Can a Larger Proper Filter Contain an Ultrafilter in a Boolean Algebra?

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In a Boolean algebra B, the set {x in B: x>=b} is identified as an ultrafilter, which is a maximal proper filter. The discussion raises questions about the existence of a larger proper filter that could contain this ultrafilter. It is suggested that a missing condition, such as "b is an atom," is crucial for the ultrafilter's properties. The need for this condition is highlighted as essential to avoid counterexamples. Overall, the conversation emphasizes the importance of specific conditions in the definition of ultrafilters within Boolean algebras.
quasar987
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Let B be a boolean algebra with smaller element 0, and let b different from 0 be in B. Apparently, {x in B: x>=b} is an ultrafilter of B.

I don't understand why a priori there could not exists a larger proper filter containing this one!
 
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I'm not sure what you you mean by that question.
 
Well, an ultrafilter in a boolean algebra (B,u,n,C,0,1) is as far as I understand, a proper filter that is maximal, in the sense that there are no filter containing it other than B itself.

Why is the set {x in B: x>=b} from post #1 an ultrafilter. Why can't there be a filter containing it other than B?
 
Have you tried constructing a counterexample?



Anyways, are you sure you're not missing a condition? e.g. is b supposed to be an atom?
 
I had vaguely found a counterexample, that's why I asked the question. Now it's clear that a condition such as "b is an atom" is missing. Thanks.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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