Let $A$ be the free Boolean algebra on denumerably many generators. How many ultrafilter does $A$ contain? How to prove it?
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What are your thoughts about that? What have you tried so far? – Taroccoesbrocco Jul 28 '18 at 10:58
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Hint: remember that an ultrafilter on $B$ is essentially the same as a boolean algebra morphism $B\to 2$. How can you characterize morphisms from a free algebra to $2$ ?
Maxime Ramzi
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Well, since $A$ has denumerably many elements and an ultrafilters is essentially the same as a Boolean epimorphism from $A$ to $2$, then $A$ has $2^\omega$ ultrafilters. – Beginner Jul 28 '18 at 11:45
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You have to compute $|Hom_{Bool}(A, 2)|$, which doesn't only depend on the cardinal of $A$ a priori. But here it does, because $A$ is free on countably many generators, so a morphism $A\to 2$ is essentially an application (set-theoretic) $\omega\to 2$. – Maxime Ramzi Jul 28 '18 at 12:35
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