Subclass of Set is Set

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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.


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.