What is the consistency strength of Ackermann + the following cardinals to ordinals isomorphism?
$begingroup$
This is a try to salvage the attempt written in the posting:
"Can we derive a large cardinal axiom by a principle of isomorphism between cardinals and ordinals?"
Here we change the base theory to Ackermann's set theory +Choice + Foundation over all classes
To restate the principle: for any property $psi$ that is parameter free definable in the langauge of Ackermann set theory that doesn't use the symbol $V$ [standing for the class of all sets], such that for every set ordinal $x$ there is class cardinal $y$ that satisfy $psi$ such that $x$ is strictly smaller than $y$, then the class of all sets (i.e. elements of $V$) that are cardinals, is isomorphic to the class of all ordinals that are sets.
Note: to avoid confusion, the terms 'cardinal' and "ordinal" are defined in the usual manner following Von Neumann's, but they are not restricted to elements of $V$.
Formally this is:
Cardinals to Ordinals isomprhism: if $psi(Y)$ is a formula that doesn't use the symbol $V$, in which only the symbol $``Y"$ appear free [and only free], and the symbol $``x"$ never occurring, and $psi(x)$ is the formula obtained from $psi(Y)$ by merely replacing each occurrence of the symbol $``Y"$ by an occurrence of the symbol $``x"$, then:
$$forall x in V (x text{ is an ordinal } to exists Y[psi(Y) wedge Y text{ is a cardinal } wedge x < Y]) \ to {x in V | x text{ is a cardinal } wedge psi(x) } text{ is order isomorphic to } {x in V| x text{ is an ordinal}} $$
is an axiom.
We also add to this theory the axiom that $V$ is of regular cardinality.
The idea is that "regular" is definable in a parameter free manner without using $V$, and so it would fulfill the antecedent of that principle, so by this principle the class of all regular cardinals in $V$ must be order isomorphic to the class of all ordinals in $V$, this is stronger than the usual Ackermann set theory, and also stronger than both $ZF$ and $MK$.
So the question is about the consistency strength of this theory?
set-theory first-order-logic large-cardinals
asked Dec 10 '18 at 14:16
ZuhairZuhair
294112
$endgroup$
add a comment |
$begingroup$
This is a try to salvage the attempt written in the posting:
"Can we derive a large cardinal axiom by a principle of isomorphism between cardinals and ordinals?"
Here we change the base theory to Ackermann's set theory +Choice + Foundation over all classes
To restate the principle: for any property $psi$ that is parameter free definable in the langauge of Ackermann set theory that doesn't use the symbol $V$ [standing for the class of all sets], such that for every set ordinal $x$ there is class cardinal $y$ that satisfy $psi$ such that $x$ is strictly smaller than $y$, then the class of all sets (i.e. elements of $V$) that are cardinals, is isomorphic to the class of all ordinals that are sets.
Note: to avoid confusion, the terms 'cardinal' and "ordinal" are defined in the usual manner following Von Neumann's, but they are not restricted to elements of $V$.
Formally this is:
Cardinals to Ordinals isomprhism: if $psi(Y)$ is a formula that doesn't use the symbol $V$, in which only the symbol $``Y"$ appear free [and only free], and the symbol $``x"$ never occurring, and $psi(x)$ is the formula obtained from $psi(Y)$ by merely replacing each occurrence of the symbol $``Y"$ by an occurrence of the symbol $``x"$, then:
$$forall x in V (x text{ is an ordinal } to exists Y[psi(Y) wedge Y text{ is a cardinal } wedge x < Y]) \ to {x in V | x text{ is a cardinal } wedge psi(x) } text{ is order isomorphic to } {x in V| x text{ is an ordinal}} $$
is an axiom.
We also add to this theory the axiom that $V$ is of regular cardinality.
The idea is that "regular" is definable in a parameter free manner without using $V$, and so it would fulfill the antecedent of that principle, so by this principle the class of all regular cardinals in $V$ must be order isomorphic to the class of all ordinals in $V$, this is stronger than the usual Ackermann set theory, and also stronger than both $ZF$ and $MK$.
So the question is about the consistency strength of this theory?
set-theory first-order-logic large-cardinals
asked Dec 10 '18 at 14:16
ZuhairZuhair
294112
$endgroup$
add a comment |
$begingroup$
This is a try to salvage the attempt written in the posting:
"Can we derive a large cardinal axiom by a principle of isomorphism between cardinals and ordinals?"
Here we change the base theory to Ackermann's set theory +Choice + Foundation over all classes
To restate the principle: for any property $psi$ that is parameter free definable in the langauge of Ackermann set theory that doesn't use the symbol $V$ [standing for the class of all sets], such that for every set ordinal $x$ there is class cardinal $y$ that satisfy $psi$ such that $x$ is strictly smaller than $y$, then the class of all sets (i.e. elements of $V$) that are cardinals, is isomorphic to the class of all ordinals that are sets.
Note: to avoid confusion, the terms 'cardinal' and "ordinal" are defined in the usual manner following Von Neumann's, but they are not restricted to elements of $V$.
Formally this is:
Cardinals to Ordinals isomprhism: if $psi(Y)$ is a formula that doesn't use the symbol $V$, in which only the symbol $``Y"$ appear free [and only free], and the symbol $``x"$ never occurring, and $psi(x)$ is the formula obtained from $psi(Y)$ by merely replacing each occurrence of the symbol $``Y"$ by an occurrence of the symbol $``x"$, then:
$$forall x in V (x text{ is an ordinal } to exists Y[psi(Y) wedge Y text{ is a cardinal } wedge x < Y]) \ to {x in V | x text{ is a cardinal } wedge psi(x) } text{ is order isomorphic to } {x in V| x text{ is an ordinal}} $$
is an axiom.
We also add to this theory the axiom that $V$ is of regular cardinality.
The idea is that "regular" is definable in a parameter free manner without using $V$, and so it would fulfill the antecedent of that principle, so by this principle the class of all regular cardinals in $V$ must be order isomorphic to the class of all ordinals in $V$, this is stronger than the usual Ackermann set theory, and also stronger than both $ZF$ and $MK$.
So the question is about the consistency strength of this theory?
set-theory first-order-logic large-cardinals
asked Dec 10 '18 at 14:16
ZuhairZuhair
294112
$endgroup$
This is a try to salvage the attempt written in the posting:
"Can we derive a large cardinal axiom by a principle of isomorphism between cardinals and ordinals?"
Here we change the base theory to Ackermann's set theory +Choice + Foundation over all classes
To restate the principle: for any property $psi$ that is parameter free definable in the langauge of Ackermann set theory that doesn't use the symbol $V$ [standing for the class of all sets], such that for every set ordinal $x$ there is class cardinal $y$ that satisfy $psi$ such that $x$ is strictly smaller than $y$, then the class of all sets (i.e. elements of $V$) that are cardinals, is isomorphic to the class of all ordinals that are sets.
Note: to avoid confusion, the terms 'cardinal' and "ordinal" are defined in the usual manner following Von Neumann's, but they are not restricted to elements of $V$.
Formally this is:
Cardinals to Ordinals isomprhism: if $psi(Y)$ is a formula that doesn't use the symbol $V$, in which only the symbol $``Y"$ appear free [and only free], and the symbol $``x"$ never occurring, and $psi(x)$ is the formula obtained from $psi(Y)$ by merely replacing each occurrence of the symbol $``Y"$ by an occurrence of the symbol $``x"$, then:
$$forall x in V (x text{ is an ordinal } to exists Y[psi(Y) wedge Y text{ is a cardinal } wedge x < Y]) \ to {x in V | x text{ is a cardinal } wedge psi(x) } text{ is order isomorphic to } {x in V| x text{ is an ordinal}} $$
is an axiom.
We also add to this theory the axiom that $V$ is of regular cardinality.
The idea is that "regular" is definable in a parameter free manner without using $V$, and so it would fulfill the antecedent of that principle, so by this principle the class of all regular cardinals in $V$ must be order isomorphic to the class of all ordinals in $V$, this is stronger than the usual Ackermann set theory, and also stronger than both $ZF$ and $MK$.
So the question is about the consistency strength of this theory?
set-theory first-order-logic large-cardinals
set-theory first-order-logic large-cardinals
asked Dec 10 '18 at 14:16
ZuhairZuhair
294112
asked Dec 10 '18 at 14:16
ZuhairZuhair
294112
edited Dec 10 '18 at 18:39
Zuhair
asked Dec 10 '18 at 14:16
ZuhairZuhair
294112
asked Dec 10 '18 at 14:16
ZuhairZuhair
294112
asked Dec 10 '18 at 14:16
ZuhairZuhair
294112
294112
add a comment |
add a comment |
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