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?

  • 1
    $\begingroup$ For #2, are you sure that whoever gave you this problem didn't mean "for every nonzero $b$ contained in $B$"? For #1, you answered yes - what was your example? Is it also an example for #2? $\endgroup$ – Carl Mummert Nov 6 '14 at 1:51
  • $\begingroup$ For question 2, it has been specified that b is nonzero. $\endgroup$ – Blackgirl5 Nov 6 '14 at 1:58
  • $\begingroup$ For question 1, I just guessed the answer...I just didn't understand well this part. $\endgroup$ – Blackgirl5 Nov 6 '14 at 2:03
  • $\begingroup$ The topic of infinite atomless Boolean algebras has come up here previously. In large part the meaningful parts of this Question were settled there. $\endgroup$ – hardmath Nov 6 '14 at 3:21

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."

  • $\begingroup$ Using your hint for question 2, I assumed that there is no such infinite boolean algebra...Right? $\endgroup$ – Blackgirl5 Nov 6 '14 at 1:51
  • $\begingroup$ Sorry, but I cannot understand the hint... $\endgroup$ – Blackgirl5 Nov 6 '14 at 1:53
  • $\begingroup$ 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 $B$). $\endgroup$ – Cameron Buie Nov 6 '14 at 4:31
  • $\begingroup$ 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. $\endgroup$ – Blackgirl5 Nov 11 '14 at 2:41
  • $\begingroup$ 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. $\endgroup$ – Blackgirl5 Nov 11 '14 at 2:47

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