Class has Subclass which is not Element

Theorem

Let $A$ be a class.

Then $A$ has at least one subclass $B$ which is not an element of $A$.

Proof

Let a set $x$ be defined as ordinary if and only if $x \notin x$.

Let $\map \phi x$ be the set property defined as:

$\map \pi x := \neg \paren {x \in x}$

Then by the axiom of specification there exists a class, which can be denoted $B$, such that:

$B = \set {x \in A: \neg \paren {x \in x} }$

That is, $B$ contains as elements of all the ordinary sets of $A$.

By the axiom of extension $B$ is unique.

Thus $B$ is the class of all ordinary sets of $A$.

$x \in B \iff \paren {x \in A \land x \notin x}$

Hence, for all $x$ which happen to be elements of $A$, we have:

$x \in B \iff x \notin x$

Aiming for a contradiction, suppose $B \in A$.

Then we could set $A$ for $x$, and so obtain:

$B \in B \iff B \notin B$

This is a contradiction.

Hence by Proof by Contradiction $B$ cannot be an element of $A$.

That is:

$B \notin V$

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 1$ Extensionality and separation: Theorem $1.2$