebra with) two algebras of sets with disjoint universes?

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Hint: Suppose $A$ and $B$ are two Boolean Algebras, and consider their product $A \times B$. Then the atoms of $A \times B$ are exactly those elements of the form either $( a , \mathbf{0}_B )$ (for $a$ an atom in $A$) or $( \mathbf{0}_A , b )$ (for $b$ an atom in $B$).

(This is to be used in conjunction with my comment above... not the one about innards, the other one.)

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Isn't it the same thing as the one I suggested? :) (Product and coproduct being the same...) –  tomasz Feb 6 '13 at 11:59
@tomasz: You might be right. My interpretation was that your answer concentrated more on the specific representation of the BA as an algebra of sets, and mine more on the pure algebra without much care about the representation. –  Arthur Fischer Feb 6 '13 at 12:38

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