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I got two questions:

1) Does there exist an infinite Boolean algebra which contains an atom?

I answered yes.

2) Does there exist an infinite Boolean algebra B such that for every b contained in B there is an atom a contained in B with a is smaller or equal than b?

I answered no.

I just cannot figure out what's the difference between these two questions. Can someone help please?

Blackgirl5
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1 Answers1

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Edit: I see from the comments that $b$ in part 2 is restricted so that it can't be the least element of $B$. That's fairly important information, and changes my answer.

One hint will apply to both parts equally well. The most familiar form of a Boolean Algebra is the power set of a set. Consider the power set of an infinite set to answer both of your questions.

Added: Your example--letting $B$ be the power set of the natural numbers--works as an example for both. As you pointed out, its atoms are precisely the singleton subsets of the natural numbers. Hence, $B$ has an atom--in fact, infinitely-many, but it has at least one, which is what matters--and so the answer to Question 1 is "yes." On the other hand, given any non-least element $b$ of $B$ (that is, any non-empty subset of the natural numbers), there is at least one atom less than or equal to it--for example, the singleton containing only the least element of $b$. Hence, the answer to Question 2 is also "yes."

Cameron Buie
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  • Using your hint for question 2, I assumed that there is no such infinite boolean algebra...Right? – Blackgirl5 Nov 06 '14 at 01:51
  • Sorry, but I cannot understand the hint... – Blackgirl5 Nov 06 '14 at 01:53
  • Well, you can assume it (correctly). However, you should be able to prove it. In fact, you should be able to show that even if we get rid of the word "infinite," there is still no Boolean algebra that satisfies the given condition (that is, even if $B$ is finite, consider what happens if $b$ is the least element of $B$). – Cameron Buie Nov 06 '14 at 04:31
  • If I take the powerset of the the natural numbers. I would say that there exist more than one atoms beecause the number of singletons(atoms) are also infinite. So, the answer would be no for Q1. – Blackgirl5 Nov 11 '14 at 02:41
  • For question 2, I think of the Lindenbaum-Tarski algebra with a set of infinite variables. Since, the number of atoms is 2 to the n. And the number of elements is 2 to the number of atoms. So, there are more elements than atoms. The answer is no. – Blackgirl5 Nov 11 '14 at 02:47
  • Not at all. The answer is "yes" for both--"an atom" means "at least one atom," not "exactly one atom." Also, don't forget that a single atom can lie below multiple elements. Again, try considering the power set of the naturals. What are the atoms there? Given a non-empty subset of the naturals, can you find an atom below (or equal to) it? – Cameron Buie Nov 11 '14 at 02:48
  • Given a subset of the naturals, the atom is the singleton that appear in this subset...Right? – Blackgirl5 Nov 11 '14 at 03:00
  • I think again for Question 2. Why was I wrong? – Blackgirl5 Nov 11 '14 at 03:02
  • I have expanded my answer to provide a more elaborate explanation. As for the problem with your reasoning for question 2: if a Boolean algebra has $2$ or more atoms, there will *always* be more elements than atoms, regardless of whether each non-least, non-atomic element lies over some atom. – Cameron Buie Nov 11 '14 at 03:11
  • Ok!!I reread my class notes and I think I understand why the answer for question 2 is yes, but I still want to hear your explanation for sure!! Thank you! – Blackgirl5 Nov 11 '14 at 03:12
  • Let me make a slight correction to my last comment: Given *any* Boolean algebra, there will *always* be more elements than atoms. – Cameron Buie Nov 11 '14 at 14:28
  • @CameronBuie the wording of the question is throwing me off here. It reads,

    "Does there exist an infinite Boolean algebra B such that for every nonzero b ∈ B there is an atom a ∈ B with a ≤ b?"

    If we consider the power set of natural numbers as our Boolean Algebra, then there are infinitely many atoms. How then, can we claim there is "an atom" such that a ≤ b, if we take b to be another atom?

     – User38 Dec 14 '14 at 16:08
  • @User38: The kicker is in the phrasing. We are not saying "there is an atom $a\in B$ such that for every non-zero element $b\in B$...." That would mean that there is one $a$ that works for every non-zero $b\in B,$ which, as you point out, is clearly not the case. Rather, we are saying that "for every non-zero $b\in B,$ there is an atom $a\in B$ such that...." This means that, regardless of any particular choice of a non-zero $b\in B,$ we can find some $a\in B$ such that.... (cont'd) – Cameron Buie Dec 15 '14 at 00:49
  • For another (perhaps clearer) example: for every $b\in\Bbb R,$ there exists some $a\in\Bbb R$ such that $a<b.$ This doesn't mean that there is an $a\in\Bbb R$ less than every $a\in\Bbb R$. If that were true, then we would have $a<a,$ which is nonsense! Instead, we mean that, for any given $b\in\Bbb R,$ there is some $a\in\Bbb R$ such that $a<b.$ That is certainly true--just put $a=b-1$. – Cameron Buie Dec 15 '14 at 00:52
  • Back to the original phrasing, and the example of the power set of the natural numbers, take any non-zero $b\in B.$ Since $b$ is non-zero, then there is some $n\in b.$ Put $a={n}.$ Then $a$ is an atom of $B$ and $a\le b,$ as desired. – Cameron Buie Dec 15 '14 at 00:54
  • @CameronBuie Ohh!! I finally understand now. Thanks so much! We had that question on assignment a while back and I never understood why I got it wrong.. – User38 Dec 15 '14 at 10:51
  • @User38: Glad I could help! – Cameron Buie Dec 15 '14 at 12:18