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?










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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


















1












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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












  • $begingroup$
    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
















1












1








1





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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edited Jan 9 at 15:20









Alex Kruckman

26.9k22556




26.9k22556










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




















  • $begingroup$
    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












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).






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    1 Answer
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    1 Answer
    1






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    active

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    4












    $begingroup$

    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).






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      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).






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        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).






        share|cite|improve this answer









        $endgroup$



        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).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 20:19









        A. PongráczA. Pongrácz

        5,9831929




        5,9831929






























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