Intersection of Non-Empty Class is Set/Proof 2

Theorem

Let $A$ be a non-empty class.

Let $\bigcap A$ denote the intersection of $A$.

Then $\bigcap A$ is a set.

Proof

Since $A$ is a non-empty class, there exists $S \in A$.

Since $S$ is an element of a class, it is not a proper class, and is thus a set.

By definition of class intersection:

$x \in \bigcap A \implies x \in S$

By the subclass definition:

$\bigcap A \subseteq S$

By Subclass of Set is Set, $\bigcap A$ is a set.

$\blacksquare$

Sources

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