Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
@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
1  
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 1 '13 at 15:09

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.