Proving that $langlemathbb{R}-{7},<rangle$ is an elementary substructure of $langlemathbb{R},<rangle$
$begingroup$
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
edited Jan 5 at 13:23
6005
37k752127
asked Jan 5 at 11:57
GytGyt
562519
$endgroup$
add a comment |
$begingroup$
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
edited Jan 5 at 13:23
6005
37k752127
asked Jan 5 at 11:57
GytGyt
562519
$endgroup$
$begingroup$
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
$endgroup$
– tomasz
Jan 7 at 10:36
add a comment |
$begingroup$
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
edited Jan 5 at 13:23
6005
37k752127
asked Jan 5 at 11:57
GytGyt
562519
$endgroup$
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
edited Jan 5 at 13:23
6005
37k752127
asked Jan 5 at 11:57
GytGyt
562519
edited Jan 5 at 13:23
6005
37k752127
asked Jan 5 at 11:57
GytGyt
562519
edited Jan 5 at 13:23
6005
37k752127
edited Jan 5 at 13:23
6005
37k752127
edited Jan 5 at 13:23
6005
37k752127
37k752127
asked Jan 5 at 11:57
GytGyt
562519
asked Jan 5 at 11:57
GytGyt
562519
asked Jan 5 at 11:57
GytGyt
562519
562519
$begingroup$
If you know that dense linear orderings have quantifier elimination, then it is an immediate consequence.
$endgroup$
– tomasz
Jan 7 at 10:36
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 5 at 12:35
metamorphymetamorphy
3,8721721
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 5 at 12:35
metamorphymetamorphy
3,8721721
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 12:35
metamorphymetamorphy
3,8721721
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 12:35
metamorphymetamorphy
3,8721721
$endgroup$
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.
answered Jan 5 at 12:35
metamorphymetamorphy
3,8721721
edited Jan 5 at 12:40
answered Jan 5 at 12:35
metamorphymetamorphy
3,8721721
answered Jan 5 at 12:35
metamorphymetamorphy
3,8721721
answered Jan 5 at 12:35
metamorphymetamorphy
3,8721721
3,8721721
add a comment |
add a comment |
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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