Subclass of Set is Set
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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