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Can a definite condition be considered as an object in $ZF$?
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Can a definite condition be considered as an object in $ZF$?
This question arose from the following question:
For every class $A$, prove that $A$ is a set if for some class $B, A in B$
Where $A in B$ is defined to be equivalent to the fact that $B$ is a set and $A$ is a member of $B$, or (if $B$ is a unary definite condition) $B(A)$.
If anyone can answer both that'd be great
logic set-theory terminology
asked Jul 24 at 23:56
BlueRoses
833314
add a comment |Â
up vote
1
down vote
favorite
Can a definite condition be considered as an object in $ZF$?
This question arose from the following question:
For every class $A$, prove that $A$ is a set if for some class $B, A in B$
Where $A in B$ is defined to be equivalent to the fact that $B$ is a set and $A$ is a member of $B$, or (if $B$ is a unary definite condition) $B(A)$.
If anyone can answer both that'd be great
logic set-theory terminology
asked Jul 24 at 23:56
BlueRoses
833314
ZF has no notion of class. You can't "prove that $A$ is a set" in ZF because every object in ZF is a set.
– Rob Arthan
Jul 26 at 22:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Can a definite condition be considered as an object in $ZF$?
This question arose from the following question:
For every class $A$, prove that $A$ is a set if for some class $B, A in B$
Where $A in B$ is defined to be equivalent to the fact that $B$ is a set and $A$ is a member of $B$, or (if $B$ is a unary definite condition) $B(A)$.
If anyone can answer both that'd be great
logic set-theory terminology
asked Jul 24 at 23:56
BlueRoses
833314
Can a definite condition be considered as an object in $ZF$?
This question arose from the following question:
For every class $A$, prove that $A$ is a set if for some class $B, A in B$
Where $A in B$ is defined to be equivalent to the fact that $B$ is a set and $A$ is a member of $B$, or (if $B$ is a unary definite condition) $B(A)$.
If anyone can answer both that'd be great
logic set-theory terminology
asked Jul 24 at 23:56
BlueRoses
833314
asked Jul 24 at 23:56
BlueRoses
833314
asked Jul 24 at 23:56
BlueRoses
833314
asked Jul 24 at 23:56
BlueRoses
833314
833314
ZF has no notion of class. You can't "prove that $A$ is a set" in ZF because every object in ZF is a set.
– Rob Arthan
Jul 26 at 22:59
add a comment |Â
ZF has no notion of class. You can't "prove that $A$ is a set" in ZF because every object in ZF is a set.
– Rob Arthan
Jul 26 at 22:59
ZF has no notion of class. You can't "prove that $A$ is a set" in ZF because every object in ZF is a set.
– Rob Arthan
Jul 26 at 22:59
ZF has no notion of class. You can't "prove that $A$ is a set" in ZF because every object in ZF is a set.
– Rob Arthan
Jul 26 at 22:59
add a comment |Â
1 Answer
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Well. Not quite exactly, but almost.
For example, the empty set is really the class $xmid xneq x$. But really there is an axiom which states $exists x(forall y(yin xleftrightarrow yneq y))$.
So a condition which defines a class is not a set per se, but it can be extensionally equivalent to a set.
For another example, if $x$ is a set, then the class $ymid yin x$ is equivalent to the set $x$.
answered Jul 25 at 0:01
Asaf Karagila
291k31402732
Thanks, I get that a class of a condition $P$ is either the (unique) set coextensive with $P$ or $P$ itself. Now could you answer my second question?
– BlueRoses
Jul 25 at 0:17
ZF has no notion of class. The only "objects" in ZF are sets. So "not quite exactly, but almost" should read "no" and your explanation should carry on from there.
– Rob Arthan
Jul 26 at 23:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Well. Not quite exactly, but almost.
For example, the empty set is really the class $xmid xneq x$. But really there is an axiom which states $exists x(forall y(yin xleftrightarrow yneq y))$.
So a condition which defines a class is not a set per se, but it can be extensionally equivalent to a set.
For another example, if $x$ is a set, then the class $ymid yin x$ is equivalent to the set $x$.
answered Jul 25 at 0:01
Asaf Karagila
291k31402732
Thanks, I get that a class of a condition $P$ is either the (unique) set coextensive with $P$ or $P$ itself. Now could you answer my second question?
– BlueRoses
Jul 25 at 0:17
ZF has no notion of class. The only "objects" in ZF are sets. So "not quite exactly, but almost" should read "no" and your explanation should carry on from there.
– Rob Arthan
Jul 26 at 23:50
add a comment |Â
up vote
3
down vote
Well. Not quite exactly, but almost.
For example, the empty set is really the class $xmid xneq x$. But really there is an axiom which states $exists x(forall y(yin xleftrightarrow yneq y))$.
So a condition which defines a class is not a set per se, but it can be extensionally equivalent to a set.
For another example, if $x$ is a set, then the class $ymid yin x$ is equivalent to the set $x$.
answered Jul 25 at 0:01
Asaf Karagila
291k31402732
Thanks, I get that a class of a condition $P$ is either the (unique) set coextensive with $P$ or $P$ itself. Now could you answer my second question?
– BlueRoses
Jul 25 at 0:17
ZF has no notion of class. The only "objects" in ZF are sets. So "not quite exactly, but almost" should read "no" and your explanation should carry on from there.
– Rob Arthan
Jul 26 at 23:50
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Well. Not quite exactly, but almost.
For example, the empty set is really the class $xmid xneq x$. But really there is an axiom which states $exists x(forall y(yin xleftrightarrow yneq y))$.
So a condition which defines a class is not a set per se, but it can be extensionally equivalent to a set.
For another example, if $x$ is a set, then the class $ymid yin x$ is equivalent to the set $x$.
answered Jul 25 at 0:01
Asaf Karagila
291k31402732
Well. Not quite exactly, but almost.
For example, the empty set is really the class $xmid xneq x$. But really there is an axiom which states $exists x(forall y(yin xleftrightarrow yneq y))$.
So a condition which defines a class is not a set per se, but it can be extensionally equivalent to a set.
For another example, if $x$ is a set, then the class $ymid yin x$ is equivalent to the set $x$.
answered Jul 25 at 0:01
Asaf Karagila
291k31402732
answered Jul 25 at 0:01
Asaf Karagila
291k31402732
answered Jul 25 at 0:01
Asaf Karagila
291k31402732
answered Jul 25 at 0:01
Asaf Karagila
291k31402732
291k31402732
Thanks, I get that a class of a condition $P$ is either the (unique) set coextensive with $P$ or $P$ itself. Now could you answer my second question?
– BlueRoses
Jul 25 at 0:17
ZF has no notion of class. The only "objects" in ZF are sets. So "not quite exactly, but almost" should read "no" and your explanation should carry on from there.
– Rob Arthan
Jul 26 at 23:50
add a comment |Â
Thanks, I get that a class of a condition $P$ is either the (unique) set coextensive with $P$ or $P$ itself. Now could you answer my second question?
– BlueRoses
Jul 25 at 0:17
ZF has no notion of class. The only "objects" in ZF are sets. So "not quite exactly, but almost" should read "no" and your explanation should carry on from there.
– Rob Arthan
Jul 26 at 23:50
Thanks, I get that a class of a condition $P$ is either the (unique) set coextensive with $P$ or $P$ itself. Now could you answer my second question?
– BlueRoses
Jul 25 at 0:17
Thanks, I get that a class of a condition $P$ is either the (unique) set coextensive with $P$ or $P$ itself. Now could you answer my second question?
– BlueRoses
Jul 25 at 0:17
ZF has no notion of class. The only "objects" in ZF are sets. So "not quite exactly, but almost" should read "no" and your explanation should carry on from there.
– Rob Arthan
Jul 26 at 23:50
ZF has no notion of class. The only "objects" in ZF are sets. So "not quite exactly, but almost" should read "no" and your explanation should carry on from there.
– Rob Arthan
Jul 26 at 23:50
add a comment |Â
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ZF has no notion of class. You can't "prove that $A$ is a set" in ZF because every object in ZF is a set.
– Rob Arthan
Jul 26 at 22:59