Why does the compactness theorem not apply to infinite languages?
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I am trying to show that compactness doesn't apply to infinite languages like $L_{omega_1,omega}$ that allow infinite FOL sentences like $forall x, bigvee_{nin omega} x approx S^n(0)$, i.e. $forall x, (x approx 0 vee x approx S(0) vee xapprox S(S(0)) dots)$.
I already picked a language $L=(0,S)$, gave an example of a set of sentences which is satisfiable, since every subset $Gamma_m$ of $Gamma_ncolon xapprox 0 vee dots vee sapprox S^n(0)$, has a finite model for every $m$ smaller than $n$.
I am trying to show that nonetheless $Gamma_n$ has no model. I can't get my head around how to perform this last step. Maybe my example of a set of sentences $Gamma_n$ which is supposed to be finitely satisfiable but not satisfiable is not the right one?
model-theory
edited Jan 9 at 15:20
Alex Kruckman
26.9k22556
asked Jan 8 at 19:46
JanikG.JanikG.
62
$endgroup$
add a comment |
$begingroup$
I am trying to show that compactness doesn't apply to infinite languages like $L_{omega_1,omega}$ that allow infinite FOL sentences like $forall x, bigvee_{nin omega} x approx S^n(0)$, i.e. $forall x, (x approx 0 vee x approx S(0) vee xapprox S(S(0)) dots)$.
I already picked a language $L=(0,S)$, gave an example of a set of sentences which is satisfiable, since every subset $Gamma_m$ of $Gamma_ncolon xapprox 0 vee dots vee sapprox S^n(0)$, has a finite model for every $m$ smaller than $n$.
I am trying to show that nonetheless $Gamma_n$ has no model. I can't get my head around how to perform this last step. Maybe my example of a set of sentences $Gamma_n$ which is supposed to be finitely satisfiable but not satisfiable is not the right one?
model-theory
edited Jan 9 at 15:20
Alex Kruckman
26.9k22556
asked Jan 8 at 19:46
JanikG.JanikG.
62
$endgroup$
$begingroup$
What does $W$ represent? I guess $0$ is a constant symbol and $S$ is a function? Please explain all your notations (in general).
$endgroup$
– A. Pongrácz
Jan 8 at 19:49
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Sorry, I am new to formatting. W with a subscript n is supposed to represent the infinite disjunction.
$endgroup$
– JanikG.
Jan 8 at 19:52
$begingroup$
And yes, 0 is a constant symbol and S is a function. Sorry again for ugly formatting.
$endgroup$
– JanikG.
Jan 8 at 19:53
$begingroup$
I edited your post to include MathJax formatting. You should use this formatting in future posts on this site. Here's a tutorial / reference on how to do this: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Alex Kruckman
Jan 9 at 15:21
$begingroup$
"Every subset $Gamma_m$ of $Gamma_n$... has a finite model for every $m$ smaller than $n$" doesn't make any sense to me. It would make more sense notationally to let $Gamma = {Gamma_nmid nin omega}$, and show that $Gamma$ is finitely satisfiable by arguing that for every $n$, the set ${Gamma_mmid m<n}$ has a model. (Whether this model is finite is irrelevant.) But in any case, your choice of $Gamma_n$ doesn't work.
$endgroup$
– Alex Kruckman
Jan 9 at 15:24
add a comment |
$begingroup$
I am trying to show that compactness doesn't apply to infinite languages like $L_{omega_1,omega}$ that allow infinite FOL sentences like $forall x, bigvee_{nin omega} x approx S^n(0)$, i.e. $forall x, (x approx 0 vee x approx S(0) vee xapprox S(S(0)) dots)$.
I already picked a language $L=(0,S)$, gave an example of a set of sentences which is satisfiable, since every subset $Gamma_m$ of $Gamma_ncolon xapprox 0 vee dots vee sapprox S^n(0)$, has a finite model for every $m$ smaller than $n$.
I am trying to show that nonetheless $Gamma_n$ has no model. I can't get my head around how to perform this last step. Maybe my example of a set of sentences $Gamma_n$ which is supposed to be finitely satisfiable but not satisfiable is not the right one?
model-theory
edited Jan 9 at 15:20
Alex Kruckman
26.9k22556
asked Jan 8 at 19:46
JanikG.JanikG.
62
$endgroup$
I am trying to show that compactness doesn't apply to infinite languages like $L_{omega_1,omega}$ that allow infinite FOL sentences like $forall x, bigvee_{nin omega} x approx S^n(0)$, i.e. $forall x, (x approx 0 vee x approx S(0) vee xapprox S(S(0)) dots)$.
I already picked a language $L=(0,S)$, gave an example of a set of sentences which is satisfiable, since every subset $Gamma_m$ of $Gamma_ncolon xapprox 0 vee dots vee sapprox S^n(0)$, has a finite model for every $m$ smaller than $n$.
I am trying to show that nonetheless $Gamma_n$ has no model. I can't get my head around how to perform this last step. Maybe my example of a set of sentences $Gamma_n$ which is supposed to be finitely satisfiable but not satisfiable is not the right one?
model-theory
model-theory
edited Jan 9 at 15:20
Alex Kruckman
26.9k22556
asked Jan 8 at 19:46
JanikG.JanikG.
62
edited Jan 9 at 15:20
Alex Kruckman
26.9k22556
asked Jan 8 at 19:46
JanikG.JanikG.
62
edited Jan 9 at 15:20
Alex Kruckman
26.9k22556
edited Jan 9 at 15:20
Alex Kruckman
26.9k22556
edited Jan 9 at 15:20
Alex Kruckman
26.9k22556
26.9k22556
asked Jan 8 at 19:46
JanikG.JanikG.
62
asked Jan 8 at 19:46
JanikG.JanikG.
62
asked Jan 8 at 19:46
JanikG.JanikG.
62
62
$begingroup$
What does $W$ represent? I guess $0$ is a constant symbol and $S$ is a function? Please explain all your notations (in general).
$endgroup$
– A. Pongrácz
Jan 8 at 19:49
$begingroup$
Sorry, I am new to formatting. W with a subscript n is supposed to represent the infinite disjunction.
$endgroup$
– JanikG.
Jan 8 at 19:52
$begingroup$
And yes, 0 is a constant symbol and S is a function. Sorry again for ugly formatting.
$endgroup$
– JanikG.
Jan 8 at 19:53
$begingroup$
I edited your post to include MathJax formatting. You should use this formatting in future posts on this site. Here's a tutorial / reference on how to do this: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Alex Kruckman
Jan 9 at 15:21
$begingroup$
"Every subset $Gamma_m$ of $Gamma_n$... has a finite model for every $m$ smaller than $n$" doesn't make any sense to me. It would make more sense notationally to let $Gamma = {Gamma_nmid nin omega}$, and show that $Gamma$ is finitely satisfiable by arguing that for every $n$, the set ${Gamma_mmid m<n}$ has a model. (Whether this model is finite is irrelevant.) But in any case, your choice of $Gamma_n$ doesn't work.
$endgroup$
– Alex Kruckman
Jan 9 at 15:24
add a comment |
$begingroup$
What does $W$ represent? I guess $0$ is a constant symbol and $S$ is a function? Please explain all your notations (in general).
$endgroup$
– A. Pongrácz
Jan 8 at 19:49
$begingroup$
Sorry, I am new to formatting. W with a subscript n is supposed to represent the infinite disjunction.
$endgroup$
– JanikG.
Jan 8 at 19:52
$begingroup$
And yes, 0 is a constant symbol and S is a function. Sorry again for ugly formatting.
$endgroup$
– JanikG.
Jan 8 at 19:53
$begingroup$
I edited your post to include MathJax formatting. You should use this formatting in future posts on this site. Here's a tutorial / reference on how to do this: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Alex Kruckman
Jan 9 at 15:21
$begingroup$
"Every subset $Gamma_m$ of $Gamma_n$... has a finite model for every $m$ smaller than $n$" doesn't make any sense to me. It would make more sense notationally to let $Gamma = {Gamma_nmid nin omega}$, and show that $Gamma$ is finitely satisfiable by arguing that for every $n$, the set ${Gamma_mmid m<n}$ has a model. (Whether this model is finite is irrelevant.) But in any case, your choice of $Gamma_n$ doesn't work.
$endgroup$
– Alex Kruckman
Jan 9 at 15:24
$begingroup$
What does $W$ represent? I guess $0$ is a constant symbol and $S$ is a function? Please explain all your notations (in general).
$endgroup$
– A. Pongrácz
Jan 8 at 19:49
$begingroup$
What does $W$ represent? I guess $0$ is a constant symbol and $S$ is a function? Please explain all your notations (in general).
$endgroup$
– A. Pongrácz
Jan 8 at 19:49
$begingroup$
Sorry, I am new to formatting. W with a subscript n is supposed to represent the infinite disjunction.
$endgroup$
– JanikG.
Jan 8 at 19:52
$begingroup$
Sorry, I am new to formatting. W with a subscript n is supposed to represent the infinite disjunction.
$endgroup$
– JanikG.
Jan 8 at 19:52
$begingroup$
And yes, 0 is a constant symbol and S is a function. Sorry again for ugly formatting.
$endgroup$
– JanikG.
Jan 8 at 19:53
$begingroup$
And yes, 0 is a constant symbol and S is a function. Sorry again for ugly formatting.
$endgroup$
– JanikG.
Jan 8 at 19:53
$begingroup$
I edited your post to include MathJax formatting. You should use this formatting in future posts on this site. Here's a tutorial / reference on how to do this: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Alex Kruckman
Jan 9 at 15:21
$begingroup$
I edited your post to include MathJax formatting. You should use this formatting in future posts on this site. Here's a tutorial / reference on how to do this: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Alex Kruckman
Jan 9 at 15:21
$begingroup$
"Every subset $Gamma_m$ of $Gamma_n$... has a finite model for every $m$ smaller than $n$" doesn't make any sense to me. It would make more sense notationally to let $Gamma = {Gamma_nmid nin omega}$, and show that $Gamma$ is finitely satisfiable by arguing that for every $n$, the set ${Gamma_mmid m<n}$ has a model. (Whether this model is finite is irrelevant.) But in any case, your choice of $Gamma_n$ doesn't work.
$endgroup$
– Alex Kruckman
Jan 9 at 15:24
$begingroup$
"Every subset $Gamma_m$ of $Gamma_n$... has a finite model for every $m$ smaller than $n$" doesn't make any sense to me. It would make more sense notationally to let $Gamma = {Gamma_nmid nin omega}$, and show that $Gamma$ is finitely satisfiable by arguing that for every $n$, the set ${Gamma_mmid m<n}$ has a model. (Whether this model is finite is irrelevant.) But in any case, your choice of $Gamma_n$ doesn't work.
$endgroup$
– Alex Kruckman
Jan 9 at 15:24
add a comment |
1 Answer
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$begingroup$
Actually, you don't need any signature: $L={=}$ is completely fine.
Note that you can express the property for any $nin mathbb{N}$ that the model contains at least $n$ different elements. In fact, there is a finite first-order formula that expresses this. Let $phi_n$ be such a formula.
(E.g., $phi_3$ is $exists x_1,x_2,x_3 ,, x_1neq x_2 wedge x_2neq x_3 wedge x_1neq x_3$.)
The theory $T$ now consists of
$bullet$ all the $phi_n$,
$bullet$ and the formula that says that one of the $phi_n$ must be false (this is an infinite disjunction of all the negations of the $phi_n$)
Altogether, this means that the model is infinite and finite (contradiction).
But any finite subsystem is satisfiable (by some finite set).
answered Jan 8 at 20:19
A. PongráczA. Pongrácz
5,9831929
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add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Actually, you don't need any signature: $L={=}$ is completely fine.
Note that you can express the property for any $nin mathbb{N}$ that the model contains at least $n$ different elements. In fact, there is a finite first-order formula that expresses this. Let $phi_n$ be such a formula.
(E.g., $phi_3$ is $exists x_1,x_2,x_3 ,, x_1neq x_2 wedge x_2neq x_3 wedge x_1neq x_3$.)
The theory $T$ now consists of
$bullet$ all the $phi_n$,
$bullet$ and the formula that says that one of the $phi_n$ must be false (this is an infinite disjunction of all the negations of the $phi_n$)
Altogether, this means that the model is infinite and finite (contradiction).
But any finite subsystem is satisfiable (by some finite set).
answered Jan 8 at 20:19
A. PongráczA. Pongrácz
5,9831929
$endgroup$
add a comment |
$begingroup$
Actually, you don't need any signature: $L={=}$ is completely fine.
Note that you can express the property for any $nin mathbb{N}$ that the model contains at least $n$ different elements. In fact, there is a finite first-order formula that expresses this. Let $phi_n$ be such a formula.
(E.g., $phi_3$ is $exists x_1,x_2,x_3 ,, x_1neq x_2 wedge x_2neq x_3 wedge x_1neq x_3$.)
The theory $T$ now consists of
$bullet$ all the $phi_n$,
$bullet$ and the formula that says that one of the $phi_n$ must be false (this is an infinite disjunction of all the negations of the $phi_n$)
Altogether, this means that the model is infinite and finite (contradiction).
But any finite subsystem is satisfiable (by some finite set).
answered Jan 8 at 20:19
A. PongráczA. Pongrácz
5,9831929
$endgroup$
add a comment |
$begingroup$
Actually, you don't need any signature: $L={=}$ is completely fine.
Note that you can express the property for any $nin mathbb{N}$ that the model contains at least $n$ different elements. In fact, there is a finite first-order formula that expresses this. Let $phi_n$ be such a formula.
(E.g., $phi_3$ is $exists x_1,x_2,x_3 ,, x_1neq x_2 wedge x_2neq x_3 wedge x_1neq x_3$.)
The theory $T$ now consists of
$bullet$ all the $phi_n$,
$bullet$ and the formula that says that one of the $phi_n$ must be false (this is an infinite disjunction of all the negations of the $phi_n$)
Altogether, this means that the model is infinite and finite (contradiction).
But any finite subsystem is satisfiable (by some finite set).
answered Jan 8 at 20:19
A. PongráczA. Pongrácz
5,9831929
$endgroup$
Actually, you don't need any signature: $L={=}$ is completely fine.
Note that you can express the property for any $nin mathbb{N}$ that the model contains at least $n$ different elements. In fact, there is a finite first-order formula that expresses this. Let $phi_n$ be such a formula.
(E.g., $phi_3$ is $exists x_1,x_2,x_3 ,, x_1neq x_2 wedge x_2neq x_3 wedge x_1neq x_3$.)
The theory $T$ now consists of
$bullet$ all the $phi_n$,
$bullet$ and the formula that says that one of the $phi_n$ must be false (this is an infinite disjunction of all the negations of the $phi_n$)
Altogether, this means that the model is infinite and finite (contradiction).
But any finite subsystem is satisfiable (by some finite set).
answered Jan 8 at 20:19
A. PongráczA. Pongrácz
5,9831929
answered Jan 8 at 20:19
A. PongráczA. Pongrácz
5,9831929
answered Jan 8 at 20:19
A. PongráczA. Pongrácz
5,9831929
answered Jan 8 at 20:19
A. PongráczA. Pongrácz
5,9831929
5,9831929
add a comment |
add a comment |
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$begingroup$
What does $W$ represent? I guess $0$ is a constant symbol and $S$ is a function? Please explain all your notations (in general).
$endgroup$
– A. Pongrácz
Jan 8 at 19:49
$begingroup$
Sorry, I am new to formatting. W with a subscript n is supposed to represent the infinite disjunction.
$endgroup$
– JanikG.
Jan 8 at 19:52
$begingroup$
And yes, 0 is a constant symbol and S is a function. Sorry again for ugly formatting.
$endgroup$
– JanikG.
Jan 8 at 19:53
$begingroup$
I edited your post to include MathJax formatting. You should use this formatting in future posts on this site. Here's a tutorial / reference on how to do this: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Alex Kruckman
Jan 9 at 15:21
$begingroup$
"Every subset $Gamma_m$ of $Gamma_n$... has a finite model for every $m$ smaller than $n$" doesn't make any sense to me. It would make more sense notationally to let $Gamma = {Gamma_nmid nin omega}$, and show that $Gamma$ is finitely satisfiable by arguing that for every $n$, the set ${Gamma_mmid m<n}$ has a model. (Whether this model is finite is irrelevant.) But in any case, your choice of $Gamma_n$ doesn't work.
$endgroup$
– Alex Kruckman
Jan 9 at 15:24