Not Every Class is a Set

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Theorem

Let $A$ be a class.

Then it is not necessarily the case that $A$ is also a set.


Proof 1

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 $A$, such that:

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

By the axiom of extension $A$ is unique.

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

Hence we have:

$x \in A \iff x \notin x$


Aiming for a contradiction, suppose $A$ were a set.

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

$A \in A \iff A \notin A$

This is a contradiction.

Hence by Proof by Contradiction $A$ cannot be a set.

But as $A$ is a subclass of the universal class $V$, $A$ is a class.

So the class $A$ of ordinary sets is not a set.

That is:

$A \notin V$

$\blacksquare$


Proof 2

Consider the universal class $V$.

From Class has Subclass which is not Element, $V$ has a subclass $A$ which is not an element of $V$.

But $V$ contains as elements all the sets.

It follows that $A$ is a class which is not a set.

$\blacksquare$