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












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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?






set-theory infinity axiom-of-choice axioms







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  • $begingroup$
    No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
    $endgroup$
    – Andrés E. Caicedo
    Jan 5 at 23:48






  • 1




    $begingroup$
    @AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
    $endgroup$
    – Gödel
    Jan 5 at 23:52










  • $begingroup$
    math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 0:31












  • $begingroup$
    I dk why you got a down-vote. I countered it.
    $endgroup$
    – DanielWainfleet
    Jan 14 at 1:34
















0












$begingroup$


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?






set-theory infinity axiom-of-choice axioms







share|cite|improve this question









$endgroup$












  • $begingroup$
    No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
    $endgroup$
    – Andrés E. Caicedo
    Jan 5 at 23:48






  • 1




    $begingroup$
    @AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
    $endgroup$
    – Gödel
    Jan 5 at 23:52










  • $begingroup$
    math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 0:31












  • $begingroup$
    I dk why you got a down-vote. I countered it.
    $endgroup$
    – DanielWainfleet
    Jan 14 at 1:34














0












0








0





$begingroup$


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?






set-theory infinity axiom-of-choice axioms







share|cite|improve this question









$endgroup$




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?






set-theory infinity axiom-of-choice axioms




set-theory infinity axiom-of-choice axioms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 23:43









GödelGödel

1,416319




1,416319












  • $begingroup$
    No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
    $endgroup$
    – Andrés E. Caicedo
    Jan 5 at 23:48






  • 1




    $begingroup$
    @AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
    $endgroup$
    – Gödel
    Jan 5 at 23:52










  • $begingroup$
    math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 0:31












  • $begingroup$
    I dk why you got a down-vote. I countered it.
    $endgroup$
    – DanielWainfleet
    Jan 14 at 1:34


















  • $begingroup$
    No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
    $endgroup$
    – Andrés E. Caicedo
    Jan 5 at 23:48






  • 1




    $begingroup$
    @AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
    $endgroup$
    – Gödel
    Jan 5 at 23:52










  • $begingroup$
    math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 0:31












  • $begingroup$
    I dk why you got a down-vote. I countered it.
    $endgroup$
    – DanielWainfleet
    Jan 14 at 1:34
















$begingroup$
No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
$endgroup$
– Andrés E. Caicedo
Jan 5 at 23:48




$begingroup$
No, the negation of choice does not imply the existence of infinite Dedekind-finite sets.
$endgroup$
– Andrés E. Caicedo
Jan 5 at 23:48




1




1




$begingroup$
@AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
$endgroup$
– Gödel
Jan 5 at 23:52




$begingroup$
@AndrésE.Caicedo So, $ZF+(X text{is infinite and Dedekind-finite})$ is consistent.
$endgroup$
– Gödel
Jan 5 at 23:52












$begingroup$
math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:31






$begingroup$
math.stackexchange.com/questions/1017361/… math.stackexchange.com/questions/343923/… math.stackexchange.com/questions/1395029/… and probably the most relevant, math.stackexchange.com/questions/2473059/… and math.stackexchange.com/questions/199087/…
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:31














$begingroup$
I dk why you got a down-vote. I countered it.
$endgroup$
– DanielWainfleet
Jan 14 at 1:34




$begingroup$
I dk why you got a down-vote. I countered it.
$endgroup$
– DanielWainfleet
Jan 14 at 1:34










2 Answers
2






active

oldest

votes


















5












$begingroup$

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" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. 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.







share|cite|improve this answer









$endgroup$













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



















1












$begingroup$

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.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    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" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. 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.







    share|cite|improve this answer









    $endgroup$













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
















    5












    $begingroup$

    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" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. 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.







    share|cite|improve this answer









    $endgroup$













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














    5












    5








    5





    $begingroup$

    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" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. 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.







    share|cite|improve this answer









    $endgroup$



    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" $Mmodels$ ZFC and add a generic sequence of Cohen reals $mathcal{G}=(g_i)_{iinomega}$. 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.








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 6 at 0:00









    Noah SchweberNoah Schweber

    124k10150287




    124k10150287












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


















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
















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




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











    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 0:51









        spaceisdarkgreenspaceisdarkgreen

        32.8k21753




        32.8k21753






























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