# Example of a set of real numbers that is Dedekind-finite but not finite

Without assuming $$AC$$, can we find an explicit example of a subset of $$\mathbb{R}$$ such that it is not finite but it is Dedekind-finite?

• No, the negation of choice does not imply the existence of infinite Dedekind-finite sets. – Andrés E. Caicedo Jan 5 '19 at 23:48
• @AndrésE.Caicedo So, $ZF+(X\ \text{is infinite and Dedekind-finite})$ is consistent. – Gödel Jan 5 '19 at 23:52
• – Asaf Karagila Jan 6 '19 at 0:31
• I dk why you got a down-vote. I countered it. – DanielWainfleet Jan 14 '19 at 1:34

Obviously it depends what you mean by "explicit," but here are a couple weak positive comments:

• In the usual Cohen construction of a model of ZF+$$\neg$$AC, we take a (countable transitie) "ground model" $$M\models$$ ZFC and add a generic sequence of Cohen reals $$\mathcal{G}=(g_i)_{i\in\omega}$$. The resulting generic extension $$M[\mathcal{G}]$$ is still a model of ZFC; to kill choice, we (in a precise sense) throw out the ordering on the elements of $$\mathcal{G}$$, adding only the set $$G=ran(\mathcal{G})$$. In the resulting inner model $$N$$, that set $$G$$ is a Dedekind-finite infinite set of reals. So that's explicit relative to the original construction of the model.

• A more satisfying answer might be given by this construction of Arnie Miller, who builds a model of ZF in which there is an infinite Dedekind-finite set of reals of low Borel rank.

• Thanks for your answer, It's so useful to me! – Gödel Jan 6 '19 at 0:12

Your question makes it sound like we could do so if we assumed AC. But AC implies we can't find any Dedekind-finite infinite set. And choice is consistent with ZF, so, as Andrés said in the comments, it is consistent with ZF that every infinite subset of the reals is Dedekind-infinite.

But, as you say, it is also consistent with ZF that there is a Dedekind finite, infinite set of reals. But, the fact that the negation is also consistent means you cannot exhibit such a set directly, constructively or non-constructively, without more assumptions than ZF, and these additional assumptions must negate AC.

What you can do in ZF(C) alone is show that if there is a model of ZF, then there is a model of ZF in which such a set exists (and the demonstration that this set exists in this model may be more-or-less explicit). Noah ascertains that this is what you must really want, and he's no doubt right, but I think there's some value in being pedantic here.