Proving that $langlemathbb{R}-{7},<rangle$ is an elementary substructure of $langlemathbb{R},<rangle$












1












$begingroup$



Prove that $langlemathbb{R}-{7},<rangle$ is an elementary substructure of $langlemathbb{R},<rangle$.




What I thought I should do is to use induction on statements to prove that. Any easier way to acomplish that?






logic model-theory







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  • $begingroup$
    If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
    $endgroup$
    – tomasz
    Jan 7 at 10:36
















1












$begingroup$



Prove that $langlemathbb{R}-{7},<rangle$ is an elementary substructure of $langlemathbb{R},<rangle$.




What I thought I should do is to use induction on statements to prove that. Any easier way to acomplish that?






logic model-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
    $endgroup$
    – tomasz
    Jan 7 at 10:36














1












1








1





$begingroup$



Prove that $langlemathbb{R}-{7},<rangle$ is an elementary substructure of $langlemathbb{R},<rangle$.




What I thought I should do is to use induction on statements to prove that. Any easier way to acomplish that?






logic model-theory







share|cite|improve this question











$endgroup$





Prove that $langlemathbb{R}-{7},<rangle$ is an elementary substructure of $langlemathbb{R},<rangle$.




What I thought I should do is to use induction on statements to prove that. Any easier way to acomplish that?






logic model-theory




logic model-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 13:23









6005

36.1k751125




36.1k751125










asked Jan 5 at 11:57









GytGyt

585319




585319












  • $begingroup$
    If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
    $endgroup$
    – tomasz
    Jan 7 at 10:36


















  • $begingroup$
    If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
    $endgroup$
    – tomasz
    Jan 7 at 10:36
















$begingroup$
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
$endgroup$
– tomasz
Jan 7 at 10:36




$begingroup$
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
$endgroup$
– tomasz
Jan 7 at 10:36










1 Answer
1






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$begingroup$

An easier way is an application of the following "sufficient condition"-style test.




If $mathfrak{A}=langle A,ldotsrangle$ is a substructure of $mathfrak{B}=langle B,ldotsrangle$ and, for any finite subset $A'$ of $A$ and any $bin B$, there is an automorphism $f$ of $mathfrak{B}$ such that $f(a)=a$ for any $ain A'$, and $f(b)in A$, then $mathfrak{A}$ is an elementary substructure of $mathfrak{B}$.




(Which may appear in your prerequisite course, maybe in somewhat stronger form, and is usually proven using Tarski–Vaught criterion). It is easy to exhibit an $f$ in your setup.






share|cite|improve this answer











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    $begingroup$

    An easier way is an application of the following "sufficient condition"-style test.




    If $mathfrak{A}=langle A,ldotsrangle$ is a substructure of $mathfrak{B}=langle B,ldotsrangle$ and, for any finite subset $A'$ of $A$ and any $bin B$, there is an automorphism $f$ of $mathfrak{B}$ such that $f(a)=a$ for any $ain A'$, and $f(b)in A$, then $mathfrak{A}$ is an elementary substructure of $mathfrak{B}$.




    (Which may appear in your prerequisite course, maybe in somewhat stronger form, and is usually proven using Tarski–Vaught criterion). It is easy to exhibit an $f$ in your setup.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      An easier way is an application of the following "sufficient condition"-style test.




      If $mathfrak{A}=langle A,ldotsrangle$ is a substructure of $mathfrak{B}=langle B,ldotsrangle$ and, for any finite subset $A'$ of $A$ and any $bin B$, there is an automorphism $f$ of $mathfrak{B}$ such that $f(a)=a$ for any $ain A'$, and $f(b)in A$, then $mathfrak{A}$ is an elementary substructure of $mathfrak{B}$.




      (Which may appear in your prerequisite course, maybe in somewhat stronger form, and is usually proven using Tarski–Vaught criterion). It is easy to exhibit an $f$ in your setup.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        An easier way is an application of the following "sufficient condition"-style test.




        If $mathfrak{A}=langle A,ldotsrangle$ is a substructure of $mathfrak{B}=langle B,ldotsrangle$ and, for any finite subset $A'$ of $A$ and any $bin B$, there is an automorphism $f$ of $mathfrak{B}$ such that $f(a)=a$ for any $ain A'$, and $f(b)in A$, then $mathfrak{A}$ is an elementary substructure of $mathfrak{B}$.




        (Which may appear in your prerequisite course, maybe in somewhat stronger form, and is usually proven using Tarski–Vaught criterion). It is easy to exhibit an $f$ in your setup.






        share|cite|improve this answer











        $endgroup$



        An easier way is an application of the following "sufficient condition"-style test.




        If $mathfrak{A}=langle A,ldotsrangle$ is a substructure of $mathfrak{B}=langle B,ldotsrangle$ and, for any finite subset $A'$ of $A$ and any $bin B$, there is an automorphism $f$ of $mathfrak{B}$ such that $f(a)=a$ for any $ain A'$, and $f(b)in A$, then $mathfrak{A}$ is an elementary substructure of $mathfrak{B}$.




        (Which may appear in your prerequisite course, maybe in somewhat stronger form, and is usually proven using Tarski–Vaught criterion). It is easy to exhibit an $f$ in your setup.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 12:40

























        answered Jan 5 at 12:35









        metamorphymetamorphy

        3,6821621




        3,6821621






























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