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Zuhair Al-Johar
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In a theory having proper classes it won't be that easy to phrase how many classes we have in comparison to the number of elements of some setclass. Here, I'll adopt the following method:

We'd say that: there are no more "classes satisfying a unary predicate $\phi$" than "the elements of some class of $X$", if and only if, there exists a class $F$ of ordered pairs such that for each class $C$ satisfying $\phi$ there is a unique element $m$ of $X$ such that: $$ C= \{c \mid (c,m) \in F \}$$

We say that there are only countably many classes to mean there are no more classes than the elements of $\omega$.

Concerning theory $\sf ZFC + Classes + Definability \ rule$, if we add to it the axiom that each class has countably many elements, would it follow that there are only countably many classes?

In a theory having proper classes it won't be that easy to phrase how many classes we have in comparison to the number of elements of some set. Here, I'll adopt the following method:

We'd say that: there are no more "classes satisfying a unary predicate $\phi$" than "the elements of some class of $X$", if and only if, there exists a class $F$ of ordered pairs such that for each class $C$ satisfying $\phi$ there is a unique element $m$ of $X$ such that: $$ C= \{c \mid (c,m) \in F \}$$

We say that there are only countably many classes to mean there are no more classes than the elements of $\omega$.

Concerning theory $\sf ZFC + Classes + Definability \ rule$, if we add to it the axiom that each class has countably many elements, would it follow that there are only countably many classes?

In a theory having proper classes it won't be that easy to phrase how many classes we have in comparison to the number of elements of some class. Here, I'll adopt the following method:

We'd say that: there are no more "classes satisfying a unary predicate $\phi$" than "the elements of some class $X$", if and only if, there exists a class $F$ of ordered pairs such that for each class $C$ satisfying $\phi$ there is a unique element $m$ of $X$ such that: $$ C= \{c \mid (c,m) \in F \}$$

We say that there are only countably many classes to mean there are no more classes than the elements of $\omega$.

Concerning theory $\sf ZFC + Classes + Definability \ rule$, if we add to it the axiom that each class has countably many elements, would it follow that there are only countably many classes?

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Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47

Can having no more than countably many classes, be inferred from, having every class being countable?

In a theory having proper classes it won't be that easy to phrase how many classes we have in comparison to the number of elements of some set. Here, I'll adopt the following method:

We'd say that: there are no more "classes satisfying a unary predicate $\phi$" than "the elements of some class of $X$", if and only if, there exists a class $F$ of ordered pairs such that for each class $C$ satisfying $\phi$ there is a unique element $m$ of $X$ such that: $$ C= \{c \mid (c,m) \in F \}$$

We say that there are only countably many classes to mean there are no more classes than the elements of $\omega$.

Concerning theory $\sf ZFC + Classes + Definability \ rule$, if we add to it the axiom that each class has countably many elements, would it follow that there are only countably many classes?