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Proof check: Prove that if there is a class $B$ such that $A in B$ then $A$ is a set

Why is the collection of all groups a proper class rather than a set?Question about set theory class notationA set is not an element of itself.Similar proof to the class of all sets is a proper classHow to prove that if $Asubseteq Atimes A$, then $A=varnothing$?Axiomatic set theoryHow can a class not be a set?Is there a set for which no defining formula can be found?Is the Russell set a proper class in $ZFC$About the definition of a class

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I am reading through Notes on Set Theory, 2nd ed. by Moschovakis, and this is part of an exercise on classes, which the author defines it as "either a set or a unary definite condition which is not coextensive with a set." Here, we use the definition that "a unary definite condition $P$ is coextensive with a set $A$ if the objects which satisfy it are precisely the members of $A$." We use the convention of the notation $x in P iff P(x)$.

Let $A$ be a class. Prove that, if there exists a class $B$ such that $A in B$, then $A$ is a set.

If $B$ is a set, then ${A}$ is a subset of $B$, and hence by the Unionset axiom $A$ is a set. Consider the case when $B$ is a proper class, i.e. an unary definite condition not coextensive with a set. In that case, we would have that $B(A)$ is true. If $A$ were a proper class, that is, an unary definite condition not coextensive with set, this would result in a contradiction because in an axiomatic set theory an unary definite conditions is not an object (at least Moschvakis defines it this way) and hence not in the domain of $B$.Thus $A$ is not a proper class, and hence a set.

Is this proof correct? If not, any help woth solving the problem would bw greatly appreciated.

elementary-set-theory

share|cite|improve this question

edited Mar 19 at 4:28

Andrés E. Caicedo

65.8k8160252

asked Mar 19 at 3:11

Cute BrownieCute Brownie

1,048417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No. Classes cannot be the element of a singleton.
  $endgroup$
  – William Elliot
  Mar 19 at 4:36
  
  
  

  
  
  
  


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  @WilliamElliot They can if they’re sets.
  $endgroup$
  – spaceisdarkgreen
  Mar 19 at 5:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This looks like a correct solution in the context of the slightly strange treatment of the book. The initial part “If B is a set...” can be simplified and merged with the second if you just pass to the coextensive condition.
  $endgroup$
  – spaceisdarkgreen
  Mar 19 at 5:18
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I am reading through Notes on Set Theory, 2nd ed. by Moschovakis, and this is part of an exercise on classes, which the author defines it as "either a set or a unary definite condition which is not coextensive with a set." Here, we use the definition that "a unary definite condition $P$ is coextensive with a set $A$ if the objects which satisfy it are precisely the members of $A$." We use the convention of the notation $x in P iff P(x)$.

Let $A$ be a class. Prove that, if there exists a class $B$ such that $A in B$, then $A$ is a set.

If $B$ is a set, then ${A}$ is a subset of $B$, and hence by the Unionset axiom $A$ is a set. Consider the case when $B$ is a proper class, i.e. an unary definite condition not coextensive with a set. In that case, we would have that $B(A)$ is true. If $A$ were a proper class, that is, an unary definite condition not coextensive with set, this would result in a contradiction because in an axiomatic set theory an unary definite conditions is not an object (at least Moschvakis defines it this way) and hence not in the domain of $B$.Thus $A$ is not a proper class, and hence a set.

Is this proof correct? If not, any help woth solving the problem would bw greatly appreciated.

elementary-set-theory

share|cite|improve this question

edited Mar 19 at 4:28

Andrés E. Caicedo

65.8k8160252

asked Mar 19 at 3:11

Cute BrownieCute Brownie

1,048417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No. Classes cannot be the element of a singleton.
  $endgroup$
  – William Elliot
  Mar 19 at 4:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WilliamElliot They can if they’re sets.
  $endgroup$
  – spaceisdarkgreen
  Mar 19 at 5:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This looks like a correct solution in the context of the slightly strange treatment of the book. The initial part “If B is a set...” can be simplified and merged with the second if you just pass to the coextensive condition.
  $endgroup$
  – spaceisdarkgreen
  Mar 19 at 5:18
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I am reading through Notes on Set Theory, 2nd ed. by Moschovakis, and this is part of an exercise on classes, which the author defines it as "either a set or a unary definite condition which is not coextensive with a set." Here, we use the definition that "a unary definite condition $P$ is coextensive with a set $A$ if the objects which satisfy it are precisely the members of $A$." We use the convention of the notation $x in P iff P(x)$.

Let $A$ be a class. Prove that, if there exists a class $B$ such that $A in B$, then $A$ is a set.

If $B$ is a set, then ${A}$ is a subset of $B$, and hence by the Unionset axiom $A$ is a set. Consider the case when $B$ is a proper class, i.e. an unary definite condition not coextensive with a set. In that case, we would have that $B(A)$ is true. If $A$ were a proper class, that is, an unary definite condition not coextensive with set, this would result in a contradiction because in an axiomatic set theory an unary definite conditions is not an object (at least Moschvakis defines it this way) and hence not in the domain of $B$.Thus $A$ is not a proper class, and hence a set.

Is this proof correct? If not, any help woth solving the problem would bw greatly appreciated.

elementary-set-theory

share|cite|improve this question

edited Mar 19 at 4:28

Andrés E. Caicedo

65.8k8160252

asked Mar 19 at 3:11

Cute BrownieCute Brownie

1,048417

$endgroup$

share|cite|improve this question

edited Mar 19 at 4:28

Andrés E. Caicedo

65.8k8160252

asked Mar 19 at 3:11

Cute BrownieCute Brownie

1,048417

share|cite|improve this question

edited Mar 19 at 4:28

Andrés E. Caicedo

65.8k8160252

asked Mar 19 at 3:11

Cute BrownieCute Brownie

1,048417

edited Mar 19 at 4:28

Andrés E. Caicedo

65.8k8160252

edited Mar 19 at 4:28

Andrés E. Caicedo

65.8k8160252

asked Mar 19 at 3:11

Cute BrownieCute Brownie

1,048417

asked Mar 19 at 3:11

Cute BrownieCute Brownie

1,048417