Is ZFC a geometric theory? Does it have a classifying topos?












8












$begingroup$


Given a language $L$, the set of geometric formulae in $L$ is the smallest set of formulae containing atomic formulae and closed by finite conjunction, arbitrary disjunction, and existential quantification.



Recall that a theory is geometric if it can be axiomatized by sentences of the form $forall mathbf x (phi(mathbf x)rightarrow psi(mathbf x)$ where $phi$ and $psi$ are geometric formulae and $mathbf x=(x_1,dots,x_n))$ for some $n$.



1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?



2) Is ZFC a geometric theory?
I suspect it is not, since it would seem to me that implication and double implication cannot be axiomatized with geometric formulae, and therefore we have problems since $$forall x forall y (forall z (zin xleftrightarrow zin y)rightarrow (x=y))$$ (extensionality) doesn't seem to be re-expressible by means of geometric formulae (in part., I have trouble in converting $forall z( zin xleftrightarrow zin y)$ into a geometric formula.)
Or am I missing something? Mind that I am just at the beginning of my studies in this topic.



3) Is it true that $Mod(mathcal E, ZFC)=mathcal E$? The only reason I have to believe this is that it should be true (correct?) if $mathcal E=Sets$, so I wouldn't be surprised if the answer is no in general.



Also, if point (1) is true and point (2) is false, also point (3) should be false, since if it were true we would have $$Geo(mathcal E, S(U))=mathcal E=Mod(mathcal E, ZFC)$$ where $Geo$ means topos morphisms and $S(U)$ is the object classifier; and hence $ZFC$ would have a classifying topos, contradiction.






logic set-theory topos-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If it's reasonably short, could you maybe add a definition of "geometric formula"?
    $endgroup$
    – Stefan Mesken
    Oct 12 '18 at 16:38






  • 1




    $begingroup$
    Of course: I have just edited.
    $endgroup$
    – W. Rether
    Oct 12 '18 at 16:41






  • 2




    $begingroup$
    ZFC is not geometric in its usual signature, but every first order theory is equivalent over classical logic to a geometric theory in an extended language. And since Choice means that every intuitionistic model of ZFC is also a classical model, it seems like the corresponding geometric theory should classify exactly the same models.
    $endgroup$
    – Malice Vidrine
    Oct 12 '18 at 17:24






  • 4




    $begingroup$
    The class of formulas you described are called the coherent formulas (or positive existential formulas, by model-theorists). Geometric formula, on the other hand, are closed under arbitrary disjunctions, in addition to finite conjunctions.
    $endgroup$
    – Alex Kruckman
    Oct 12 '18 at 19:25


















8












$begingroup$


Given a language $L$, the set of geometric formulae in $L$ is the smallest set of formulae containing atomic formulae and closed by finite conjunction, arbitrary disjunction, and existential quantification.



Recall that a theory is geometric if it can be axiomatized by sentences of the form $forall mathbf x (phi(mathbf x)rightarrow psi(mathbf x)$ where $phi$ and $psi$ are geometric formulae and $mathbf x=(x_1,dots,x_n))$ for some $n$.



1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?



2) Is ZFC a geometric theory?
I suspect it is not, since it would seem to me that implication and double implication cannot be axiomatized with geometric formulae, and therefore we have problems since $$forall x forall y (forall z (zin xleftrightarrow zin y)rightarrow (x=y))$$ (extensionality) doesn't seem to be re-expressible by means of geometric formulae (in part., I have trouble in converting $forall z( zin xleftrightarrow zin y)$ into a geometric formula.)
Or am I missing something? Mind that I am just at the beginning of my studies in this topic.



3) Is it true that $Mod(mathcal E, ZFC)=mathcal E$? The only reason I have to believe this is that it should be true (correct?) if $mathcal E=Sets$, so I wouldn't be surprised if the answer is no in general.



Also, if point (1) is true and point (2) is false, also point (3) should be false, since if it were true we would have $$Geo(mathcal E, S(U))=mathcal E=Mod(mathcal E, ZFC)$$ where $Geo$ means topos morphisms and $S(U)$ is the object classifier; and hence $ZFC$ would have a classifying topos, contradiction.






logic set-theory topos-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If it's reasonably short, could you maybe add a definition of "geometric formula"?
    $endgroup$
    – Stefan Mesken
    Oct 12 '18 at 16:38






  • 1




    $begingroup$
    Of course: I have just edited.
    $endgroup$
    – W. Rether
    Oct 12 '18 at 16:41






  • 2




    $begingroup$
    ZFC is not geometric in its usual signature, but every first order theory is equivalent over classical logic to a geometric theory in an extended language. And since Choice means that every intuitionistic model of ZFC is also a classical model, it seems like the corresponding geometric theory should classify exactly the same models.
    $endgroup$
    – Malice Vidrine
    Oct 12 '18 at 17:24






  • 4




    $begingroup$
    The class of formulas you described are called the coherent formulas (or positive existential formulas, by model-theorists). Geometric formula, on the other hand, are closed under arbitrary disjunctions, in addition to finite conjunctions.
    $endgroup$
    – Alex Kruckman
    Oct 12 '18 at 19:25
















8












8








8


5



$begingroup$


Given a language $L$, the set of geometric formulae in $L$ is the smallest set of formulae containing atomic formulae and closed by finite conjunction, arbitrary disjunction, and existential quantification.



Recall that a theory is geometric if it can be axiomatized by sentences of the form $forall mathbf x (phi(mathbf x)rightarrow psi(mathbf x)$ where $phi$ and $psi$ are geometric formulae and $mathbf x=(x_1,dots,x_n))$ for some $n$.



1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?



2) Is ZFC a geometric theory?
I suspect it is not, since it would seem to me that implication and double implication cannot be axiomatized with geometric formulae, and therefore we have problems since $$forall x forall y (forall z (zin xleftrightarrow zin y)rightarrow (x=y))$$ (extensionality) doesn't seem to be re-expressible by means of geometric formulae (in part., I have trouble in converting $forall z( zin xleftrightarrow zin y)$ into a geometric formula.)
Or am I missing something? Mind that I am just at the beginning of my studies in this topic.



3) Is it true that $Mod(mathcal E, ZFC)=mathcal E$? The only reason I have to believe this is that it should be true (correct?) if $mathcal E=Sets$, so I wouldn't be surprised if the answer is no in general.



Also, if point (1) is true and point (2) is false, also point (3) should be false, since if it were true we would have $$Geo(mathcal E, S(U))=mathcal E=Mod(mathcal E, ZFC)$$ where $Geo$ means topos morphisms and $S(U)$ is the object classifier; and hence $ZFC$ would have a classifying topos, contradiction.






logic set-theory topos-theory







share|cite|improve this question











$endgroup$




Given a language $L$, the set of geometric formulae in $L$ is the smallest set of formulae containing atomic formulae and closed by finite conjunction, arbitrary disjunction, and existential quantification.



Recall that a theory is geometric if it can be axiomatized by sentences of the form $forall mathbf x (phi(mathbf x)rightarrow psi(mathbf x)$ where $phi$ and $psi$ are geometric formulae and $mathbf x=(x_1,dots,x_n))$ for some $n$.



1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?



2) Is ZFC a geometric theory?
I suspect it is not, since it would seem to me that implication and double implication cannot be axiomatized with geometric formulae, and therefore we have problems since $$forall x forall y (forall z (zin xleftrightarrow zin y)rightarrow (x=y))$$ (extensionality) doesn't seem to be re-expressible by means of geometric formulae (in part., I have trouble in converting $forall z( zin xleftrightarrow zin y)$ into a geometric formula.)
Or am I missing something? Mind that I am just at the beginning of my studies in this topic.



3) Is it true that $Mod(mathcal E, ZFC)=mathcal E$? The only reason I have to believe this is that it should be true (correct?) if $mathcal E=Sets$, so I wouldn't be surprised if the answer is no in general.



Also, if point (1) is true and point (2) is false, also point (3) should be false, since if it were true we would have $$Geo(mathcal E, S(U))=mathcal E=Mod(mathcal E, ZFC)$$ where $Geo$ means topos morphisms and $S(U)$ is the object classifier; and hence $ZFC$ would have a classifying topos, contradiction.






logic set-theory topos-theory




logic set-theory topos-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 14 '18 at 15:39







W. Rether

















asked Oct 12 '18 at 16:29









W. RetherW. Rether

748417




748417








  • 1




    $begingroup$
    If it's reasonably short, could you maybe add a definition of "geometric formula"?
    $endgroup$
    – Stefan Mesken
    Oct 12 '18 at 16:38






  • 1




    $begingroup$
    Of course: I have just edited.
    $endgroup$
    – W. Rether
    Oct 12 '18 at 16:41






  • 2




    $begingroup$
    ZFC is not geometric in its usual signature, but every first order theory is equivalent over classical logic to a geometric theory in an extended language. And since Choice means that every intuitionistic model of ZFC is also a classical model, it seems like the corresponding geometric theory should classify exactly the same models.
    $endgroup$
    – Malice Vidrine
    Oct 12 '18 at 17:24






  • 4




    $begingroup$
    The class of formulas you described are called the coherent formulas (or positive existential formulas, by model-theorists). Geometric formula, on the other hand, are closed under arbitrary disjunctions, in addition to finite conjunctions.
    $endgroup$
    – Alex Kruckman
    Oct 12 '18 at 19:25
















  • 1




    $begingroup$
    If it's reasonably short, could you maybe add a definition of "geometric formula"?
    $endgroup$
    – Stefan Mesken
    Oct 12 '18 at 16:38






  • 1




    $begingroup$
    Of course: I have just edited.
    $endgroup$
    – W. Rether
    Oct 12 '18 at 16:41






  • 2




    $begingroup$
    ZFC is not geometric in its usual signature, but every first order theory is equivalent over classical logic to a geometric theory in an extended language. And since Choice means that every intuitionistic model of ZFC is also a classical model, it seems like the corresponding geometric theory should classify exactly the same models.
    $endgroup$
    – Malice Vidrine
    Oct 12 '18 at 17:24






  • 4




    $begingroup$
    The class of formulas you described are called the coherent formulas (or positive existential formulas, by model-theorists). Geometric formula, on the other hand, are closed under arbitrary disjunctions, in addition to finite conjunctions.
    $endgroup$
    – Alex Kruckman
    Oct 12 '18 at 19:25










1




1




$begingroup$
If it's reasonably short, could you maybe add a definition of "geometric formula"?
$endgroup$
– Stefan Mesken
Oct 12 '18 at 16:38




$begingroup$
If it's reasonably short, could you maybe add a definition of "geometric formula"?
$endgroup$
– Stefan Mesken
Oct 12 '18 at 16:38




1




1




$begingroup$
Of course: I have just edited.
$endgroup$
– W. Rether
Oct 12 '18 at 16:41




$begingroup$
Of course: I have just edited.
$endgroup$
– W. Rether
Oct 12 '18 at 16:41




2




2




$begingroup$
ZFC is not geometric in its usual signature, but every first order theory is equivalent over classical logic to a geometric theory in an extended language. And since Choice means that every intuitionistic model of ZFC is also a classical model, it seems like the corresponding geometric theory should classify exactly the same models.
$endgroup$
– Malice Vidrine
Oct 12 '18 at 17:24




$begingroup$
ZFC is not geometric in its usual signature, but every first order theory is equivalent over classical logic to a geometric theory in an extended language. And since Choice means that every intuitionistic model of ZFC is also a classical model, it seems like the corresponding geometric theory should classify exactly the same models.
$endgroup$
– Malice Vidrine
Oct 12 '18 at 17:24




4




4




$begingroup$
The class of formulas you described are called the coherent formulas (or positive existential formulas, by model-theorists). Geometric formula, on the other hand, are closed under arbitrary disjunctions, in addition to finite conjunctions.
$endgroup$
– Alex Kruckman
Oct 12 '18 at 19:25






$begingroup$
The class of formulas you described are called the coherent formulas (or positive existential formulas, by model-theorists). Geometric formula, on the other hand, are closed under arbitrary disjunctions, in addition to finite conjunctions.
$endgroup$
– Alex Kruckman
Oct 12 '18 at 19:25












1 Answer
1






active

oldest

votes


















2












$begingroup$


1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?




Yes, in the following sense: If $T$ is a first-order theory which happens to have a classifying topos, then it is Morita-equivalent (has naturally equivalent categories of models in any Grothendieck topos) to a geometric theory. This is because any Grothendieck topos is the classifying topos of some geometric theory. (Here I'm following the usual terminology in the topos theory community to allow arbitrary disjunctions in geometric formulas, else the latter statement is wrong.)



(Since the term "classifying topos" is not usually defined for first-order theories, let me be more specific: Suppose that, unusually for first-order theories, pullbacks of models of $T$ along geometric morphisms are again models of $T$. Then we can setup a (pseudo-)functor associating to any Grothendieck topos $mathcal{E}$ the category of $T$-models in $mathcal{E}$. We say that $T$ has a classifying topos if and only if this functor is representable.)




2) Is ZFC a geometric theory?




Not as usually stated, for exactly the reason you mention; same with IZF (intuitionistic Zermelo–Fraenkel).



But I believe that Malice's observation in the comments is right: The Morleyization of ZFC is a geometric theory, and ZFC is Morita-equivalent to its Morleyization.



However, note that in many toposes, there will not even be a single model of ZFC (or its Morleyization): From such a model we can construct a model of PA. But many toposes don't contain a model of PA, for instance the effective topos doesn't (assuming any metatheory) and sheaf toposes over overt spaces don't (assuming a metatheory which doesn't provide for a model of PA). Also, assuming ZFC as the metatheory, not even the topos of sets can be verified to contain a model of ZFC.




3) Is it true that $mathrm{Mod}(mathcal{E},mathrm{ZFC})=mathcal{E}$?




No, that's false, even in the case that $mathcal{E}$ is the topos of sets. It would be true if instead of $mathrm{ZFC}$ we put $mathbb{O}$, the theory with exactly one sort and no atomic propositions, relations or axioms whatsoever.



A model of ZFC (in the topos of sets or in an arbitrary topos) is not (what the topos believes to be) a single set, but (what the topos believes to be) a full set-theoretic universe.






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    2












    $begingroup$


    1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?




    Yes, in the following sense: If $T$ is a first-order theory which happens to have a classifying topos, then it is Morita-equivalent (has naturally equivalent categories of models in any Grothendieck topos) to a geometric theory. This is because any Grothendieck topos is the classifying topos of some geometric theory. (Here I'm following the usual terminology in the topos theory community to allow arbitrary disjunctions in geometric formulas, else the latter statement is wrong.)



    (Since the term "classifying topos" is not usually defined for first-order theories, let me be more specific: Suppose that, unusually for first-order theories, pullbacks of models of $T$ along geometric morphisms are again models of $T$. Then we can setup a (pseudo-)functor associating to any Grothendieck topos $mathcal{E}$ the category of $T$-models in $mathcal{E}$. We say that $T$ has a classifying topos if and only if this functor is representable.)




    2) Is ZFC a geometric theory?




    Not as usually stated, for exactly the reason you mention; same with IZF (intuitionistic Zermelo–Fraenkel).



    But I believe that Malice's observation in the comments is right: The Morleyization of ZFC is a geometric theory, and ZFC is Morita-equivalent to its Morleyization.



    However, note that in many toposes, there will not even be a single model of ZFC (or its Morleyization): From such a model we can construct a model of PA. But many toposes don't contain a model of PA, for instance the effective topos doesn't (assuming any metatheory) and sheaf toposes over overt spaces don't (assuming a metatheory which doesn't provide for a model of PA). Also, assuming ZFC as the metatheory, not even the topos of sets can be verified to contain a model of ZFC.




    3) Is it true that $mathrm{Mod}(mathcal{E},mathrm{ZFC})=mathcal{E}$?




    No, that's false, even in the case that $mathcal{E}$ is the topos of sets. It would be true if instead of $mathrm{ZFC}$ we put $mathbb{O}$, the theory with exactly one sort and no atomic propositions, relations or axioms whatsoever.



    A model of ZFC (in the topos of sets or in an arbitrary topos) is not (what the topos believes to be) a single set, but (what the topos believes to be) a full set-theoretic universe.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$


      1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?




      Yes, in the following sense: If $T$ is a first-order theory which happens to have a classifying topos, then it is Morita-equivalent (has naturally equivalent categories of models in any Grothendieck topos) to a geometric theory. This is because any Grothendieck topos is the classifying topos of some geometric theory. (Here I'm following the usual terminology in the topos theory community to allow arbitrary disjunctions in geometric formulas, else the latter statement is wrong.)



      (Since the term "classifying topos" is not usually defined for first-order theories, let me be more specific: Suppose that, unusually for first-order theories, pullbacks of models of $T$ along geometric morphisms are again models of $T$. Then we can setup a (pseudo-)functor associating to any Grothendieck topos $mathcal{E}$ the category of $T$-models in $mathcal{E}$. We say that $T$ has a classifying topos if and only if this functor is representable.)




      2) Is ZFC a geometric theory?




      Not as usually stated, for exactly the reason you mention; same with IZF (intuitionistic Zermelo–Fraenkel).



      But I believe that Malice's observation in the comments is right: The Morleyization of ZFC is a geometric theory, and ZFC is Morita-equivalent to its Morleyization.



      However, note that in many toposes, there will not even be a single model of ZFC (or its Morleyization): From such a model we can construct a model of PA. But many toposes don't contain a model of PA, for instance the effective topos doesn't (assuming any metatheory) and sheaf toposes over overt spaces don't (assuming a metatheory which doesn't provide for a model of PA). Also, assuming ZFC as the metatheory, not even the topos of sets can be verified to contain a model of ZFC.




      3) Is it true that $mathrm{Mod}(mathcal{E},mathrm{ZFC})=mathcal{E}$?




      No, that's false, even in the case that $mathcal{E}$ is the topos of sets. It would be true if instead of $mathrm{ZFC}$ we put $mathbb{O}$, the theory with exactly one sort and no atomic propositions, relations or axioms whatsoever.



      A model of ZFC (in the topos of sets or in an arbitrary topos) is not (what the topos believes to be) a single set, but (what the topos believes to be) a full set-theoretic universe.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$


        1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?




        Yes, in the following sense: If $T$ is a first-order theory which happens to have a classifying topos, then it is Morita-equivalent (has naturally equivalent categories of models in any Grothendieck topos) to a geometric theory. This is because any Grothendieck topos is the classifying topos of some geometric theory. (Here I'm following the usual terminology in the topos theory community to allow arbitrary disjunctions in geometric formulas, else the latter statement is wrong.)



        (Since the term "classifying topos" is not usually defined for first-order theories, let me be more specific: Suppose that, unusually for first-order theories, pullbacks of models of $T$ along geometric morphisms are again models of $T$. Then we can setup a (pseudo-)functor associating to any Grothendieck topos $mathcal{E}$ the category of $T$-models in $mathcal{E}$. We say that $T$ has a classifying topos if and only if this functor is representable.)




        2) Is ZFC a geometric theory?




        Not as usually stated, for exactly the reason you mention; same with IZF (intuitionistic Zermelo–Fraenkel).



        But I believe that Malice's observation in the comments is right: The Morleyization of ZFC is a geometric theory, and ZFC is Morita-equivalent to its Morleyization.



        However, note that in many toposes, there will not even be a single model of ZFC (or its Morleyization): From such a model we can construct a model of PA. But many toposes don't contain a model of PA, for instance the effective topos doesn't (assuming any metatheory) and sheaf toposes over overt spaces don't (assuming a metatheory which doesn't provide for a model of PA). Also, assuming ZFC as the metatheory, not even the topos of sets can be verified to contain a model of ZFC.




        3) Is it true that $mathrm{Mod}(mathcal{E},mathrm{ZFC})=mathcal{E}$?




        No, that's false, even in the case that $mathcal{E}$ is the topos of sets. It would be true if instead of $mathrm{ZFC}$ we put $mathbb{O}$, the theory with exactly one sort and no atomic propositions, relations or axioms whatsoever.



        A model of ZFC (in the topos of sets or in an arbitrary topos) is not (what the topos believes to be) a single set, but (what the topos believes to be) a full set-theoretic universe.






        share|cite|improve this answer









        $endgroup$




        1) If $T$ is a geometric theory then it has a classifying topos. Does the converse hold? I.e., does the existence of the classifying topos imply that the theory is axiomatizable as a geometric theory?




        Yes, in the following sense: If $T$ is a first-order theory which happens to have a classifying topos, then it is Morita-equivalent (has naturally equivalent categories of models in any Grothendieck topos) to a geometric theory. This is because any Grothendieck topos is the classifying topos of some geometric theory. (Here I'm following the usual terminology in the topos theory community to allow arbitrary disjunctions in geometric formulas, else the latter statement is wrong.)



        (Since the term "classifying topos" is not usually defined for first-order theories, let me be more specific: Suppose that, unusually for first-order theories, pullbacks of models of $T$ along geometric morphisms are again models of $T$. Then we can setup a (pseudo-)functor associating to any Grothendieck topos $mathcal{E}$ the category of $T$-models in $mathcal{E}$. We say that $T$ has a classifying topos if and only if this functor is representable.)




        2) Is ZFC a geometric theory?




        Not as usually stated, for exactly the reason you mention; same with IZF (intuitionistic Zermelo–Fraenkel).



        But I believe that Malice's observation in the comments is right: The Morleyization of ZFC is a geometric theory, and ZFC is Morita-equivalent to its Morleyization.



        However, note that in many toposes, there will not even be a single model of ZFC (or its Morleyization): From such a model we can construct a model of PA. But many toposes don't contain a model of PA, for instance the effective topos doesn't (assuming any metatheory) and sheaf toposes over overt spaces don't (assuming a metatheory which doesn't provide for a model of PA). Also, assuming ZFC as the metatheory, not even the topos of sets can be verified to contain a model of ZFC.




        3) Is it true that $mathrm{Mod}(mathcal{E},mathrm{ZFC})=mathcal{E}$?




        No, that's false, even in the case that $mathcal{E}$ is the topos of sets. It would be true if instead of $mathrm{ZFC}$ we put $mathbb{O}$, the theory with exactly one sort and no atomic propositions, relations or axioms whatsoever.



        A model of ZFC (in the topos of sets or in an arbitrary topos) is not (what the topos believes to be) a single set, but (what the topos believes to be) a full set-theoretic universe.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 21:38









        Ingo BlechschmidtIngo Blechschmidt

        1,395815




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