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In the article on ultrafilters, Wikipedia claims that

In ZF without the axiom of choice, it is possible that every ultrafilter is principal.{see p.316, [Halbeisen, L.J.] "Combinatorial Set Theory", Springer 2012}

I assume this means that "All ultrafilters are principal" is consistent with ZF, i.e. that there exist models of ZF in which this statement holds. This is also confirmed by this question.

Now I don't know a lot about model theory and also no do not necessarily need to know the details, I was just surprised because I thought one could construct examples of non-principal ultrafilters. For example, consider the sub-Boolean-algebra $\mathcal{B}$ of $\mathcal{P}(\mathbb{Q})$ generated by the sets of the form

$$ \lbrace x \in \mathbb{Q} \mid a < x \rbrace, \lbrace x \in \mathbb{Q} \mid a > x \rbrace $$

for $a \in \mathbb{Q}$. Then surely the set

$$ \lbrace \lbrace x \in \mathbb{Q} \mid 0 < x < a \rbrace \mid a > 0 \rbrace $$

generates a non-principal ultrafilter

$$ \mathcal{U} = \lbrace U \in \mathcal{B} \mid \exists a > 0: \lbrace x \in \mathbb{Q} \mid 0 < x < a \rbrace \subseteq U \rbrace $$

in $\mathcal{B}$, and this can be concluded without using the axiom of choice?

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  • $\begingroup$ How do you know that the set you gave is contained in a ultrafilter? $\endgroup$ – zarathustra May 8 '17 at 13:45
  • $\begingroup$ I am claiming more, namely that the filter generated by this set if already an ultrafilter. Will edit to make this more clear. $\endgroup$ – Bib-lost May 8 '17 at 13:47
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    $\begingroup$ For an easier example, consider the Boolean algebra $B = \{X\subseteq \mathbb{N}\mid X \text{ is finite or cofinite}\}$. Then $U = \{X\in B\mid X\text{ is cofinite}\}$ is already a non-principal ultrafilter on this Boolean algebra. $\endgroup$ – Alex Kruckman May 8 '17 at 15:46
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It is easy to construct filters which have no principal ultrafilters which extend them. For example, any filter extending the cofinite filter.

But in order to prove that there is any filter extending them to begin with, you need to use the axiom of choice.

Also, note that there is a difference between an ultrafilter on a set, and an ultrafilter on a Boolean algebra. It is true that you can reformulate the existence of ultrafilters as prime ideals on Boolean algebras, but there is something to be said there.

It might well be the case that your defined filter is an ultrafilter on the subalgebra of rays. But how does that help you finding an ultrafilter on the set $\Bbb Q$?

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    $\begingroup$ You mean to say that the claim in this article ("it is possible that all ultrafilters are principal") only holds for ultrafilters on sets, rather than ultrafilters in Boolean algebras in general? $\endgroup$ – Bib-lost May 8 '17 at 13:53
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    $\begingroup$ Yes. That is indeed the standard use of the terminology. $\endgroup$ – Asaf Karagila May 8 '17 at 13:54
  • $\begingroup$ Thanks. That was not clear from the Wikipedia page, where the two concepts are used and no clear distinction is made between statements about one or the other. $\endgroup$ – Bib-lost May 8 '17 at 13:55

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