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On algebra-valued models for set theory

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$begingroup$

Consider the following bicomplemented lattice L of the form $ L=(L, wedge, vee, neg, D, 0, 1)$, where the base set, is order-isomorphic to an ordinal. Furthermore, $neg x = max y: x wedge y = 0 $ exists for every $x in L$ as well as $Dx = min y in L: x vee y =1 $ exists for every $x in L$. Notice that we are only considering linear L. Now we can define a negation $sim x =_df Dx wedge (x vee neg x)$ and an implication $ xleadsto y =_df neg(x wedge neg y)$, that we will interpret respectively as negation and implication in our L-valued models.

Now we can define by transfinite recursion a L-valued model of set theory $V^L$. Take the following particular case $L_4$= $(1,b,a,0, wedge, vee, neg, D, 0, 1)$ where $1 geq b geq a geq 0$. Then we define the corresponding model $V^L_4$ as usual. Now, is it possible to define a sentence $varphi$ in the language of set theory (without parameters ) such that $[varphi]^L=a$ ? If you answer is negative, how would you prove it ? By induction on the complexity of sentences (so you take $Pi_1$ and $Sigma_1$ sentences as base case and then you consider more complex sentences for your step)?

set-theory

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asked Mar 29 at 16:15

Jesus MartinezJesus Martinez

355

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is this difference from the question you've asked before?
  $endgroup$
  – Stefan Mesken
  Mar 30 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear @Stefan , well I was not satisfied with the previous answers since we do not have a dense homogenous subalgebra in our algebra ( so it is not a homogenous forcing ). For instance take the following formula $varphi = exists x,y, z (x=y wedge z notin x wedge z in y )$ , then $[varphi]^L_4 = b$. But now is it possible to find a formula in the LANGUAGE OF SET THEORY (so without parameters) that gets values $a$ ? Please let me know if i am not being clear enough and thank you very much for your time.
  $endgroup$
  – Jesus Martinez
  Apr 2 at 22:28
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Consider the following bicomplemented lattice L of the form $ L=(L, wedge, vee, neg, D, 0, 1)$, where the base set, is order-isomorphic to an ordinal. Furthermore, $neg x = max y: x wedge y = 0 $ exists for every $x in L$ as well as $Dx = min y in L: x vee y =1 $ exists for every $x in L$. Notice that we are only considering linear L. Now we can define a negation $sim x =_df Dx wedge (x vee neg x)$ and an implication $ xleadsto y =_df neg(x wedge neg y)$, that we will interpret respectively as negation and implication in our L-valued models.

Now we can define by transfinite recursion a L-valued model of set theory $V^L$. Take the following particular case $L_4$= $(1,b,a,0, wedge, vee, neg, D, 0, 1)$ where $1 geq b geq a geq 0$. Then we define the corresponding model $V^L_4$ as usual. Now, is it possible to define a sentence $varphi$ in the language of set theory (without parameters ) such that $[varphi]^L=a$ ? If you answer is negative, how would you prove it ? By induction on the complexity of sentences (so you take $Pi_1$ and $Sigma_1$ sentences as base case and then you consider more complex sentences for your step)?

set-theory

share|cite|improve this question

asked Mar 29 at 16:15

Jesus MartinezJesus Martinez

355

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is this difference from the question you've asked before?
  $endgroup$
  – Stefan Mesken
  Mar 30 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear @Stefan , well I was not satisfied with the previous answers since we do not have a dense homogenous subalgebra in our algebra ( so it is not a homogenous forcing ). For instance take the following formula $varphi = exists x,y, z (x=y wedge z notin x wedge z in y )$ , then $[varphi]^L_4 = b$. But now is it possible to find a formula in the LANGUAGE OF SET THEORY (so without parameters) that gets values $a$ ? Please let me know if i am not being clear enough and thank you very much for your time.
  $endgroup$
  – Jesus Martinez
  Apr 2 at 22:28
  
  

  
  
  
  

add a comment | 

$begingroup$

Consider the following bicomplemented lattice L of the form $ L=(L, wedge, vee, neg, D, 0, 1)$, where the base set, is order-isomorphic to an ordinal. Furthermore, $neg x = max y: x wedge y = 0 $ exists for every $x in L$ as well as $Dx = min y in L: x vee y =1 $ exists for every $x in L$. Notice that we are only considering linear L. Now we can define a negation $sim x =_df Dx wedge (x vee neg x)$ and an implication $ xleadsto y =_df neg(x wedge neg y)$, that we will interpret respectively as negation and implication in our L-valued models.

Now we can define by transfinite recursion a L-valued model of set theory $V^L$. Take the following particular case $L_4$= $(1,b,a,0, wedge, vee, neg, D, 0, 1)$ where $1 geq b geq a geq 0$. Then we define the corresponding model $V^L_4$ as usual. Now, is it possible to define a sentence $varphi$ in the language of set theory (without parameters ) such that $[varphi]^L=a$ ? If you answer is negative, how would you prove it ? By induction on the complexity of sentences (so you take $Pi_1$ and $Sigma_1$ sentences as base case and then you consider more complex sentences for your step)?

set-theory

share|cite|improve this question

asked Mar 29 at 16:15

Jesus MartinezJesus Martinez

355

$endgroup$

share|cite|improve this question

asked Mar 29 at 16:15

Jesus MartinezJesus Martinez

355

share|cite|improve this question

asked Mar 29 at 16:15

Jesus MartinezJesus Martinez

355

asked Mar 29 at 16:15

Jesus MartinezJesus Martinez

355

asked Mar 29 at 16:15

Jesus MartinezJesus Martinez

355