The Boolean-valued model $V^B$ is full.

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I was reading this lemma from Jech's tome and came across a "clearly, ..." so i tried to prove it on my own but i'm failing miserably. To give context:

Lemma 14.19. $V^B$ is full. Given a formula $varphi(x, ...)$, there exists some $a in V^B$ such that, $$ |varphi(a, ...)| = |exists xvarphi(x, ...)|$$ Proof. $le$ holds for every $a$ because of the definition of $|exists xvarphi(x, ...)|$, we wish to find $a in V^B$ such that $ge$ holds. Let $u_0 = |exists xvarphi(x, ...)|$. Let $$ D = u in B: textthere is some a_u text such that u le .$$
It is clear that $D$ is open and dense below $u_0$. Let W be a maximal set of pairwise disjoint elements of $D$; clearly, $sumu: uin W ge u_0$. By lemma 14.18 we are done.$hspace3cmBox$

---

Now if $u in D$ we know that there exists some $a_u$ such that $u le |varphi(a_u, ...)|$, so $u le sum_a in V^B|varphi(a, ...)| = |exists xvarphi(x, ...)| = u_0$ and again for each $u in W$ we have $u le u_0$ and in order to prove $sumu: uin W ge u_0$ we have to prove $sumu: uin W = u_0$. And this is where i'm stuck. I have tried reaching a contradiction by assuming that equality doesn't hold but i got no results. Any hints or answers are appreciated.

Thanks for your patience.

set-theory boolean-algebra

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asked Jul 20 at 9:52

Shervin Sorouri

334111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just recall the definition of the value of the existential formula. $bigvee D=|exists xvarphi(x, ...)|$ by definition. In particular $|varphi(a, ...)|in D$ for every $a$.
  – Apostolos
  Jul 20 at 10:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Apostolos, I know but what i am struggling is linking it to $sum W$.
  – Shervin Sorouri
  Jul 20 at 10:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum W=sum D$ since it's maximal, no?
  – Apostolos
  Jul 20 at 10:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Assume $sum W<sum D$ hence there is a $uin D$ such that $(sum W)+u>sum W$. Then $u-sum Wleq u$ so $u-sum Win D$ and thus $Wcupu-sum W$ is a set of pairwise disjoint elements of $D$ extending $W$ contradicting the maximality of $W$.
  – Apostolos
  Jul 20 at 11:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I mean the $ucdot(-sum W)$, yes.
  – Apostolos
  Jul 20 at 12:11
  
  

  
  
  
  

 | 
show 2 more comments

up vote
3
down vote

favorite

I was reading this lemma from Jech's tome and came across a "clearly, ..." so i tried to prove it on my own but i'm failing miserably. To give context:

Lemma 14.19. $V^B$ is full. Given a formula $varphi(x, ...)$, there exists some $a in V^B$ such that, $$ |varphi(a, ...)| = |exists xvarphi(x, ...)|$$ Proof. $le$ holds for every $a$ because of the definition of $|exists xvarphi(x, ...)|$, we wish to find $a in V^B$ such that $ge$ holds. Let $u_0 = |exists xvarphi(x, ...)|$. Let $$ D = u in B: textthere is some a_u text such that u le .$$
It is clear that $D$ is open and dense below $u_0$. Let W be a maximal set of pairwise disjoint elements of $D$; clearly, $sumu: uin W ge u_0$. By lemma 14.18 we are done.$hspace3cmBox$

---

Now if $u in D$ we know that there exists some $a_u$ such that $u le |varphi(a_u, ...)|$, so $u le sum_a in V^B|varphi(a, ...)| = |exists xvarphi(x, ...)| = u_0$ and again for each $u in W$ we have $u le u_0$ and in order to prove $sumu: uin W ge u_0$ we have to prove $sumu: uin W = u_0$. And this is where i'm stuck. I have tried reaching a contradiction by assuming that equality doesn't hold but i got no results. Any hints or answers are appreciated.

Thanks for your patience.

set-theory boolean-algebra

share|cite|improve this question

asked Jul 20 at 9:52

Shervin Sorouri

334111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just recall the definition of the value of the existential formula. $bigvee D=|exists xvarphi(x, ...)|$ by definition. In particular $|varphi(a, ...)|in D$ for every $a$.
  – Apostolos
  Jul 20 at 10:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Apostolos, I know but what i am struggling is linking it to $sum W$.
  – Shervin Sorouri
  Jul 20 at 10:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum W=sum D$ since it's maximal, no?
  – Apostolos
  Jul 20 at 10:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Assume $sum W<sum D$ hence there is a $uin D$ such that $(sum W)+u>sum W$. Then $u-sum Wleq u$ so $u-sum Win D$ and thus $Wcupu-sum W$ is a set of pairwise disjoint elements of $D$ extending $W$ contradicting the maximality of $W$.
  – Apostolos
  Jul 20 at 11:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I mean the $ucdot(-sum W)$, yes.
  – Apostolos
  Jul 20 at 12:11
  
  

  
  
  
  

 | 
show 2 more comments

up vote
3
down vote

favorite

up vote
3
down vote

favorite

I was reading this lemma from Jech's tome and came across a "clearly, ..." so i tried to prove it on my own but i'm failing miserably. To give context:

Lemma 14.19. $V^B$ is full. Given a formula $varphi(x, ...)$, there exists some $a in V^B$ such that, $$ |varphi(a, ...)| = |exists xvarphi(x, ...)|$$ Proof. $le$ holds for every $a$ because of the definition of $|exists xvarphi(x, ...)|$, we wish to find $a in V^B$ such that $ge$ holds. Let $u_0 = |exists xvarphi(x, ...)|$. Let $$ D = u in B: textthere is some a_u text such that u le .$$
It is clear that $D$ is open and dense below $u_0$. Let W be a maximal set of pairwise disjoint elements of $D$; clearly, $sumu: uin W ge u_0$. By lemma 14.18 we are done.$hspace3cmBox$

---

Now if $u in D$ we know that there exists some $a_u$ such that $u le |varphi(a_u, ...)|$, so $u le sum_a in V^B|varphi(a, ...)| = |exists xvarphi(x, ...)| = u_0$ and again for each $u in W$ we have $u le u_0$ and in order to prove $sumu: uin W ge u_0$ we have to prove $sumu: uin W = u_0$. And this is where i'm stuck. I have tried reaching a contradiction by assuming that equality doesn't hold but i got no results. Any hints or answers are appreciated.

Thanks for your patience.

set-theory boolean-algebra

share|cite|improve this question

asked Jul 20 at 9:52

Shervin Sorouri

334111

I was reading this lemma from Jech's tome and came across a "clearly, ..." so i tried to prove it on my own but i'm failing miserably. To give context:

Lemma 14.19. $V^B$ is full. Given a formula $varphi(x, ...)$, there exists some $a in V^B$ such that, $$ |varphi(a, ...)| = |exists xvarphi(x, ...)|$$ Proof. $le$ holds for every $a$ because of the definition of $|exists xvarphi(x, ...)|$, we wish to find $a in V^B$ such that $ge$ holds. Let $u_0 = |exists xvarphi(x, ...)|$. Let $$ D = u in B: textthere is some a_u text such that u le .$$
It is clear that $D$ is open and dense below $u_0$. Let W be a maximal set of pairwise disjoint elements of $D$; clearly, $sumu: uin W ge u_0$. By lemma 14.18 we are done.$hspace3cmBox$

---

Now if $u in D$ we know that there exists some $a_u$ such that $u le |varphi(a_u, ...)|$, so $u le sum_a in V^B|varphi(a, ...)| = |exists xvarphi(x, ...)| = u_0$ and again for each $u in W$ we have $u le u_0$ and in order to prove $sumu: uin W ge u_0$ we have to prove $sumu: uin W = u_0$. And this is where i'm stuck. I have tried reaching a contradiction by assuming that equality doesn't hold but i got no results. Any hints or answers are appreciated.

Thanks for your patience.

set-theory boolean-algebra

share|cite|improve this question

asked Jul 20 at 9:52

Shervin Sorouri

334111

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 9:52

Shervin Sorouri

334111

asked Jul 20 at 9:52

Shervin Sorouri

334111

asked Jul 20 at 9:52

Shervin Sorouri

334111

334111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just recall the definition of the value of the existential formula. $bigvee D=|exists xvarphi(x, ...)|$ by definition. In particular $|varphi(a, ...)|in D$ for every $a$.
  – Apostolos
  Jul 20 at 10:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Apostolos, I know but what i am struggling is linking it to $sum W$.
  – Shervin Sorouri
  Jul 20 at 10:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum W=sum D$ since it's maximal, no?
  – Apostolos
  Jul 20 at 10:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Assume $sum W<sum D$ hence there is a $uin D$ such that $(sum W)+u>sum W$. Then $u-sum Wleq u$ so $u-sum Win D$ and thus $Wcupu-sum W$ is a set of pairwise disjoint elements of $D$ extending $W$ contradicting the maximality of $W$.
  – Apostolos
  Jul 20 at 11:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I mean the $ucdot(-sum W)$, yes.
  – Apostolos
  Jul 20 at 12:11
  
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just recall the definition of the value of the existential formula. $bigvee D=|exists xvarphi(x, ...)|$ by definition. In particular $|varphi(a, ...)|in D$ for every $a$.
  – Apostolos
  Jul 20 at 10:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Apostolos, I know but what i am struggling is linking it to $sum W$.
  – Shervin Sorouri
  Jul 20 at 10:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum W=sum D$ since it's maximal, no?
  – Apostolos
  Jul 20 at 10:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Assume $sum W<sum D$ hence there is a $uin D$ such that $(sum W)+u>sum W$. Then $u-sum Wleq u$ so $u-sum Win D$ and thus $Wcupu-sum W$ is a set of pairwise disjoint elements of $D$ extending $W$ contradicting the maximality of $W$.
  – Apostolos
  Jul 20 at 11:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I mean the $ucdot(-sum W)$, yes.
  – Apostolos
  Jul 20 at 12:11
  
  

  
  
  
  

1

1

Just recall the definition of the value of the existential formula. $bigvee D=|exists xvarphi(x, ...)|$ by definition. In particular $|varphi(a, ...)|in D$ for every $a$.
– Apostolos
Jul 20 at 10:04

Just recall the definition of the value of the existential formula. $bigvee D=|exists xvarphi(x, ...)|$ by definition. In particular $|varphi(a, ...)|in D$ for every $a$.
– Apostolos
Jul 20 at 10:04

@Apostolos, I know but what i am struggling is linking it to $sum W$.
– Shervin Sorouri
Jul 20 at 10:07

@Apostolos, I know but what i am struggling is linking it to $sum W$.
– Shervin Sorouri
Jul 20 at 10:07

1

1

$sum W=sum D$ since it's maximal, no?
– Apostolos
Jul 20 at 10:09

$sum W=sum D$ since it's maximal, no?
– Apostolos
Jul 20 at 10:09

1

1

Assume $sum W<sum D$ hence there is a $uin D$ such that $(sum W)+u>sum W$. Then $u-sum Wleq u$ so $u-sum Win D$ and thus $Wcupu-sum W$ is a set of pairwise disjoint elements of $D$ extending $W$ contradicting the maximality of $W$.
– Apostolos
Jul 20 at 11:35

Assume $sum W<sum D$ hence there is a $uin D$ such that $(sum W)+u>sum W$. Then $u-sum Wleq u$ so $u-sum Win D$ and thus $Wcupu-sum W$ is a set of pairwise disjoint elements of $D$ extending $W$ contradicting the maximality of $W$.
– Apostolos
Jul 20 at 11:35

1

1

I mean the $ucdot(-sum W)$, yes.
– Apostolos
Jul 20 at 12:11

I mean the $ucdot(-sum W)$, yes.
– Apostolos
Jul 20 at 12:11

 | 
show 2 more comments

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