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Why do the hereditarily finite sets model ZF-Infinity, if we need Infinity to talk about them?

0

$begingroup$

As is standard in Kunen, we can talk about $WF$ by working up to it through the "sets of rank $n$" function, $R$, defined in the usual way by iterating collection.

In particular, for limit ordinals $gamma$ we take

$R(gamma)= bigcup_{alpha < gamma} R(alpha)$.

It is then standard to take the hereditarily finite sets $HF = V_omega$ to be $R(omega)$.

My question, more specifically, pertains to the axiom of infinity and how $HF$ models $ZF wedge neg Infty$. I understand that the gist of the proof is that it only contains finite elements, but its construction involves $omega$ by the definition of the $R$ function, and $omega$ being a set necessitates the axiom of infinity. How then does $HF$ model $ZFC wedge neg Infty$ if we need an infinite set to exist in order to even talk about it?

Additionally, we would need $V_{omega}$ to be a proper class, so my language ("contains" finite elements...) is imprecise, but this seems less conceptually difficult in comparison to my previous issue. Nevertheless some clarity on how to deal with this issue more precisely would be good.

(Since logicians seem to get by I assume my issue is either misinformed or unimportant insofar as model theory is concerned. I'd like the explanation regardless.)

logic set-theory axioms

share|cite|improve this question

edited Mar 24 at 20:16

Cameron Buie

87k773161

asked Mar 24 at 18:53

BigSocksBigSocks

31

$endgroup$

add a comment | 

0

$begingroup$

As is standard in Kunen, we can talk about $WF$ by working up to it through the "sets of rank $n$" function, $R$, defined in the usual way by iterating collection.

In particular, for limit ordinals $gamma$ we take

$R(gamma)= bigcup_{alpha < gamma} R(alpha)$.

It is then standard to take the hereditarily finite sets $HF = V_omega$ to be $R(omega)$.

My question, more specifically, pertains to the axiom of infinity and how $HF$ models $ZF wedge neg Infty$. I understand that the gist of the proof is that it only contains finite elements, but its construction involves $omega$ by the definition of the $R$ function, and $omega$ being a set necessitates the axiom of infinity. How then does $HF$ model $ZFC wedge neg Infty$ if we need an infinite set to exist in order to even talk about it?

Additionally, we would need $V_{omega}$ to be a proper class, so my language ("contains" finite elements...) is imprecise, but this seems less conceptually difficult in comparison to my previous issue. Nevertheless some clarity on how to deal with this issue more precisely would be good.

(Since logicians seem to get by I assume my issue is either misinformed or unimportant insofar as model theory is concerned. I'd like the explanation regardless.)

logic set-theory axioms

share|cite|improve this question

edited Mar 24 at 20:16

Cameron Buie

87k773161

asked Mar 24 at 18:53

BigSocksBigSocks

31

$endgroup$

add a comment | 

$begingroup$

As is standard in Kunen, we can talk about $WF$ by working up to it through the "sets of rank $n$" function, $R$, defined in the usual way by iterating collection.

In particular, for limit ordinals $gamma$ we take

$R(gamma)= bigcup_{alpha < gamma} R(alpha)$.

It is then standard to take the hereditarily finite sets $HF = V_omega$ to be $R(omega)$.

My question, more specifically, pertains to the axiom of infinity and how $HF$ models $ZF wedge neg Infty$. I understand that the gist of the proof is that it only contains finite elements, but its construction involves $omega$ by the definition of the $R$ function, and $omega$ being a set necessitates the axiom of infinity. How then does $HF$ model $ZFC wedge neg Infty$ if we need an infinite set to exist in order to even talk about it?

Additionally, we would need $V_{omega}$ to be a proper class, so my language ("contains" finite elements...) is imprecise, but this seems less conceptually difficult in comparison to my previous issue. Nevertheless some clarity on how to deal with this issue more precisely would be good.

(Since logicians seem to get by I assume my issue is either misinformed or unimportant insofar as model theory is concerned. I'd like the explanation regardless.)

logic set-theory axioms

share|cite|improve this question

edited Mar 24 at 20:16

Cameron Buie

87k773161

asked Mar 24 at 18:53

BigSocksBigSocks

31

$endgroup$

share|cite|improve this question

edited Mar 24 at 20:16

Cameron Buie

87k773161

asked Mar 24 at 18:53

BigSocksBigSocks

31

share|cite|improve this question

edited Mar 24 at 20:16

Cameron Buie

87k773161

asked Mar 24 at 18:53

BigSocksBigSocks

31

edited Mar 24 at 20:16

Cameron Buie

87k773161

edited Mar 24 at 20:16

Cameron Buie

87k773161

asked Mar 24 at 18:53

BigSocksBigSocks

31

asked Mar 24 at 18:53

BigSocksBigSocks

31

1 Answer
1

active

oldest

votes

3

$begingroup$

The kicker here is that $omega$ is external to the model, since $omeganotin R(omega).$ This means that, as far as the model is concerned, $omega$ is a proper class--the proper class of all ordinals. Furthermore, since $V_omeganotin V_omega,$ then $V_omega$ is also a proper class, as far as the model is concerned--namely, the proper class of all sets.

What we're doing here is constructing an inner model. Assuming that $mathsf{ZF}$ holds, then $V_omega$ exists, and it can be shown that all axioms of $mathsf{ZF}$ (except the Axiom of Infinity, which fails) still hold if we restrict our sets to those in $V_omega$.

share|cite|improve this answer

answered Mar 24 at 19:18

Cameron BuieCameron Buie

87k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this. The word $textit{external}$ helped a bit, as well as the last sentence, focusing on the bit about restriction. It seems like cheating, but when you put it that way it makes sense.
  $endgroup$
  – BigSocks
  Mar 24 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To the proposer: We can define $xin omega$ as an abbreviation for $[, x $ is an ordinal and $forall yin xcup {x},(y=cup yimplies y=emptyset,) ,]$... And "$ x $ is an ordinal " is an abbreviation for "$ x $ is a transitive set , well-ordered by $ in $ ". It is just terribly inconvenient not to use names (i.e. abbreviations) for classes. And when expressed unabbreviated, we must not add any new assumptions or new axioms.
  $endgroup$
  – DanielWainfleet
  Apr 3 at 19:40
  
  
  

  
  
  
  

add a comment | 

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3

$begingroup$

The kicker here is that $omega$ is external to the model, since $omeganotin R(omega).$ This means that, as far as the model is concerned, $omega$ is a proper class--the proper class of all ordinals. Furthermore, since $V_omeganotin V_omega,$ then $V_omega$ is also a proper class, as far as the model is concerned--namely, the proper class of all sets.

What we're doing here is constructing an inner model. Assuming that $mathsf{ZF}$ holds, then $V_omega$ exists, and it can be shown that all axioms of $mathsf{ZF}$ (except the Axiom of Infinity, which fails) still hold if we restrict our sets to those in $V_omega$.

share|cite|improve this answer

answered Mar 24 at 19:18

Cameron BuieCameron Buie

87k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this. The word $textit{external}$ helped a bit, as well as the last sentence, focusing on the bit about restriction. It seems like cheating, but when you put it that way it makes sense.
  $endgroup$
  – BigSocks
  Mar 24 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To the proposer: We can define $xin omega$ as an abbreviation for $[, x $ is an ordinal and $forall yin xcup {x},(y=cup yimplies y=emptyset,) ,]$... And "$ x $ is an ordinal " is an abbreviation for "$ x $ is a transitive set , well-ordered by $ in $ ". It is just terribly inconvenient not to use names (i.e. abbreviations) for classes. And when expressed unabbreviated, we must not add any new assumptions or new axioms.
  $endgroup$
  – DanielWainfleet
  Apr 3 at 19:40
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

The kicker here is that $omega$ is external to the model, since $omeganotin R(omega).$ This means that, as far as the model is concerned, $omega$ is a proper class--the proper class of all ordinals. Furthermore, since $V_omeganotin V_omega,$ then $V_omega$ is also a proper class, as far as the model is concerned--namely, the proper class of all sets.

What we're doing here is constructing an inner model. Assuming that $mathsf{ZF}$ holds, then $V_omega$ exists, and it can be shown that all axioms of $mathsf{ZF}$ (except the Axiom of Infinity, which fails) still hold if we restrict our sets to those in $V_omega$.

share|cite|improve this answer

answered Mar 24 at 19:18

Cameron BuieCameron Buie

87k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this. The word $textit{external}$ helped a bit, as well as the last sentence, focusing on the bit about restriction. It seems like cheating, but when you put it that way it makes sense.
  $endgroup$
  – BigSocks
  Mar 24 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To the proposer: We can define $xin omega$ as an abbreviation for $[, x $ is an ordinal and $forall yin xcup {x},(y=cup yimplies y=emptyset,) ,]$... And "$ x $ is an ordinal " is an abbreviation for "$ x $ is a transitive set , well-ordered by $ in $ ". It is just terribly inconvenient not to use names (i.e. abbreviations) for classes. And when expressed unabbreviated, we must not add any new assumptions or new axioms.
  $endgroup$
  – DanielWainfleet
  Apr 3 at 19:40
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The kicker here is that $omega$ is external to the model, since $omeganotin R(omega).$ This means that, as far as the model is concerned, $omega$ is a proper class--the proper class of all ordinals. Furthermore, since $V_omeganotin V_omega,$ then $V_omega$ is also a proper class, as far as the model is concerned--namely, the proper class of all sets.

What we're doing here is constructing an inner model. Assuming that $mathsf{ZF}$ holds, then $V_omega$ exists, and it can be shown that all axioms of $mathsf{ZF}$ (except the Axiom of Infinity, which fails) still hold if we restrict our sets to those in $V_omega$.

share|cite|improve this answer

answered Mar 24 at 19:18

Cameron BuieCameron Buie

87k773161

$endgroup$

share|cite|improve this answer

answered Mar 24 at 19:18

Cameron BuieCameron Buie

87k773161

answered Mar 24 at 19:18

Cameron BuieCameron Buie

87k773161

answered Mar 24 at 19:18

Cameron BuieCameron Buie

87k773161