Proving a class satisfies axioms of ZF-C
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We were given an exercise, to prove that for every regular $theta$, the class ${x||tc(x)|<theta}$ satisfies ZF-C without the powerset axiom. We ran into some difficulties proving the replacement axiom.
So, assuming all elements of the range are in $H_theta$, we have no problem. But what guarantees us that this is the case? Can we nor work with functions that send elements in $H_theta$ out of range?
set-theory
asked Dec 22 '18 at 10:19
Uri George PeterzilUri George Peterzil
10610
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add a comment |
$begingroup$
We were given an exercise, to prove that for every regular $theta$, the class ${x||tc(x)|<theta}$ satisfies ZF-C without the powerset axiom. We ran into some difficulties proving the replacement axiom.
So, assuming all elements of the range are in $H_theta$, we have no problem. But what guarantees us that this is the case? Can we nor work with functions that send elements in $H_theta$ out of range?
set-theory
asked Dec 22 '18 at 10:19
Uri George PeterzilUri George Peterzil
10610
$endgroup$
$begingroup$
What does ZF-C mean? It's either ZFC or ZF.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 13:25
add a comment |
$begingroup$
We were given an exercise, to prove that for every regular $theta$, the class ${x||tc(x)|<theta}$ satisfies ZF-C without the powerset axiom. We ran into some difficulties proving the replacement axiom.
So, assuming all elements of the range are in $H_theta$, we have no problem. But what guarantees us that this is the case? Can we nor work with functions that send elements in $H_theta$ out of range?
set-theory
asked Dec 22 '18 at 10:19
Uri George PeterzilUri George Peterzil
10610
$endgroup$
We were given an exercise, to prove that for every regular $theta$, the class ${x||tc(x)|<theta}$ satisfies ZF-C without the powerset axiom. We ran into some difficulties proving the replacement axiom.
So, assuming all elements of the range are in $H_theta$, we have no problem. But what guarantees us that this is the case? Can we nor work with functions that send elements in $H_theta$ out of range?
set-theory
set-theory
asked Dec 22 '18 at 10:19
Uri George PeterzilUri George Peterzil
10610
asked Dec 22 '18 at 10:19
Uri George PeterzilUri George Peterzil
10610
asked Dec 22 '18 at 10:19
Uri George PeterzilUri George Peterzil
10610
asked Dec 22 '18 at 10:19
Uri George PeterzilUri George Peterzil
10610
asked Dec 22 '18 at 10:19
Uri George PeterzilUri George Peterzil
10610
10610
$begingroup$
What does ZF-C mean? It's either ZFC or ZF.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 13:25
add a comment |
$begingroup$
What does ZF-C mean? It's either ZFC or ZF.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 13:25
$begingroup$
What does ZF-C mean? It's either ZFC or ZF.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 13:25
$begingroup$
What does ZF-C mean? It's either ZFC or ZF.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 13:25
add a comment |
1 Answer
1
active
oldest
votes
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The axiom schema of replacement (e.g. as described here) only applies to those $phi$ such that for all $x$ we can find a unique $y$ such that $phi(x,y)$ holds (omitting parameters $w_i$ for simplicity). This $y$ must thus exist and be unique in the model $V$ we are considering (so in particular it must be in $V$ or $H_theta$), or we don't even need to consider this $phi$.
answered Dec 22 '18 at 10:38
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
It's basically a refomulation of what Arthur said, but maybe a reformulation makes it clearer for you?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:39
$begingroup$
Thanks, we were simply using a formulation of what formula satisfies the function property, that is incorrect when working with several models.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 10:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The axiom schema of replacement (e.g. as described here) only applies to those $phi$ such that for all $x$ we can find a unique $y$ such that $phi(x,y)$ holds (omitting parameters $w_i$ for simplicity). This $y$ must thus exist and be unique in the model $V$ we are considering (so in particular it must be in $V$ or $H_theta$), or we don't even need to consider this $phi$.
answered Dec 22 '18 at 10:38
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
It's basically a refomulation of what Arthur said, but maybe a reformulation makes it clearer for you?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:39
$begingroup$
Thanks, we were simply using a formulation of what formula satisfies the function property, that is incorrect when working with several models.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 10:56
add a comment |
$begingroup$
The axiom schema of replacement (e.g. as described here) only applies to those $phi$ such that for all $x$ we can find a unique $y$ such that $phi(x,y)$ holds (omitting parameters $w_i$ for simplicity). This $y$ must thus exist and be unique in the model $V$ we are considering (so in particular it must be in $V$ or $H_theta$), or we don't even need to consider this $phi$.
answered Dec 22 '18 at 10:38
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
It's basically a refomulation of what Arthur said, but maybe a reformulation makes it clearer for you?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:39
$begingroup$
Thanks, we were simply using a formulation of what formula satisfies the function property, that is incorrect when working with several models.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 10:56
add a comment |
$begingroup$
The axiom schema of replacement (e.g. as described here) only applies to those $phi$ such that for all $x$ we can find a unique $y$ such that $phi(x,y)$ holds (omitting parameters $w_i$ for simplicity). This $y$ must thus exist and be unique in the model $V$ we are considering (so in particular it must be in $V$ or $H_theta$), or we don't even need to consider this $phi$.
answered Dec 22 '18 at 10:38
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
The axiom schema of replacement (e.g. as described here) only applies to those $phi$ such that for all $x$ we can find a unique $y$ such that $phi(x,y)$ holds (omitting parameters $w_i$ for simplicity). This $y$ must thus exist and be unique in the model $V$ we are considering (so in particular it must be in $V$ or $H_theta$), or we don't even need to consider this $phi$.
answered Dec 22 '18 at 10:38
Henno BrandsmaHenno Brandsma
114k348124
answered Dec 22 '18 at 10:38
Henno BrandsmaHenno Brandsma
114k348124
answered Dec 22 '18 at 10:38
Henno BrandsmaHenno Brandsma
114k348124
answered Dec 22 '18 at 10:38
Henno BrandsmaHenno Brandsma
114k348124
114k348124
$begingroup$
It's basically a refomulation of what Arthur said, but maybe a reformulation makes it clearer for you?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:39
$begingroup$
Thanks, we were simply using a formulation of what formula satisfies the function property, that is incorrect when working with several models.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 10:56
add a comment |
$begingroup$
It's basically a refomulation of what Arthur said, but maybe a reformulation makes it clearer for you?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:39
$begingroup$
Thanks, we were simply using a formulation of what formula satisfies the function property, that is incorrect when working with several models.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 10:56
$begingroup$
It's basically a refomulation of what Arthur said, but maybe a reformulation makes it clearer for you?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:39
$begingroup$
It's basically a refomulation of what Arthur said, but maybe a reformulation makes it clearer for you?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:39
$begingroup$
Thanks, we were simply using a formulation of what formula satisfies the function property, that is incorrect when working with several models.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 10:56
$begingroup$
Thanks, we were simply using a formulation of what formula satisfies the function property, that is incorrect when working with several models.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 10:56
add a comment |
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$begingroup$
What does ZF-C mean? It's either ZFC or ZF.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 13:25