Prove that there exist $2^{aleph_{0}}$ well-orderings of the set of all natural numbers.












1














From Chapter 6 Hrbacek and Jech (Ordinal Numbers) Problem 1.3. I don't understand how to begin. Below is the only solution I could find online, but even then I still don't understand how they arrived with those figures.
enter image description here






elementary-set-theory







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  • 2




    How would a chapter and exercise number be helpful without the name of the reference?
    – spaceisdarkgreen
    Nov 19 '18 at 5:34






  • 1




    Of course there are only $aleph_1$ distinct types of well-orderings on $mathbb{N}$.
    – Henno Brandsma
    Nov 19 '18 at 22:18










  • Any bijection of $mathbb{N}$ can be declared an order isomorphism with another well-order (of type $omega$), e.g.
    – Henno Brandsma
    Nov 19 '18 at 22:20
















1














From Chapter 6 Hrbacek and Jech (Ordinal Numbers) Problem 1.3. I don't understand how to begin. Below is the only solution I could find online, but even then I still don't understand how they arrived with those figures.
enter image description here






elementary-set-theory







share|cite|improve this question




















  • 2




    How would a chapter and exercise number be helpful without the name of the reference?
    – spaceisdarkgreen
    Nov 19 '18 at 5:34






  • 1




    Of course there are only $aleph_1$ distinct types of well-orderings on $mathbb{N}$.
    – Henno Brandsma
    Nov 19 '18 at 22:18










  • Any bijection of $mathbb{N}$ can be declared an order isomorphism with another well-order (of type $omega$), e.g.
    – Henno Brandsma
    Nov 19 '18 at 22:20














1












1








1







From Chapter 6 Hrbacek and Jech (Ordinal Numbers) Problem 1.3. I don't understand how to begin. Below is the only solution I could find online, but even then I still don't understand how they arrived with those figures.
enter image description here






elementary-set-theory







share|cite|improve this question















From Chapter 6 Hrbacek and Jech (Ordinal Numbers) Problem 1.3. I don't understand how to begin. Below is the only solution I could find online, but even then I still don't understand how they arrived with those figures.
enter image description here






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








edited Nov 20 '18 at 4:22

























asked Nov 19 '18 at 4:40









Barycentric_Bash

40139




40139








  • 2




    How would a chapter and exercise number be helpful without the name of the reference?
    – spaceisdarkgreen
    Nov 19 '18 at 5:34






  • 1




    Of course there are only $aleph_1$ distinct types of well-orderings on $mathbb{N}$.
    – Henno Brandsma
    Nov 19 '18 at 22:18










  • Any bijection of $mathbb{N}$ can be declared an order isomorphism with another well-order (of type $omega$), e.g.
    – Henno Brandsma
    Nov 19 '18 at 22:20














  • 2




    How would a chapter and exercise number be helpful without the name of the reference?
    – spaceisdarkgreen
    Nov 19 '18 at 5:34






  • 1




    Of course there are only $aleph_1$ distinct types of well-orderings on $mathbb{N}$.
    – Henno Brandsma
    Nov 19 '18 at 22:18










  • Any bijection of $mathbb{N}$ can be declared an order isomorphism with another well-order (of type $omega$), e.g.
    – Henno Brandsma
    Nov 19 '18 at 22:20








2




2




How would a chapter and exercise number be helpful without the name of the reference?
– spaceisdarkgreen
Nov 19 '18 at 5:34




How would a chapter and exercise number be helpful without the name of the reference?
– spaceisdarkgreen
Nov 19 '18 at 5:34




1




1




Of course there are only $aleph_1$ distinct types of well-orderings on $mathbb{N}$.
– Henno Brandsma
Nov 19 '18 at 22:18




Of course there are only $aleph_1$ distinct types of well-orderings on $mathbb{N}$.
– Henno Brandsma
Nov 19 '18 at 22:18












Any bijection of $mathbb{N}$ can be declared an order isomorphism with another well-order (of type $omega$), e.g.
– Henno Brandsma
Nov 19 '18 at 22:20




Any bijection of $mathbb{N}$ can be declared an order isomorphism with another well-order (of type $omega$), e.g.
– Henno Brandsma
Nov 19 '18 at 22:20










1 Answer
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Given the usual order on $Bbb N$, you can show there are at least $2^{aleph_0}$ because you can break $Bbb N$ into $aleph_0$ pairs $(2n, 2n+1)$. You can then invert each pair or not, which is $aleph_0$ binary choices. There are at most $aleph_0^{aleph_0}$ because there are $aleph_0$ choices for the first element of the well order, and again for the next, and so on. If you know $2^{aleph_0}=aleph_0^{aleph_0}=mathfrak c$ you are done.






share|cite|improve this answer

















  • 2




    I think your upper bound argument is kind of vague (and might give the misleading impression that the order type is necessarily $omega$). A more simple way is just that the well orders are all subsets of $mathbb Ntimes mathbb N.$
    – spaceisdarkgreen
    Nov 19 '18 at 5:46










  • @spaceisdarkgreen: good point. When I first read it I thought it claimed the well order was a set of pairs $(a,b)$ where $ain Bbb N$ and $bin Bbb N$ was the position of $a$ in the list, which again claims that the order type is $omega$, but the well order can just be all the pairs $(a,b)$ where $a$ comes before $b$.
    – Ross Millikan
    Nov 19 '18 at 6:04











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Given the usual order on $Bbb N$, you can show there are at least $2^{aleph_0}$ because you can break $Bbb N$ into $aleph_0$ pairs $(2n, 2n+1)$. You can then invert each pair or not, which is $aleph_0$ binary choices. There are at most $aleph_0^{aleph_0}$ because there are $aleph_0$ choices for the first element of the well order, and again for the next, and so on. If you know $2^{aleph_0}=aleph_0^{aleph_0}=mathfrak c$ you are done.






share|cite|improve this answer

















  • 2




    I think your upper bound argument is kind of vague (and might give the misleading impression that the order type is necessarily $omega$). A more simple way is just that the well orders are all subsets of $mathbb Ntimes mathbb N.$
    – spaceisdarkgreen
    Nov 19 '18 at 5:46










  • @spaceisdarkgreen: good point. When I first read it I thought it claimed the well order was a set of pairs $(a,b)$ where $ain Bbb N$ and $bin Bbb N$ was the position of $a$ in the list, which again claims that the order type is $omega$, but the well order can just be all the pairs $(a,b)$ where $a$ comes before $b$.
    – Ross Millikan
    Nov 19 '18 at 6:04
















2














Given the usual order on $Bbb N$, you can show there are at least $2^{aleph_0}$ because you can break $Bbb N$ into $aleph_0$ pairs $(2n, 2n+1)$. You can then invert each pair or not, which is $aleph_0$ binary choices. There are at most $aleph_0^{aleph_0}$ because there are $aleph_0$ choices for the first element of the well order, and again for the next, and so on. If you know $2^{aleph_0}=aleph_0^{aleph_0}=mathfrak c$ you are done.






share|cite|improve this answer

















  • 2




    I think your upper bound argument is kind of vague (and might give the misleading impression that the order type is necessarily $omega$). A more simple way is just that the well orders are all subsets of $mathbb Ntimes mathbb N.$
    – spaceisdarkgreen
    Nov 19 '18 at 5:46










  • @spaceisdarkgreen: good point. When I first read it I thought it claimed the well order was a set of pairs $(a,b)$ where $ain Bbb N$ and $bin Bbb N$ was the position of $a$ in the list, which again claims that the order type is $omega$, but the well order can just be all the pairs $(a,b)$ where $a$ comes before $b$.
    – Ross Millikan
    Nov 19 '18 at 6:04














2












2








2






Given the usual order on $Bbb N$, you can show there are at least $2^{aleph_0}$ because you can break $Bbb N$ into $aleph_0$ pairs $(2n, 2n+1)$. You can then invert each pair or not, which is $aleph_0$ binary choices. There are at most $aleph_0^{aleph_0}$ because there are $aleph_0$ choices for the first element of the well order, and again for the next, and so on. If you know $2^{aleph_0}=aleph_0^{aleph_0}=mathfrak c$ you are done.






share|cite|improve this answer












Given the usual order on $Bbb N$, you can show there are at least $2^{aleph_0}$ because you can break $Bbb N$ into $aleph_0$ pairs $(2n, 2n+1)$. You can then invert each pair or not, which is $aleph_0$ binary choices. There are at most $aleph_0^{aleph_0}$ because there are $aleph_0$ choices for the first element of the well order, and again for the next, and so on. If you know $2^{aleph_0}=aleph_0^{aleph_0}=mathfrak c$ you are done.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 19 '18 at 5:32









Ross Millikan

292k23196371




292k23196371








  • 2




    I think your upper bound argument is kind of vague (and might give the misleading impression that the order type is necessarily $omega$). A more simple way is just that the well orders are all subsets of $mathbb Ntimes mathbb N.$
    – spaceisdarkgreen
    Nov 19 '18 at 5:46










  • @spaceisdarkgreen: good point. When I first read it I thought it claimed the well order was a set of pairs $(a,b)$ where $ain Bbb N$ and $bin Bbb N$ was the position of $a$ in the list, which again claims that the order type is $omega$, but the well order can just be all the pairs $(a,b)$ where $a$ comes before $b$.
    – Ross Millikan
    Nov 19 '18 at 6:04














  • 2




    I think your upper bound argument is kind of vague (and might give the misleading impression that the order type is necessarily $omega$). A more simple way is just that the well orders are all subsets of $mathbb Ntimes mathbb N.$
    – spaceisdarkgreen
    Nov 19 '18 at 5:46










  • @spaceisdarkgreen: good point. When I first read it I thought it claimed the well order was a set of pairs $(a,b)$ where $ain Bbb N$ and $bin Bbb N$ was the position of $a$ in the list, which again claims that the order type is $omega$, but the well order can just be all the pairs $(a,b)$ where $a$ comes before $b$.
    – Ross Millikan
    Nov 19 '18 at 6:04








2




2




I think your upper bound argument is kind of vague (and might give the misleading impression that the order type is necessarily $omega$). A more simple way is just that the well orders are all subsets of $mathbb Ntimes mathbb N.$
– spaceisdarkgreen
Nov 19 '18 at 5:46




I think your upper bound argument is kind of vague (and might give the misleading impression that the order type is necessarily $omega$). A more simple way is just that the well orders are all subsets of $mathbb Ntimes mathbb N.$
– spaceisdarkgreen
Nov 19 '18 at 5:46












@spaceisdarkgreen: good point. When I first read it I thought it claimed the well order was a set of pairs $(a,b)$ where $ain Bbb N$ and $bin Bbb N$ was the position of $a$ in the list, which again claims that the order type is $omega$, but the well order can just be all the pairs $(a,b)$ where $a$ comes before $b$.
– Ross Millikan
Nov 19 '18 at 6:04




@spaceisdarkgreen: good point. When I first read it I thought it claimed the well order was a set of pairs $(a,b)$ where $ain Bbb N$ and $bin Bbb N$ was the position of $a$ in the list, which again claims that the order type is $omega$, but the well order can just be all the pairs $(a,b)$ where $a$ comes before $b$.
– Ross Millikan
Nov 19 '18 at 6:04


















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