Can we take the intersection of ALL successor(inductive) sets?












0












$begingroup$


In Halmos Naive set theory, there is the following excerpt introducing natural numbers :




In this language the axiom of infinity simply says that there exists a successor [inductive] set A. Since the intersection of every (non-empty) family of successor sets is a successor set itself (proof?), the intersection of all the successor sets included in A is a successor set $omega$.




I can prove the first part(given that the family is finite), however, I'm not sure why we can extend the conclusion to the intersection of all the successor sets(bolded), which may be an intersection of infinitely many sets, due to the Axiom of Infinity. However, in Halmos book infinite intersection is not yet defined... To be succinct, my problem is focused on the word "all", why it could be used here?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    See page 15 for the definition of the intersection $cap mathcal C$ of a (non empty) collection $mathcal C$ of sets : there is no restrictions, provided that the collection is not empty.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 28 at 8:08








  • 1




    $begingroup$
    Thus, the argument of page 44 is : we have a successor set by Axiom of Infinity. Thus, consider the collection of all successor sstes (that is not empty by the axiom) and apply the operation of intersection above.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 28 at 8:11










  • $begingroup$
    @MauroALLEGRANZA Thank you. I checked that page and persuaded myself indeed there's no limitation on the "finiteness" of the collection for intersection. I should have read the text more carefully. Please, add this comment as an answer and I should accept it, as it's the crux of my question here.
    $endgroup$
    – Macrophage
    Jan 28 at 8:17
















0












$begingroup$


In Halmos Naive set theory, there is the following excerpt introducing natural numbers :




In this language the axiom of infinity simply says that there exists a successor [inductive] set A. Since the intersection of every (non-empty) family of successor sets is a successor set itself (proof?), the intersection of all the successor sets included in A is a successor set $omega$.




I can prove the first part(given that the family is finite), however, I'm not sure why we can extend the conclusion to the intersection of all the successor sets(bolded), which may be an intersection of infinitely many sets, due to the Axiom of Infinity. However, in Halmos book infinite intersection is not yet defined... To be succinct, my problem is focused on the word "all", why it could be used here?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    See page 15 for the definition of the intersection $cap mathcal C$ of a (non empty) collection $mathcal C$ of sets : there is no restrictions, provided that the collection is not empty.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 28 at 8:08








  • 1




    $begingroup$
    Thus, the argument of page 44 is : we have a successor set by Axiom of Infinity. Thus, consider the collection of all successor sstes (that is not empty by the axiom) and apply the operation of intersection above.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 28 at 8:11










  • $begingroup$
    @MauroALLEGRANZA Thank you. I checked that page and persuaded myself indeed there's no limitation on the "finiteness" of the collection for intersection. I should have read the text more carefully. Please, add this comment as an answer and I should accept it, as it's the crux of my question here.
    $endgroup$
    – Macrophage
    Jan 28 at 8:17














0












0








0





$begingroup$


In Halmos Naive set theory, there is the following excerpt introducing natural numbers :




In this language the axiom of infinity simply says that there exists a successor [inductive] set A. Since the intersection of every (non-empty) family of successor sets is a successor set itself (proof?), the intersection of all the successor sets included in A is a successor set $omega$.




I can prove the first part(given that the family is finite), however, I'm not sure why we can extend the conclusion to the intersection of all the successor sets(bolded), which may be an intersection of infinitely many sets, due to the Axiom of Infinity. However, in Halmos book infinite intersection is not yet defined... To be succinct, my problem is focused on the word "all", why it could be used here?










share|cite|improve this question









$endgroup$




In Halmos Naive set theory, there is the following excerpt introducing natural numbers :




In this language the axiom of infinity simply says that there exists a successor [inductive] set A. Since the intersection of every (non-empty) family of successor sets is a successor set itself (proof?), the intersection of all the successor sets included in A is a successor set $omega$.




I can prove the first part(given that the family is finite), however, I'm not sure why we can extend the conclusion to the intersection of all the successor sets(bolded), which may be an intersection of infinitely many sets, due to the Axiom of Infinity. However, in Halmos book infinite intersection is not yet defined... To be succinct, my problem is focused on the word "all", why it could be used here?







elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 28 at 7:59









MacrophageMacrophage

1,191115




1,191115








  • 1




    $begingroup$
    See page 15 for the definition of the intersection $cap mathcal C$ of a (non empty) collection $mathcal C$ of sets : there is no restrictions, provided that the collection is not empty.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 28 at 8:08








  • 1




    $begingroup$
    Thus, the argument of page 44 is : we have a successor set by Axiom of Infinity. Thus, consider the collection of all successor sstes (that is not empty by the axiom) and apply the operation of intersection above.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 28 at 8:11










  • $begingroup$
    @MauroALLEGRANZA Thank you. I checked that page and persuaded myself indeed there's no limitation on the "finiteness" of the collection for intersection. I should have read the text more carefully. Please, add this comment as an answer and I should accept it, as it's the crux of my question here.
    $endgroup$
    – Macrophage
    Jan 28 at 8:17














  • 1




    $begingroup$
    See page 15 for the definition of the intersection $cap mathcal C$ of a (non empty) collection $mathcal C$ of sets : there is no restrictions, provided that the collection is not empty.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 28 at 8:08








  • 1




    $begingroup$
    Thus, the argument of page 44 is : we have a successor set by Axiom of Infinity. Thus, consider the collection of all successor sstes (that is not empty by the axiom) and apply the operation of intersection above.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 28 at 8:11










  • $begingroup$
    @MauroALLEGRANZA Thank you. I checked that page and persuaded myself indeed there's no limitation on the "finiteness" of the collection for intersection. I should have read the text more carefully. Please, add this comment as an answer and I should accept it, as it's the crux of my question here.
    $endgroup$
    – Macrophage
    Jan 28 at 8:17








1




1




$begingroup$
See page 15 for the definition of the intersection $cap mathcal C$ of a (non empty) collection $mathcal C$ of sets : there is no restrictions, provided that the collection is not empty.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 8:08






$begingroup$
See page 15 for the definition of the intersection $cap mathcal C$ of a (non empty) collection $mathcal C$ of sets : there is no restrictions, provided that the collection is not empty.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 8:08






1




1




$begingroup$
Thus, the argument of page 44 is : we have a successor set by Axiom of Infinity. Thus, consider the collection of all successor sstes (that is not empty by the axiom) and apply the operation of intersection above.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 8:11




$begingroup$
Thus, the argument of page 44 is : we have a successor set by Axiom of Infinity. Thus, consider the collection of all successor sstes (that is not empty by the axiom) and apply the operation of intersection above.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 8:11












$begingroup$
@MauroALLEGRANZA Thank you. I checked that page and persuaded myself indeed there's no limitation on the "finiteness" of the collection for intersection. I should have read the text more carefully. Please, add this comment as an answer and I should accept it, as it's the crux of my question here.
$endgroup$
– Macrophage
Jan 28 at 8:17




$begingroup$
@MauroALLEGRANZA Thank you. I checked that page and persuaded myself indeed there's no limitation on the "finiteness" of the collection for intersection. I should have read the text more carefully. Please, add this comment as an answer and I should accept it, as it's the crux of my question here.
$endgroup$
– Macrophage
Jan 28 at 8:17










1 Answer
1






active

oldest

votes


















1












$begingroup$

The key point is the point right after your bold text. Included in $A$. Simply consider



$${ain Amidforall Bsubseteq A:Btext{ is a successor set}to ain B},$$



Or, in simpler terms, $bigcap{Bsubseteq Amid Btext{ is a successor set}}$. Separation is enough here, of course.



What is important for this proof, as a whole, is that $A$ is any successor set. It does not matter which one you take, so as long as you have at least one of them, you define the same $omega$.






share|cite|improve this answer









$endgroup$













    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%2f3090592%2fcan-we-take-the-intersection-of-all-successorinductive-sets%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The key point is the point right after your bold text. Included in $A$. Simply consider



    $${ain Amidforall Bsubseteq A:Btext{ is a successor set}to ain B},$$



    Or, in simpler terms, $bigcap{Bsubseteq Amid Btext{ is a successor set}}$. Separation is enough here, of course.



    What is important for this proof, as a whole, is that $A$ is any successor set. It does not matter which one you take, so as long as you have at least one of them, you define the same $omega$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The key point is the point right after your bold text. Included in $A$. Simply consider



      $${ain Amidforall Bsubseteq A:Btext{ is a successor set}to ain B},$$



      Or, in simpler terms, $bigcap{Bsubseteq Amid Btext{ is a successor set}}$. Separation is enough here, of course.



      What is important for this proof, as a whole, is that $A$ is any successor set. It does not matter which one you take, so as long as you have at least one of them, you define the same $omega$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The key point is the point right after your bold text. Included in $A$. Simply consider



        $${ain Amidforall Bsubseteq A:Btext{ is a successor set}to ain B},$$



        Or, in simpler terms, $bigcap{Bsubseteq Amid Btext{ is a successor set}}$. Separation is enough here, of course.



        What is important for this proof, as a whole, is that $A$ is any successor set. It does not matter which one you take, so as long as you have at least one of them, you define the same $omega$.






        share|cite|improve this answer









        $endgroup$



        The key point is the point right after your bold text. Included in $A$. Simply consider



        $${ain Amidforall Bsubseteq A:Btext{ is a successor set}to ain B},$$



        Or, in simpler terms, $bigcap{Bsubseteq Amid Btext{ is a successor set}}$. Separation is enough here, of course.



        What is important for this proof, as a whole, is that $A$ is any successor set. It does not matter which one you take, so as long as you have at least one of them, you define the same $omega$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 8:10









        Asaf KaragilaAsaf Karagila

        307k33439770




        307k33439770






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090592%2fcan-we-take-the-intersection-of-all-successorinductive-sets%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

            android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

            SQL update select statement

            'app-layout' is not a known element: how to share Component with different Modules