Is there such a thing as "completion with respect to the axiom of choice?"

Question

Completions abound in mathematics. For example:

1. the completion a metric space with respect to Cauchy sequences
2. the algebraic closure of a field
3. the Stone-Čech compactification of a topological space.

In all of the above examples, we start of with a model $M$ that is "missing" some elements, and find a way to add them.

Now suppose $M$ is a model of $\mathrm{ZF}$. The way I see it, "AC fails for $M$" just means that "$M$ is missing functions." In particular, it means that some non-empty family of non-empty sets $\mathcal{F} \in M$ has the property that $\prod \mathcal{F}$ is empty (rather than being packed to the brim full of functions). So I ask:

Question. Is there notion of "the completion of $M$ with respect to the axiom of choice?"

Answer 1

No. There is no such notion.

It is sometimes possible to extend $M$ to a model of $\sf ZFC$ without adding ordinals. But this extension is not always canonical, and certainly not unique.

But there are models of $\sf ZF$ which cannot be extended to models of $\sf ZFC$ without adding ordinals. For example, Gitik's model where every limit ordinal has cofinality $\omega$. If you add back choice, then by induction we can show that every ordinal in the model is countable (the model itself is constructed by destroying the power set axiom using forcing).

This is a bit of a big deal, in this sense, because much like in a completion of a metric space the original space is dense, we expect that the ordinals of $M$ are the ordinals of the larger model. Otherwise we may end up changing the model too much.

It should be remarked, though, that given any family of non-empty sets $\cal F$, we can always force a choice function from that particular family. Often in the cost of making $\bigcup\cal F$ countable (or otherwise well-orderable), which may or may not collapse cardinals.

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It should be remarked, regardless to the above, that sometimes there is an extension which does add ordinals:

Jon Barwise, Admissible Sets and Structures: An Approach to Definability Theory (Berlin: Springer-Verlag, 1975), 365-379.

Every countable transitive model of $\sf ZF$ has an end-extension satisfying $\sf ZF+\it V=L$. Note, however, that more often than not this is a non-standard model.