# Intersection of Non-Empty Class is Set/Proof 2

## Theorem

Let $A$ be a non-empty class.

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

Then $\ds \bigcap A$ 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:

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

By the subclass definition:

$\ds \bigcap A \subseteq S$

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

$\blacksquare$