2

Suppose that $A$ is a free $\sigma$-algebra on a countable set of generators. Does $A$ have a $\sigma$-ultrafilter?

I have no difficulty in proving that $A$ contains at least an ultrafilter. For example, consider the set $F$ of all non-zero elements of $A$. Then

(1) $1$ (the top element of $A$) is non-empty, so $1\in F$.

(2) Suppose that $a\in F$ and that there exists $b\in A$ such that $a\leq b$. Then $b$ is also non-empty because it contains $a$, so $b\in F$.

(3) If $a,b\in F$, then $a\wedge b$ is also non-empty, so $a\wedge b\in F$.

This proves that $F$ is a filter.

(4) Suppose now that both $a,-a\in F$. Then so does their meet, which is a contradiction because $a\wedge -a$ is empty. Hence, $F$ is an ultrafilter.

Consider now a sequence $a_1,a_2,...$ of non-empty elements of $A$. Does $a_1\wedge a_2 \wedge ...\in F$?

As for me, I don't see why it should not be the case. After all, the only one way for $a_1\wedge a_2 \wedge ...$ to be empty is if one of $a_1,a_2,...$ is, but all of them are by definition non-empty, so it must be the case that $a_1\wedge a_2 \wedge ...$ is in $F$. Yet I learnt to be cautious with free $\sigma$-algebras. Could someone please confirm (or repudiate) that $F$ is a $\sigma$-ultrafilter of $A$?

Beginner
  • 574
  • 1
    Your argument is mistaken. Firstly, the set $F$ of non-zero elements of any Boolean algebra (other than the two-element one) is not a filter, since $a \in F$ and $\neg a \in F$ for each $0 < a < 1$, but $a \wedge \neg a \notin F$. Secondly, your item (4) does not show that $F$ is an ultrafilter. To show that a filter $F$ is an ultrafilter, you need to show that either $a \in F$ or $\neg a \in F$ for each $a$. What you show is simply that it is not the case that both $a \in F$ and $\neg a \in F$, which in fact holds for any filter (although again, your set $F$ is not a filter). – Pilcrow Jun 27 '22 at 19:42
  • @Pilcrow Thank you for pointing out my mistake. I am a bit confused. Typically, the principal filter of $A$ generated by $x$ is defined as $F_x={a\in A|x\leq a}$ (e.g. Wikipedia). Shouldn't we add, therefore, that $x\not=0$ or instead write $F_x={a\in A|x<a}$? – Beginner Jun 29 '22 at 12:27
  • The set ${ a \in A \mid x < a }$ is not a filter in any Boolean algebra (unless $x$ happens to be a coatom). If you want to exclude the total filter, then yes, you can require that $x \neq 0$. – Pilcrow Jun 29 '22 at 20:43

0 Answers0