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I'm trying to solve the next exercise:

Construct a sequence $\mathcal{B}_0,\mathcal{B}_1, \ldots$ of countable Boolean algebras such that for all $m \neq n$ then $\mathcal{B}_m \ncong \mathcal{B}_n$.

I know that two countable atomless Boolean algebras are isomorphic, so I guess it has something to do with the number of atoms?! But what are examples of these countable Boolean algebras and how can I construct them?

PatrickR
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natural
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  • I suppose you mean $\cal B_m\neq B_n$? – Asaf Karagila Jan 30 '13 at 16:28
  • Just to clarify, do you mean countably infinite when you say countable? – ferson2020 Jan 30 '13 at 16:28
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    @ferson2020 : That would have to be what he means. Notice that he said "atomless". Finite Boolean algebras are never atomless, so if a BA is countable and atomless, then it's countably infinite and atomless. – Michael Hardy Jan 30 '13 at 17:01
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    As a hint, let $B_i$ be a BA with $i$ atoms, but to make it countable infinite, come up with a chain of elements for each atom. – ferson2020 Jan 30 '13 at 17:36
  • But for example two infinte chains with a common smallest and greatest element is not a distributive lattice hence no Boolean algebra. So how can I make it countably infinite with leaving it a BA? – natural Feb 01 '13 at 15:09

1 Answers1

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Take a countable atomless Boolean algebra $A$ and let $B_n$ be the finite Boolean algebra with $n$ atoms (and $2^n$ elements).

Then the product $A\times B_n$ is a countable Boolean algebra with $n$ atoms, and all the $A\times B_n$ are non-isomorphic.

PatrickR
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