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?
 A: 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.
A: 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.
