Vtyjkyuk

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 $F_k = F_k-1 cup barb_k$. 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.

The proof of the general case is similar, proceeding to define a filter by transfinite recursion rather than by recursion on $mathbbN$. (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.