Let $\mathfrak B$ be a Boolean algebra with an infinite power $\kappa$. My question is how many ultrafilters does it have? $\kappa$ or $2^\kappa$? Or even smaller?
Asked
Active
Viewed 987 times
7
Andrés E. Caicedo
- 79,201
Popopo
- 2,242
-
I know in finite case it has power $n$ where $|\mathfrak B|=2^n$, so I guess in the infinite case the cardinality should also not greater than it of $\mathfrak{B}$. – Popopo Jun 08 '12 at 19:07
1 Answers
8
It can be at least those two options:
Example I:
Consider the algebra $\{A\subseteq\mathbb R\mid A\text{ finite, or }\mathbb R\setminus A\text{ is finite}\}$, that is finite co-finite subsets of $\mathbb R$. It is not hard to verify that this is indeed a Boolean algebra of size $2^{\aleph_0}$.
Suppose that $U$ is an ultrafilter over this Boolean algebra, if it does not contain any singleton then it has to be the collection of co-finite sets; if it contains a singleton then it is principal. We have continuum many principal ultrafilters and one free. Therefore a Boolean algebra of size continuum with continuum many ultrafilters.
Example II: On the other hand consider $\mathcal P(\mathbb N)$. We know that there are $2^{2^{\aleph_0}}$ many ultrafilters over this Boolean algebra. So we have a Boolean algebra of size $2^{\aleph_0}$ with $2^{2^{\aleph_0}}$ many ultrafilters.
For a further discussion on this sort of example see: The set of ultrafilters on an infinite set is uncountable
Asaf Karagila
- 393,674
-
@Asaf_Karagila:Thank you. But Example II is not so clear for me, could you give me a reference? – Popopo Jun 08 '12 at 19:23
-
1@Popopo: I added a link with a discussion for such result. In our case we look for "ultrafilters over $\mathbb N$" which mean essentially ultrafilters in the Boolean algebra $\mathcal P(\mathbb N)$. – Asaf Karagila Jun 08 '12 at 19:34
-
-