Sets Without a Minimal Element (Axiom of Foundation)












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I’m trying to move from an initial appreciation of Cantor’s arguments concerning infinite sets to a more rigorous, axiom-based understanding. I’m having trouble with axiom 8, the Axiom of Foundation or Regularity, in Zermelo-Fraenkel set theory, however. This axiom is explained as saying that every non-empty set has a minimal member, where “minimal” means “does not contain a proper sub-set of the set”.
The other discussions I’ve seen here focus on avoiding infinite descending chains of sets, but it seems straightforward to construct non-descending counter-examples to the axiom.
In a set where the nth member is the set of positive integers starting with value n, for example,



i.e. {{1, 2, 3, 4, …}, {2, 3, 4, 5, …}, {3, 4, 5, 6, …}, …}



every member contains the next member (and all subsequent members) as a proper subset.
Isn’t this true of every set whose members are either infinite series repeated with progressively later starting values, as above, or infinitely sub-divided sets? (E.g. the set of integers contains the set of even numbers, which contains the set of multiples of 4, and so on.)
If not, why not?






elementary-set-theory







share|cite|improve this question









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  • 3




    $begingroup$
    Here minimality is with respect to $in$, not to $subseteq$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 17:41










  • $begingroup$
    No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:46










  • $begingroup$
    "Proper" should be "nonempty."
    $endgroup$
    – Noah Schweber
    Dec 31 '18 at 18:06










  • $begingroup$
    math.stackexchange.com/questions/214408/…
    $endgroup$
    – Asaf Karagila♦
    Dec 31 '18 at 18:15
















0












$begingroup$


I’m trying to move from an initial appreciation of Cantor’s arguments concerning infinite sets to a more rigorous, axiom-based understanding. I’m having trouble with axiom 8, the Axiom of Foundation or Regularity, in Zermelo-Fraenkel set theory, however. This axiom is explained as saying that every non-empty set has a minimal member, where “minimal” means “does not contain a proper sub-set of the set”.
The other discussions I’ve seen here focus on avoiding infinite descending chains of sets, but it seems straightforward to construct non-descending counter-examples to the axiom.
In a set where the nth member is the set of positive integers starting with value n, for example,



i.e. {{1, 2, 3, 4, …}, {2, 3, 4, 5, …}, {3, 4, 5, 6, …}, …}



every member contains the next member (and all subsequent members) as a proper subset.
Isn’t this true of every set whose members are either infinite series repeated with progressively later starting values, as above, or infinitely sub-divided sets? (E.g. the set of integers contains the set of even numbers, which contains the set of multiples of 4, and so on.)
If not, why not?






elementary-set-theory







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Here minimality is with respect to $in$, not to $subseteq$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 17:41










  • $begingroup$
    No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:46










  • $begingroup$
    "Proper" should be "nonempty."
    $endgroup$
    – Noah Schweber
    Dec 31 '18 at 18:06










  • $begingroup$
    math.stackexchange.com/questions/214408/…
    $endgroup$
    – Asaf Karagila♦
    Dec 31 '18 at 18:15














0












0








0





$begingroup$


I’m trying to move from an initial appreciation of Cantor’s arguments concerning infinite sets to a more rigorous, axiom-based understanding. I’m having trouble with axiom 8, the Axiom of Foundation or Regularity, in Zermelo-Fraenkel set theory, however. This axiom is explained as saying that every non-empty set has a minimal member, where “minimal” means “does not contain a proper sub-set of the set”.
The other discussions I’ve seen here focus on avoiding infinite descending chains of sets, but it seems straightforward to construct non-descending counter-examples to the axiom.
In a set where the nth member is the set of positive integers starting with value n, for example,



i.e. {{1, 2, 3, 4, …}, {2, 3, 4, 5, …}, {3, 4, 5, 6, …}, …}



every member contains the next member (and all subsequent members) as a proper subset.
Isn’t this true of every set whose members are either infinite series repeated with progressively later starting values, as above, or infinitely sub-divided sets? (E.g. the set of integers contains the set of even numbers, which contains the set of multiples of 4, and so on.)
If not, why not?






elementary-set-theory







share|cite|improve this question









$endgroup$




I’m trying to move from an initial appreciation of Cantor’s arguments concerning infinite sets to a more rigorous, axiom-based understanding. I’m having trouble with axiom 8, the Axiom of Foundation or Regularity, in Zermelo-Fraenkel set theory, however. This axiom is explained as saying that every non-empty set has a minimal member, where “minimal” means “does not contain a proper sub-set of the set”.
The other discussions I’ve seen here focus on avoiding infinite descending chains of sets, but it seems straightforward to construct non-descending counter-examples to the axiom.
In a set where the nth member is the set of positive integers starting with value n, for example,



i.e. {{1, 2, 3, 4, …}, {2, 3, 4, 5, …}, {3, 4, 5, 6, …}, …}



every member contains the next member (and all subsequent members) as a proper subset.
Isn’t this true of every set whose members are either infinite series repeated with progressively later starting values, as above, or infinitely sub-divided sets? (E.g. the set of integers contains the set of even numbers, which contains the set of multiples of 4, and so on.)
If not, why not?






elementary-set-theory




elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 17:40









Bill SwanBill Swan

31




31








  • 3




    $begingroup$
    Here minimality is with respect to $in$, not to $subseteq$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 17:41










  • $begingroup$
    No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:46










  • $begingroup$
    "Proper" should be "nonempty."
    $endgroup$
    – Noah Schweber
    Dec 31 '18 at 18:06










  • $begingroup$
    math.stackexchange.com/questions/214408/…
    $endgroup$
    – Asaf Karagila♦
    Dec 31 '18 at 18:15














  • 3




    $begingroup$
    Here minimality is with respect to $in$, not to $subseteq$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 17:41










  • $begingroup$
    No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:46










  • $begingroup$
    "Proper" should be "nonempty."
    $endgroup$
    – Noah Schweber
    Dec 31 '18 at 18:06










  • $begingroup$
    math.stackexchange.com/questions/214408/…
    $endgroup$
    – Asaf Karagila♦
    Dec 31 '18 at 18:15








3




3




$begingroup$
Here minimality is with respect to $in$, not to $subseteq$.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 17:41




$begingroup$
Here minimality is with respect to $in$, not to $subseteq$.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 17:41












$begingroup$
No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
$endgroup$
– user3482749
Dec 31 '18 at 17:46




$begingroup$
No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
$endgroup$
– user3482749
Dec 31 '18 at 17:46












$begingroup$
"Proper" should be "nonempty."
$endgroup$
– Noah Schweber
Dec 31 '18 at 18:06




$begingroup$
"Proper" should be "nonempty."
$endgroup$
– Noah Schweber
Dec 31 '18 at 18:06












$begingroup$
math.stackexchange.com/questions/214408/…
$endgroup$
– Asaf Karagila♦
Dec 31 '18 at 18:15




$begingroup$
math.stackexchange.com/questions/214408/…
$endgroup$
– Asaf Karagila♦
Dec 31 '18 at 18:15










2 Answers
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Regularity is focusing on $in$, not $subseteq$.




First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$



In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.




That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.



Regularity says:




For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.




Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
      $endgroup$
      – Bill Swan
      Jan 2 at 10:48











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Regularity is focusing on $in$, not $subseteq$.




    First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$



    In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.




    That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.



    Regularity says:




    For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.




    Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Regularity is focusing on $in$, not $subseteq$.




      First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$



      In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.




      That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.



      Regularity says:




      For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.




      Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Regularity is focusing on $in$, not $subseteq$.




        First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$



        In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.




        That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.



        Regularity says:




        For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.




        Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.






        share|cite|improve this answer









        $endgroup$



        Regularity is focusing on $in$, not $subseteq$.




        First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$



        In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.




        That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.



        Regularity says:




        For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.




        Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 18:05









        Noah SchweberNoah Schweber

        126k10150288




        126k10150288























            0












            $begingroup$

            every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
              $endgroup$
              – Bill Swan
              Jan 2 at 10:48
















            0












            $begingroup$

            every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
              $endgroup$
              – Bill Swan
              Jan 2 at 10:48














            0












            0








            0





            $begingroup$

            every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)






            share|cite|improve this answer









            $endgroup$



            every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 31 '18 at 18:04









            HoloHolo

            5,87421131




            5,87421131












            • $begingroup$
              Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
              $endgroup$
              – Bill Swan
              Jan 2 at 10:48


















            • $begingroup$
              Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
              $endgroup$
              – Bill Swan
              Jan 2 at 10:48
















            $begingroup$
            Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
            $endgroup$
            – Bill Swan
            Jan 2 at 10:48




            $begingroup$
            Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
            $endgroup$
            – Bill Swan
            Jan 2 at 10:48


















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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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            Gyllenstierna

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