“All ultrafilters are principal” consistent with ZF?

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?

• How do you know that the set you gave is contained in a ultrafilter? – zarathustra May 8 '17 at 13:45
• I am claiming more, namely that the filter generated by this set if already an ultrafilter. Will edit to make this more clear. – Bib-lost May 8 '17 at 13:47
• 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. – Alex Kruckman May 8 '17 at 15:46

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 $\Bbb Q$?