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Given a finite set $X=\{1,2,3\}$, are the following all the Boolean algebras?

$A_1=\{\{1\},\{2\}\},$ $A_2=\{\{1\},\{3\}\},$ $A_3=\{\{2\},\{3\}\},$

$A_4=\{\{1\},\{1,2\}\},$ $A_5=\{\{1\},\{1,3\}\},$ $A_6=\{\{1\},\{2,3\}\},$

$A_7=\{\{2\},\{1,2\}\},$ $A_8=\{\{2\},\{1,3\}\},$ $A_9=\{\{2\},\{2,3\}\},$

$A_{10}=\{\{3\},\{1,2\}\},$ $A_{11}=\{\{3\},\{1,3\}\},$ $A_{12}=\{\{3\},\{2,3\}\},$ $A_{13}=\{\{\},\{1,2,3\}\},$ $A_{14}=\{\}$

I can't get a practical example so please correct me if I am using the definition of a Boolean algebra on a set wrongly

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1 Answer 1

The empty set and the maximal set always have to be in the Boolean algebra, like in $A_{13}$. In the first 12 cases you just listed the atoms which is sufficient information in case of finite Boolean algebras, and these are correct.

So, fully written, $A_1=\{\{\}, \{1\},\{2\},\{1,2\}\}$, and so on... $A_{13}$ is ok, and $A_{14}=\{\{\}\}$.

Otherwise, seems more or less complete, you miss the ones with one atom, like $\{\{\},\{1\}\}$ or $\{\{\},\{1,2\}\}$, and the biggest one with 3 atoms, the full powerset $P(\{0,1,2\})$.

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Thanks Berci, I am not quiet sure of how many of these atoms we need in each Boolean algebra though. –  Jay Nov 17 '12 at 15:43

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