Is the function sending each set to its corresponding Hartogs number injective?












0














Let $V$ be the class of all sets, $text{Ord}$ be the class of all ordinals, $text{InOrd}$ be the class of all initial ordinals, and $f:V to text{Ord}$ be the function that sends each set to its corresponding Hartogs number.



As a result, $f(a)=a+1$ for all $a in Bbb N$ and thus $f restriction Bbb N$ is injective.



I find it difficult to imagine the Hartogs numbers of infinite sets.




My question: Are $f,frestriction text{Ord},frestriction text{InOrd}$ injective?




Thank you in advance for your help!






elementary-set-theory ordinals







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  • What is the reason for downvoting my question? Did I do anything wrong?
    – Le Anh Dung
    Dec 5 '18 at 7:35
















0














Let $V$ be the class of all sets, $text{Ord}$ be the class of all ordinals, $text{InOrd}$ be the class of all initial ordinals, and $f:V to text{Ord}$ be the function that sends each set to its corresponding Hartogs number.



As a result, $f(a)=a+1$ for all $a in Bbb N$ and thus $f restriction Bbb N$ is injective.



I find it difficult to imagine the Hartogs numbers of infinite sets.




My question: Are $f,frestriction text{Ord},frestriction text{InOrd}$ injective?




Thank you in advance for your help!






elementary-set-theory ordinals







share|cite|improve this question
























  • What is the reason for downvoting my question? Did I do anything wrong?
    – Le Anh Dung
    Dec 5 '18 at 7:35














0












0








0







Let $V$ be the class of all sets, $text{Ord}$ be the class of all ordinals, $text{InOrd}$ be the class of all initial ordinals, and $f:V to text{Ord}$ be the function that sends each set to its corresponding Hartogs number.



As a result, $f(a)=a+1$ for all $a in Bbb N$ and thus $f restriction Bbb N$ is injective.



I find it difficult to imagine the Hartogs numbers of infinite sets.




My question: Are $f,frestriction text{Ord},frestriction text{InOrd}$ injective?




Thank you in advance for your help!






elementary-set-theory ordinals







share|cite|improve this question















Let $V$ be the class of all sets, $text{Ord}$ be the class of all ordinals, $text{InOrd}$ be the class of all initial ordinals, and $f:V to text{Ord}$ be the function that sends each set to its corresponding Hartogs number.



As a result, $f(a)=a+1$ for all $a in Bbb N$ and thus $f restriction Bbb N$ is injective.



I find it difficult to imagine the Hartogs numbers of infinite sets.




My question: Are $f,frestriction text{Ord},frestriction text{InOrd}$ injective?




Thank you in advance for your help!






elementary-set-theory ordinals




elementary-set-theory ordinals






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share|cite|improve this question













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edited Dec 5 '18 at 3:16







Le Anh Dung

















asked Dec 5 '18 at 3:09









Le Anh DungLe Anh Dung

1,0511521




1,0511521












  • What is the reason for downvoting my question? Did I do anything wrong?
    – Le Anh Dung
    Dec 5 '18 at 7:35


















  • What is the reason for downvoting my question? Did I do anything wrong?
    – Le Anh Dung
    Dec 5 '18 at 7:35
















What is the reason for downvoting my question? Did I do anything wrong?
– Le Anh Dung
Dec 5 '18 at 7:35




What is the reason for downvoting my question? Did I do anything wrong?
– Le Anh Dung
Dec 5 '18 at 7:35










2 Answers
2






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oldest

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2














The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?



The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:




  • Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.


  • We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.


  • But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.




Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.






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  • Thank you so much for your detailed answer!
    – Le Anh Dung
    Dec 5 '18 at 16:25



















0














Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.






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    2 Answers
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    2 Answers
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    The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?



    The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:




    • Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.


    • We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.


    • But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.




    Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.






    share|cite|improve this answer





















    • Thank you so much for your detailed answer!
      – Le Anh Dung
      Dec 5 '18 at 16:25
















    2














    The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?



    The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:




    • Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.


    • We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.


    • But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.




    Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.






    share|cite|improve this answer





















    • Thank you so much for your detailed answer!
      – Le Anh Dung
      Dec 5 '18 at 16:25














    2












    2








    2






    The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?



    The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:




    • Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.


    • We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.


    • But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.




    Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.






    share|cite|improve this answer












    The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?



    The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:




    • Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.


    • We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.


    • But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.




    Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 3:24









    Noah SchweberNoah Schweber

    122k10149284




    122k10149284












    • Thank you so much for your detailed answer!
      – Le Anh Dung
      Dec 5 '18 at 16:25


















    • Thank you so much for your detailed answer!
      – Le Anh Dung
      Dec 5 '18 at 16:25
















    Thank you so much for your detailed answer!
    – Le Anh Dung
    Dec 5 '18 at 16:25




    Thank you so much for your detailed answer!
    – Le Anh Dung
    Dec 5 '18 at 16:25











    0














    Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.






    share|cite|improve this answer


























      0














      Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.






      share|cite|improve this answer
























        0












        0








        0






        Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.






        share|cite|improve this answer












        Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 3:25









        GödelGödel

        1,413319




        1,413319






























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