Extending a Filter in a Well-Ordered Boolean Algebra to an Ultrafilter WITHOUT the Axiom of Choice

Question

Hypothesis: Let $B$ be a well-ordered boolean algebra and let $F \subseteq B$ be a filter on $B$.

Goal: Show that $F$ can be extended to an ultrafilter without the axiom of choice (or any equivalent notion thereof).

I only know how to do this in the case that $B$ is countable:

1. Let $b_0, b_1, \ldots$ denote the countable elements of $B$.

2. Let $0$ denote the minimal element of $B$.

3. Then $0 \notin F$ and if $U$ is an ultrafilter on $B$ then $0 \notin U$.

4. Now begin extending $F$ to $F_1, F_2, \ldots , F_k, \ldots$ by inducting on the members of $B = \{b_0, b_1, \ldots\}$ as follows:

   Inductive Step: Suppose $b_k \notin F_{k-1}$. If $b_k \cdot a = 0$ for any $a \in F_{k-1}$, then let $F_k = F_{k-1} \cup \{\bar{b_k}\}$. Otherwise, let $F_{k} = F_{k-1} \cup \{b_k\}$.

5. Letting $U = \bigcup_{k = 1}^\infty F_k$ gives us an ultrafilter extension of $F$ s.t. $U \subseteq B$ as desired. Furthermore, we did not use the axiom of choice.

But what happens if $B$ is uncountable (but still well-ordered)? Evidently it's still possible to show.

Answer 1

The proof of the general case is similar, proceeding to define a filter by transfinite recursion rather than by recursion on $\mathbb{N}$. (I am using "recursion" to refer to a method of defining a sequence, and "induction" to refer to a method of proving that every element of the sequence has some desired property such as the finite intersection property.)

If $B$ is well-ordered then we can enumerate its elements as $B = \{b_\alpha : \alpha < \kappa\}$ for some ordinal $\kappa$. The successor step of the recursive definition is the same as in the countable case $\kappa = \omega$, although in any case I think you only want to put the new element $b_\alpha$ into $F_{\alpha+1}$ if $b_\alpha \cdot a_0 \cdots a_{n-1} \ne 0$ for every finite subset $\{a_0,\ldots,a_{n-1}\} \subset F_\alpha$, so as to preserve the finite intersection property.

The new ingredient in the proof is the limit step. If we want to define $F_\lambda$ where $\lambda < \kappa$ is a limit ordinal, then we may assume as our inductive hypothesis that we have defined $F_\alpha$ at all previous steps $\alpha < \lambda$ and that these $F_\alpha$'s form an increasing chain under inclusion. Then we simply set $F_\lambda = \bigcup_{\alpha < \lambda} F_\alpha$ and observe that the finite intersection property is preserved under unions of chains.

Remark: It might be possible to define the ultrafilter more directly from the well-ordering using some trick. For example, given a well-ordered vector space one can define a basis for that space simply by taking all vectors that are not in the span of the vectors preceding them in the well-ordering. I don't know if there is a similar trick that would work here, but in any case the advantage of transfinite recursion is that it provides straightforward solutions to a large variety of problems once you know the basics of the method.