Subclass of Set is Set

Theorem

Let $A$ be a set.

Let $\map \phi x$ be a condition in which $x$ is taken to be a set.

Then there exists a set that consists of all of the elements of $A$ that satisfies this condition.

In ZF, this result is known as the Axiom of Specification.

Proof

By the axiom of class comprehension, let $B$ be the class defined as:

\(\ds B\)

\(=\)

\(\ds \set {x: x \in A \land \map \phi x}\)

\(\ds \)

\(=\)

\(\ds \set {x \in A: \map \phi x}\)

Set-Builder Notation

Aiming for a contradiction, suppose that $B$ is not a set.

Then $B$ must be a proper class.

It is easily seen that $B \subseteq A$.

So by the Axiom of Powers:

$B \in \powerset A$

where $\powerset A$ is denotes the power set of $A$.

But by Proper Class is not Element of Class, this is a contradiction.

Therefore by Proof by Contradiction it follows that $B$ is a set.

$\blacksquare$

Sources

• 2002: Thomas Jech: Set Theory (3rd ed.) ... (previous) ... (next): Chapter $1$: Separation Schema