If $mathcal{M} equiv mathcal{N}$ then there is some $L$-structure $mathcal{R}$ and elementary embeddings...
Let $L$ be a language and $mathcal{M},mathcal{N}$ be $L$-structures. Suppose $mathcal{M}$ and $mathcal{N}$ are elementarily equivalent. Prove that there is some $L$-structure $mathcal{R}$ and elementary embeddings $f:mathcal{M} rightarrow mathcal{R}$ and $g:mathcal{N} rightarrow mathcal{R}$.
My first attempt was to take $mathcal{R}=mathcal{N}$. Then the identity map is an elementary embedding from $mathcal{N}$ to $mathcal{R}$. It remains to show there is an elementary embedding from $mathcal{M}$ to $mathcal{N}$. However, being elementarily equivalent does not imply the existence of elementary embedding. So my method wouldn't work.
Then I have to construct this $L$-structure $mathcal{R}$ so to make it satisfy the desired property. How can I construct it?
first-order-logic model-theory
asked Nov 27 '18 at 2:13
bbw
47038
|
show 2 more comments
Let $L$ be a language and $mathcal{M},mathcal{N}$ be $L$-structures. Suppose $mathcal{M}$ and $mathcal{N}$ are elementarily equivalent. Prove that there is some $L$-structure $mathcal{R}$ and elementary embeddings $f:mathcal{M} rightarrow mathcal{R}$ and $g:mathcal{N} rightarrow mathcal{R}$.
My first attempt was to take $mathcal{R}=mathcal{N}$. Then the identity map is an elementary embedding from $mathcal{N}$ to $mathcal{R}$. It remains to show there is an elementary embedding from $mathcal{M}$ to $mathcal{N}$. However, being elementarily equivalent does not imply the existence of elementary embedding. So my method wouldn't work.
Then I have to construct this $L$-structure $mathcal{R}$ so to make it satisfy the desired property. How can I construct it?
first-order-logic model-theory
asked Nov 27 '18 at 2:13
bbw
47038
1
Hint. Expand the language by giving each element of $mathcal M$ and each element of $mathcal N$ a name. Let $T_1$ and $T_2$ be the theories of $mathcal M$ and $mathcal N$ in the expanded language, and let $T=T_1cup T_2$. Use elementary equivalence of $mathcal M$ and $mathcal N$ to show that each finite subset of $T$ is satisfiable; conclude by compactness that $T$ is satisfiable; let $mathbb R$ be a model of $T$; observe that $mathcal M$ and $mathcal N$ are elementarily embeddable in $mathcal R$.
– bof
Nov 27 '18 at 3:40
@bof, I have a maybe unrelated question: does $Th(mathcal{M}) = Th(mathcal{N})$ imply $|M|=|N|$?
– bbw
Nov 27 '18 at 4:13
@bbw No, Lowenheim-Skolem says in many cases, there will be models of $Th(mathcal M)$ of any infinite cardinality and all models of a complete theory have the same theory (namely that complete theory).
– spaceisdarkgreen
Nov 27 '18 at 4:20
If $|M|$ means the cardinality of the underlying set of $mathcal M$, not necessarily.
– bof
Nov 27 '18 at 4:20
@bof, by "expand the language", do you mean taking $L_1=L cup M$ and $L_2=L cup N$ or you mean taking $L'=L cup M cup N$?
– bbw
Nov 27 '18 at 4:26
|
show 2 more comments
Let $L$ be a language and $mathcal{M},mathcal{N}$ be $L$-structures. Suppose $mathcal{M}$ and $mathcal{N}$ are elementarily equivalent. Prove that there is some $L$-structure $mathcal{R}$ and elementary embeddings $f:mathcal{M} rightarrow mathcal{R}$ and $g:mathcal{N} rightarrow mathcal{R}$.
My first attempt was to take $mathcal{R}=mathcal{N}$. Then the identity map is an elementary embedding from $mathcal{N}$ to $mathcal{R}$. It remains to show there is an elementary embedding from $mathcal{M}$ to $mathcal{N}$. However, being elementarily equivalent does not imply the existence of elementary embedding. So my method wouldn't work.
Then I have to construct this $L$-structure $mathcal{R}$ so to make it satisfy the desired property. How can I construct it?
first-order-logic model-theory
asked Nov 27 '18 at 2:13
bbw
47038
Let $L$ be a language and $mathcal{M},mathcal{N}$ be $L$-structures. Suppose $mathcal{M}$ and $mathcal{N}$ are elementarily equivalent. Prove that there is some $L$-structure $mathcal{R}$ and elementary embeddings $f:mathcal{M} rightarrow mathcal{R}$ and $g:mathcal{N} rightarrow mathcal{R}$.
My first attempt was to take $mathcal{R}=mathcal{N}$. Then the identity map is an elementary embedding from $mathcal{N}$ to $mathcal{R}$. It remains to show there is an elementary embedding from $mathcal{M}$ to $mathcal{N}$. However, being elementarily equivalent does not imply the existence of elementary embedding. So my method wouldn't work.
Then I have to construct this $L$-structure $mathcal{R}$ so to make it satisfy the desired property. How can I construct it?
first-order-logic model-theory
first-order-logic model-theory
asked Nov 27 '18 at 2:13
bbw
47038
asked Nov 27 '18 at 2:13
bbw
47038
asked Nov 27 '18 at 2:13
bbw
47038
asked Nov 27 '18 at 2:13
bbw
47038
asked Nov 27 '18 at 2:13
bbw
47038
47038
1
Hint. Expand the language by giving each element of $mathcal M$ and each element of $mathcal N$ a name. Let $T_1$ and $T_2$ be the theories of $mathcal M$ and $mathcal N$ in the expanded language, and let $T=T_1cup T_2$. Use elementary equivalence of $mathcal M$ and $mathcal N$ to show that each finite subset of $T$ is satisfiable; conclude by compactness that $T$ is satisfiable; let $mathbb R$ be a model of $T$; observe that $mathcal M$ and $mathcal N$ are elementarily embeddable in $mathcal R$.
– bof
Nov 27 '18 at 3:40
@bof, I have a maybe unrelated question: does $Th(mathcal{M}) = Th(mathcal{N})$ imply $|M|=|N|$?
– bbw
Nov 27 '18 at 4:13
@bbw No, Lowenheim-Skolem says in many cases, there will be models of $Th(mathcal M)$ of any infinite cardinality and all models of a complete theory have the same theory (namely that complete theory).
– spaceisdarkgreen
Nov 27 '18 at 4:20
If $|M|$ means the cardinality of the underlying set of $mathcal M$, not necessarily.
– bof
Nov 27 '18 at 4:20
@bof, by "expand the language", do you mean taking $L_1=L cup M$ and $L_2=L cup N$ or you mean taking $L'=L cup M cup N$?
– bbw
Nov 27 '18 at 4:26
|
show 2 more comments
1
Hint. Expand the language by giving each element of $mathcal M$ and each element of $mathcal N$ a name. Let $T_1$ and $T_2$ be the theories of $mathcal M$ and $mathcal N$ in the expanded language, and let $T=T_1cup T_2$. Use elementary equivalence of $mathcal M$ and $mathcal N$ to show that each finite subset of $T$ is satisfiable; conclude by compactness that $T$ is satisfiable; let $mathbb R$ be a model of $T$; observe that $mathcal M$ and $mathcal N$ are elementarily embeddable in $mathcal R$.
– bof
Nov 27 '18 at 3:40
@bof, I have a maybe unrelated question: does $Th(mathcal{M}) = Th(mathcal{N})$ imply $|M|=|N|$?
– bbw
Nov 27 '18 at 4:13
@bbw No, Lowenheim-Skolem says in many cases, there will be models of $Th(mathcal M)$ of any infinite cardinality and all models of a complete theory have the same theory (namely that complete theory).
– spaceisdarkgreen
Nov 27 '18 at 4:20
If $|M|$ means the cardinality of the underlying set of $mathcal M$, not necessarily.
– bof
Nov 27 '18 at 4:20
@bof, by "expand the language", do you mean taking $L_1=L cup M$ and $L_2=L cup N$ or you mean taking $L'=L cup M cup N$?
– bbw
Nov 27 '18 at 4:26
1
1
Hint. Expand the language by giving each element of $mathcal M$ and each element of $mathcal N$ a name. Let $T_1$ and $T_2$ be the theories of $mathcal M$ and $mathcal N$ in the expanded language, and let $T=T_1cup T_2$. Use elementary equivalence of $mathcal M$ and $mathcal N$ to show that each finite subset of $T$ is satisfiable; conclude by compactness that $T$ is satisfiable; let $mathbb R$ be a model of $T$; observe that $mathcal M$ and $mathcal N$ are elementarily embeddable in $mathcal R$.
– bof
Nov 27 '18 at 3:40
Hint. Expand the language by giving each element of $mathcal M$ and each element of $mathcal N$ a name. Let $T_1$ and $T_2$ be the theories of $mathcal M$ and $mathcal N$ in the expanded language, and let $T=T_1cup T_2$. Use elementary equivalence of $mathcal M$ and $mathcal N$ to show that each finite subset of $T$ is satisfiable; conclude by compactness that $T$ is satisfiable; let $mathbb R$ be a model of $T$; observe that $mathcal M$ and $mathcal N$ are elementarily embeddable in $mathcal R$.
– bof
Nov 27 '18 at 3:40
@bof, I have a maybe unrelated question: does $Th(mathcal{M}) = Th(mathcal{N})$ imply $|M|=|N|$?
– bbw
Nov 27 '18 at 4:13
@bof, I have a maybe unrelated question: does $Th(mathcal{M}) = Th(mathcal{N})$ imply $|M|=|N|$?
– bbw
Nov 27 '18 at 4:13
@bbw No, Lowenheim-Skolem says in many cases, there will be models of $Th(mathcal M)$ of any infinite cardinality and all models of a complete theory have the same theory (namely that complete theory).
– spaceisdarkgreen
Nov 27 '18 at 4:20
@bbw No, Lowenheim-Skolem says in many cases, there will be models of $Th(mathcal M)$ of any infinite cardinality and all models of a complete theory have the same theory (namely that complete theory).
– spaceisdarkgreen
Nov 27 '18 at 4:20
If $|M|$ means the cardinality of the underlying set of $mathcal M$, not necessarily.
– bof
Nov 27 '18 at 4:20
If $|M|$ means the cardinality of the underlying set of $mathcal M$, not necessarily.
– bof
Nov 27 '18 at 4:20
@bof, by "expand the language", do you mean taking $L_1=L cup M$ and $L_2=L cup N$ or you mean taking $L'=L cup M cup N$?
– bbw
Nov 27 '18 at 4:26
@bof, by "expand the language", do you mean taking $L_1=L cup M$ and $L_2=L cup N$ or you mean taking $L'=L cup M cup N$?
– bbw
Nov 27 '18 at 4:26
|
show 2 more comments
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1
Hint. Expand the language by giving each element of $mathcal M$ and each element of $mathcal N$ a name. Let $T_1$ and $T_2$ be the theories of $mathcal M$ and $mathcal N$ in the expanded language, and let $T=T_1cup T_2$. Use elementary equivalence of $mathcal M$ and $mathcal N$ to show that each finite subset of $T$ is satisfiable; conclude by compactness that $T$ is satisfiable; let $mathbb R$ be a model of $T$; observe that $mathcal M$ and $mathcal N$ are elementarily embeddable in $mathcal R$.
– bof
Nov 27 '18 at 3:40
@bof, I have a maybe unrelated question: does $Th(mathcal{M}) = Th(mathcal{N})$ imply $|M|=|N|$?
– bbw
Nov 27 '18 at 4:13
@bbw No, Lowenheim-Skolem says in many cases, there will be models of $Th(mathcal M)$ of any infinite cardinality and all models of a complete theory have the same theory (namely that complete theory).
– spaceisdarkgreen
Nov 27 '18 at 4:20
If $|M|$ means the cardinality of the underlying set of $mathcal M$, not necessarily.
– bof
Nov 27 '18 at 4:20
@bof, by "expand the language", do you mean taking $L_1=L cup M$ and $L_2=L cup N$ or you mean taking $L'=L cup M cup N$?
– bbw
Nov 27 '18 at 4:26