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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?

YCB
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  • No, the negation of choice does not imply the existence of infinite Dedekind-finite sets. – Andrés E. Caicedo Jan 05 '19 at 23:48
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    @AndrésE.Caicedo So, $ZF+(X\ \text{is infinite and Dedekind-finite})$ is consistent. – YCB Jan 05 '19 at 23:52
  • https://math.stackexchange.com/questions/1017361/what-is-the-opposite-of-the-axiom-of-choice https://math.stackexchange.com/questions/343923/failure-of-choice-only-for-sets-above-a-certain-rank https://math.stackexchange.com/questions/1395029/does-the-special-continuum-hypothesis-imply-the-axiom-of-choice and probably the most relevant, https://math.stackexchange.com/questions/2473059/elucidating-the-nature-of-infinite-dedekind-finite-subsets-in-the-reals and https://math.stackexchange.com/questions/199087/which-set-is-unwell-orderable – Asaf Karagila Jan 06 '19 at 00:31
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    I dk why you got a down-vote. I countered it. – DanielWainfleet Jan 14 '19 at 01:34
  • If someone wants one example directly, the reference link https://math.stackexchange.com/questions/2473059/elucidating-the-nature-of-infinite-dedekind-finite-subsets-in-the-reals of Asaf Karagila will help. – An5Drama May 01 '24 at 12:54

2 Answers2

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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.

Noah Schweber
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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.

spaceisdarkgreen
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