Proving that $langlemathbb{R}-{7},<rangle$ is an elementary substructure of $langlemathbb{R},<rangle$
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
edited 2 days ago
6005
35.7k751125
asked 2 days ago
GytGyt
583319
add a comment |
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
edited 2 days ago
6005
35.7k751125
asked 2 days ago
GytGyt
583319
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
– tomasz
10 hours ago
add a comment |
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
edited 2 days ago
6005
35.7k751125
asked 2 days ago
GytGyt
583319
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
edited 2 days ago
6005
35.7k751125
asked 2 days ago
GytGyt
583319
edited 2 days ago
6005
35.7k751125
asked 2 days ago
GytGyt
583319
edited 2 days ago
6005
35.7k751125
edited 2 days ago
6005
35.7k751125
edited 2 days ago
6005
35.7k751125
35.7k751125
asked 2 days ago
GytGyt
583319
asked 2 days ago
GytGyt
583319
asked 2 days ago
GytGyt
583319
583319
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
– tomasz
10 hours ago
add a comment |
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
– tomasz
10 hours ago
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
– tomasz
10 hours ago
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
– tomasz
10 hours ago
add a comment |
1 Answer
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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.
answered 2 days ago
metamorphymetamorphy
3,6571621
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 2 days ago
metamorphymetamorphy
3,6571621
add a comment |
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.
answered 2 days ago
metamorphymetamorphy
3,6571621
add a comment |
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.
answered 2 days ago
metamorphymetamorphy
3,6571621
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.
answered 2 days ago
metamorphymetamorphy
3,6571621
edited 2 days ago
answered 2 days ago
metamorphymetamorphy
3,6571621
answered 2 days ago
metamorphymetamorphy
3,6571621
answered 2 days ago
metamorphymetamorphy
3,6571621
3,6571621
add a comment |
add a comment |
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If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
– tomasz
10 hours ago