model theory - existence of a model
$begingroup$
Let $Gamma$ be a consistent set of $L$-sentences with infinite cardinality. If it has an infinite model, then there exists a model for
$$
Gamma' =Gamma cup {lnot c_a = c_b : a neq b },
$$
where $c_i$ is a constant symbol.
How would I go about proving this? I'm imagining that I can extend the model for $Gamma$ to have a larger domain (to include more constant symbols) but I don't know how to show that it is actually a model for $Gamma'$. Maybe with compactness?
logic model-theory
asked Dec 8 '18 at 20:40
bofbof
152
$endgroup$
add a comment |
$begingroup$
Let $Gamma$ be a consistent set of $L$-sentences with infinite cardinality. If it has an infinite model, then there exists a model for
$$
Gamma' =Gamma cup {lnot c_a = c_b : a neq b },
$$
where $c_i$ is a constant symbol.
How would I go about proving this? I'm imagining that I can extend the model for $Gamma$ to have a larger domain (to include more constant symbols) but I don't know how to show that it is actually a model for $Gamma'$. Maybe with compactness?
logic model-theory
asked Dec 8 '18 at 20:40
bofbof
152
$endgroup$
1
$begingroup$
Yes, compactness.
$endgroup$
– Asaf Karagila♦
Dec 8 '18 at 20:43
$begingroup$
Yes, with compactness. But it should be assumed that the constant symbols $c_i$ do not appear in $Gamma$, otherwise the statement is not true.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:44
$begingroup$
@AlexKruckman Why is that assumption necessary?
$endgroup$
– bof
Dec 8 '18 at 20:45
$begingroup$
For example, in the vocabulary $(c_a,c_b)$, $Gamma$ could be the set ${c_a = c_b}$.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:48
$begingroup$
@AlexKruckman Thanks, that makes sense. Now, compactness requires that every finite subset of $Gamma'$ has a model but I'm not sure how I would go about proving that for every subset. Any hints?
$endgroup$
– bof
Dec 8 '18 at 20:53
add a comment |
$begingroup$
Let $Gamma$ be a consistent set of $L$-sentences with infinite cardinality. If it has an infinite model, then there exists a model for
$$
Gamma' =Gamma cup {lnot c_a = c_b : a neq b },
$$
where $c_i$ is a constant symbol.
How would I go about proving this? I'm imagining that I can extend the model for $Gamma$ to have a larger domain (to include more constant symbols) but I don't know how to show that it is actually a model for $Gamma'$. Maybe with compactness?
logic model-theory
asked Dec 8 '18 at 20:40
bofbof
152
$endgroup$
Let $Gamma$ be a consistent set of $L$-sentences with infinite cardinality. If it has an infinite model, then there exists a model for
$$
Gamma' =Gamma cup {lnot c_a = c_b : a neq b },
$$
where $c_i$ is a constant symbol.
How would I go about proving this? I'm imagining that I can extend the model for $Gamma$ to have a larger domain (to include more constant symbols) but I don't know how to show that it is actually a model for $Gamma'$. Maybe with compactness?
logic model-theory
logic model-theory
asked Dec 8 '18 at 20:40
bofbof
152
asked Dec 8 '18 at 20:40
bofbof
152
asked Dec 8 '18 at 20:40
bofbof
152
asked Dec 8 '18 at 20:40
bofbof
152
asked Dec 8 '18 at 20:40
bofbof
152
152
1
$begingroup$
Yes, compactness.
$endgroup$
– Asaf Karagila♦
Dec 8 '18 at 20:43
$begingroup$
Yes, with compactness. But it should be assumed that the constant symbols $c_i$ do not appear in $Gamma$, otherwise the statement is not true.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:44
$begingroup$
@AlexKruckman Why is that assumption necessary?
$endgroup$
– bof
Dec 8 '18 at 20:45
$begingroup$
For example, in the vocabulary $(c_a,c_b)$, $Gamma$ could be the set ${c_a = c_b}$.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:48
$begingroup$
@AlexKruckman Thanks, that makes sense. Now, compactness requires that every finite subset of $Gamma'$ has a model but I'm not sure how I would go about proving that for every subset. Any hints?
$endgroup$
– bof
Dec 8 '18 at 20:53
add a comment |
1
$begingroup$
Yes, compactness.
$endgroup$
– Asaf Karagila♦
Dec 8 '18 at 20:43
$begingroup$
Yes, with compactness. But it should be assumed that the constant symbols $c_i$ do not appear in $Gamma$, otherwise the statement is not true.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:44
$begingroup$
@AlexKruckman Why is that assumption necessary?
$endgroup$
– bof
Dec 8 '18 at 20:45
$begingroup$
For example, in the vocabulary $(c_a,c_b)$, $Gamma$ could be the set ${c_a = c_b}$.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:48
$begingroup$
@AlexKruckman Thanks, that makes sense. Now, compactness requires that every finite subset of $Gamma'$ has a model but I'm not sure how I would go about proving that for every subset. Any hints?
$endgroup$
– bof
Dec 8 '18 at 20:53
1
1
$begingroup$
Yes, compactness.
$endgroup$
– Asaf Karagila♦
Dec 8 '18 at 20:43
$begingroup$
Yes, compactness.
$endgroup$
– Asaf Karagila♦
Dec 8 '18 at 20:43
$begingroup$
Yes, with compactness. But it should be assumed that the constant symbols $c_i$ do not appear in $Gamma$, otherwise the statement is not true.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:44
$begingroup$
Yes, with compactness. But it should be assumed that the constant symbols $c_i$ do not appear in $Gamma$, otherwise the statement is not true.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:44
$begingroup$
@AlexKruckman Why is that assumption necessary?
$endgroup$
– bof
Dec 8 '18 at 20:45
$begingroup$
@AlexKruckman Why is that assumption necessary?
$endgroup$
– bof
Dec 8 '18 at 20:45
$begingroup$
For example, in the vocabulary $(c_a,c_b)$, $Gamma$ could be the set ${c_a = c_b}$.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:48
$begingroup$
For example, in the vocabulary $(c_a,c_b)$, $Gamma$ could be the set ${c_a = c_b}$.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:48
$begingroup$
@AlexKruckman Thanks, that makes sense. Now, compactness requires that every finite subset of $Gamma'$ has a model but I'm not sure how I would go about proving that for every subset. Any hints?
$endgroup$
– bof
Dec 8 '18 at 20:53
$begingroup$
@AlexKruckman Thanks, that makes sense. Now, compactness requires that every finite subset of $Gamma'$ has a model but I'm not sure how I would go about proving that for every subset. Any hints?
$endgroup$
– bof
Dec 8 '18 at 20:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A finite subset of $Gamma'$ is a finite subset of $Gamma$, together with the assertions that finitely many $c_1,dots,c_n$ are pairwise distinct.
Take any infinite model of $Gamma$ (which exists by assumption), interpret $c_1,dots,c_n$ as distinct elements, and interpret the rest of the constant symbols arbitrarily (we are free to do this, since the constant symbols are not mentioned in $Gamma$). This is a model of our finite subset of $Gamma'$. That's all there is to it.
This is the standard proof of the (upwards) Löwenheim-Skolem theorem.
answered Dec 8 '18 at 20:57
Alex KruckmanAlex Kruckman
27k22556
$endgroup$
$begingroup$
Thank you. This was very helpful.
$endgroup$
– bof
Dec 8 '18 at 21:05
$begingroup$
I'm sorry but I have another question. Why can't we take any infinite model of $Gamma$ and interpret $c_i$ as distinct elements and use it for the model of $Gamma'$? What's special about finite subsets?
$endgroup$
– bof
Dec 8 '18 at 21:19
1
$begingroup$
Because there might be more constants then there are elements of the model.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:39
1
$begingroup$
The point of the Löwenheim-Skolem theorem os to take a theory that has some infinite model (say a countably infinite one) and prove that it has models of all larger cardinities.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:41
add a comment |
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$begingroup$
A finite subset of $Gamma'$ is a finite subset of $Gamma$, together with the assertions that finitely many $c_1,dots,c_n$ are pairwise distinct.
Take any infinite model of $Gamma$ (which exists by assumption), interpret $c_1,dots,c_n$ as distinct elements, and interpret the rest of the constant symbols arbitrarily (we are free to do this, since the constant symbols are not mentioned in $Gamma$). This is a model of our finite subset of $Gamma'$. That's all there is to it.
This is the standard proof of the (upwards) Löwenheim-Skolem theorem.
answered Dec 8 '18 at 20:57
Alex KruckmanAlex Kruckman
27k22556
$endgroup$
$begingroup$
Thank you. This was very helpful.
$endgroup$
– bof
Dec 8 '18 at 21:05
$begingroup$
I'm sorry but I have another question. Why can't we take any infinite model of $Gamma$ and interpret $c_i$ as distinct elements and use it for the model of $Gamma'$? What's special about finite subsets?
$endgroup$
– bof
Dec 8 '18 at 21:19
1
$begingroup$
Because there might be more constants then there are elements of the model.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:39
1
$begingroup$
The point of the Löwenheim-Skolem theorem os to take a theory that has some infinite model (say a countably infinite one) and prove that it has models of all larger cardinities.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:41
add a comment |
$begingroup$
A finite subset of $Gamma'$ is a finite subset of $Gamma$, together with the assertions that finitely many $c_1,dots,c_n$ are pairwise distinct.
Take any infinite model of $Gamma$ (which exists by assumption), interpret $c_1,dots,c_n$ as distinct elements, and interpret the rest of the constant symbols arbitrarily (we are free to do this, since the constant symbols are not mentioned in $Gamma$). This is a model of our finite subset of $Gamma'$. That's all there is to it.
This is the standard proof of the (upwards) Löwenheim-Skolem theorem.
answered Dec 8 '18 at 20:57
Alex KruckmanAlex Kruckman
27k22556
$endgroup$
$begingroup$
Thank you. This was very helpful.
$endgroup$
– bof
Dec 8 '18 at 21:05
$begingroup$
I'm sorry but I have another question. Why can't we take any infinite model of $Gamma$ and interpret $c_i$ as distinct elements and use it for the model of $Gamma'$? What's special about finite subsets?
$endgroup$
– bof
Dec 8 '18 at 21:19
1
$begingroup$
Because there might be more constants then there are elements of the model.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:39
1
$begingroup$
The point of the Löwenheim-Skolem theorem os to take a theory that has some infinite model (say a countably infinite one) and prove that it has models of all larger cardinities.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:41
add a comment |
$begingroup$
A finite subset of $Gamma'$ is a finite subset of $Gamma$, together with the assertions that finitely many $c_1,dots,c_n$ are pairwise distinct.
Take any infinite model of $Gamma$ (which exists by assumption), interpret $c_1,dots,c_n$ as distinct elements, and interpret the rest of the constant symbols arbitrarily (we are free to do this, since the constant symbols are not mentioned in $Gamma$). This is a model of our finite subset of $Gamma'$. That's all there is to it.
This is the standard proof of the (upwards) Löwenheim-Skolem theorem.
answered Dec 8 '18 at 20:57
Alex KruckmanAlex Kruckman
27k22556
$endgroup$
A finite subset of $Gamma'$ is a finite subset of $Gamma$, together with the assertions that finitely many $c_1,dots,c_n$ are pairwise distinct.
Take any infinite model of $Gamma$ (which exists by assumption), interpret $c_1,dots,c_n$ as distinct elements, and interpret the rest of the constant symbols arbitrarily (we are free to do this, since the constant symbols are not mentioned in $Gamma$). This is a model of our finite subset of $Gamma'$. That's all there is to it.
This is the standard proof of the (upwards) Löwenheim-Skolem theorem.
answered Dec 8 '18 at 20:57
Alex KruckmanAlex Kruckman
27k22556
answered Dec 8 '18 at 20:57
Alex KruckmanAlex Kruckman
27k22556
answered Dec 8 '18 at 20:57
Alex KruckmanAlex Kruckman
27k22556
answered Dec 8 '18 at 20:57
Alex KruckmanAlex Kruckman
27k22556
27k22556
$begingroup$
Thank you. This was very helpful.
$endgroup$
– bof
Dec 8 '18 at 21:05
$begingroup$
I'm sorry but I have another question. Why can't we take any infinite model of $Gamma$ and interpret $c_i$ as distinct elements and use it for the model of $Gamma'$? What's special about finite subsets?
$endgroup$
– bof
Dec 8 '18 at 21:19
1
$begingroup$
Because there might be more constants then there are elements of the model.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:39
1
$begingroup$
The point of the Löwenheim-Skolem theorem os to take a theory that has some infinite model (say a countably infinite one) and prove that it has models of all larger cardinities.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:41
add a comment |
$begingroup$
Thank you. This was very helpful.
$endgroup$
– bof
Dec 8 '18 at 21:05
$begingroup$
I'm sorry but I have another question. Why can't we take any infinite model of $Gamma$ and interpret $c_i$ as distinct elements and use it for the model of $Gamma'$? What's special about finite subsets?
$endgroup$
– bof
Dec 8 '18 at 21:19
1
$begingroup$
Because there might be more constants then there are elements of the model.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:39
1
$begingroup$
The point of the Löwenheim-Skolem theorem os to take a theory that has some infinite model (say a countably infinite one) and prove that it has models of all larger cardinities.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:41
$begingroup$
Thank you. This was very helpful.
$endgroup$
– bof
Dec 8 '18 at 21:05
$begingroup$
Thank you. This was very helpful.
$endgroup$
– bof
Dec 8 '18 at 21:05
$begingroup$
I'm sorry but I have another question. Why can't we take any infinite model of $Gamma$ and interpret $c_i$ as distinct elements and use it for the model of $Gamma'$? What's special about finite subsets?
$endgroup$
– bof
Dec 8 '18 at 21:19
$begingroup$
I'm sorry but I have another question. Why can't we take any infinite model of $Gamma$ and interpret $c_i$ as distinct elements and use it for the model of $Gamma'$? What's special about finite subsets?
$endgroup$
– bof
Dec 8 '18 at 21:19
1
1
$begingroup$
Because there might be more constants then there are elements of the model.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:39
$begingroup$
Because there might be more constants then there are elements of the model.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:39
1
1
$begingroup$
The point of the Löwenheim-Skolem theorem os to take a theory that has some infinite model (say a countably infinite one) and prove that it has models of all larger cardinities.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:41
$begingroup$
The point of the Löwenheim-Skolem theorem os to take a theory that has some infinite model (say a countably infinite one) and prove that it has models of all larger cardinities.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 21:41
add a comment |
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1
$begingroup$
Yes, compactness.
$endgroup$
– Asaf Karagila♦
Dec 8 '18 at 20:43
$begingroup$
Yes, with compactness. But it should be assumed that the constant symbols $c_i$ do not appear in $Gamma$, otherwise the statement is not true.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:44
$begingroup$
@AlexKruckman Why is that assumption necessary?
$endgroup$
– bof
Dec 8 '18 at 20:45
$begingroup$
For example, in the vocabulary $(c_a,c_b)$, $Gamma$ could be the set ${c_a = c_b}$.
$endgroup$
– Alex Kruckman
Dec 8 '18 at 20:48
$begingroup$
@AlexKruckman Thanks, that makes sense. Now, compactness requires that every finite subset of $Gamma'$ has a model but I'm not sure how I would go about proving that for every subset. Any hints?
$endgroup$
– bof
Dec 8 '18 at 20:53