Intersection of Non-Empty Class is Set

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Theorem

Let $A$ be a non-empty class.

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


Then $\ds \bigcap A$ is a set.


Corollary

Let $x$ be a non-empty set.

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


Then $\ds \bigcap x$ is a set.


Proof 1

Let $V$ denote the basic universe such that $A \subseteq V$.

We are given that $A$ is non-empty.

Then $\exists x \in A$ where $x$ is a set.

By definition of intersection of class, every element of $\ds \bigcap A$ is an element of all elements of $A$.

Thus:

$\ds \bigcap A \subseteq x$

We are given that $A$ is a subclass of the basic universe $V$.

Thus $x \in V$ by definition of basic universe.


By the Axiom of Swelledness, $V$ is a swelled class.

By definition of swelled class, every subclass of a set $x \in V$ is a set.

It follows $\ds \bigcap A$ is a set.

$\blacksquare$


Proof 2

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.

$\blacksquare$