Asaf Karagila
I don't have much choice...

We all know and love Cohen's first model where the axiom of choice fails. It is the O.G. symmetric extension. But Cohen didn't invent the idea on his own, he used Fraenkel's ideas from his work on set theory with atoms and permutation models. The two results, however, are significantly different.

Fraenkel's construction does not affect sets of ordinals, in particular the real numbers can still be well-ordered in his models. Cohen's work, however, directly breaks that. The Dedekind-finite set added is a set of reals. In particular, the reals cannot be well-ordered no more.

But this very thing has another effect. Fraenkel's construction results in an amorphous set. Namely, a set which cannot be split into two infinite sets. But as a set of reals you can always split it into infinite sets: which reals have $$0$$ and which don't, for example, is such partition in Cohen's model (here we use the fact that generic reals are very generic, of course). Another effect is that an amorphous set cannot be linearly ordered, and a set of reals, well, is linearly ordered.

Now. I am sure we all remember Kuratowski's celebrated theorem: $\aleph_0\leq^*\mathfrak p\iff\aleph_0\leq 2^\mathfrak p.$ There is a surjection onto $$\omega$$ if and only if $$\omega$$ injects into the power set. One direction is trivial, the other not so much. And how does this tie into the whole thing?

Well, an amorphous set cannot be mapped onto $$\omega$$, since that would imply it can be split into two infinite sets (the preimage of the even and preimage of the odd, for example). So its power set is Dedekind-finite. But that means that the finite/co-finite algebra is a $$\sigma$$-algebra! Indeed, that much is true for any set with a Dedekind-finite power set.

What about the converse? Will specifically, can we find a set whose power set is not Dedekind-finite, but its finite subsets are? And it hit me, that the answer is (as often is the case) right under my nose in Cohen's model. The canonical Dedekind-finite set of reals (and in fact any Dedekind-finite set in the model) can be mapped onto $$\omega$$ quite easily. So its power set is certainly Dedekind-infinite. However, since the set is linearly ordered, its finite subsets can be uniformly enumerated. In particular, if there was a countable set of finite subsets, its union would be a countable subset.

So in Cohen's model the Dedekind-finite sets are not amorphous, but their finite/co-finite still form an $$\infty$$-complete Boolean algebra!

Odd. I know.

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