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












-1














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






















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






  • 1




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










  • 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/…
    – Asaf Karagila♦
    Jan 6 at 0:31


















-1














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






















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






  • 1




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










  • 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/…
    – Asaf Karagila♦
    Jan 6 at 0:31
















-1












-1








-1







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













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

1,413319




1,413319












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






  • 1




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










  • 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/…
    – Asaf Karagila♦
    Jan 6 at 0:31




















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






  • 1




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










  • 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/…
    – Asaf Karagila♦
    Jan 6 at 0:31


















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




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




1




1




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




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












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/…
– Asaf Karagila♦
Jan 6 at 0:31






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/…
– Asaf Karagila♦
Jan 6 at 0:31












2 Answers
2






active

oldest

votes


















4














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





















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



















1














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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063329%2fexample-of-a-set-of-real-numbers-that-is-dedekind-finite-but-not-finite%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    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





















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
















    4














    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





















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














    4












    4








    4






    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












    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

    122k10149284




    122k10149284












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


















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
















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




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











    1














    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


























      1














      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
























        1












        1








        1






        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












        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

        32.5k21753




        32.5k21753






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063329%2fexample-of-a-set-of-real-numbers-that-is-dedekind-finite-but-not-finite%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
            Read more