# Intersection of Class is Subset of Intersection of Subclass

## Theorem

Let $V$ be a basic universe.

Let $A$ and $B$ be classes of $V$:

- $A \subseteq V, B \subseteq V$

such that it is not the case that $A = B = \O$.

Let $\bigcap A$ and $\bigcap B$ denote the intersection of $A$ and intersection of $B$ respectively.

Let $A$ be a subclass of $B$:

- $A \subseteq B$

Then $\bigcap B$ is a subset of $\bigcap A$:

- $\bigcap B \subseteq \bigcap A$

## Proof

First we consider the degenerate case where $A = B = \O$.

By Intersection of Empty Class we have that:

- $\bigcap \O = V$

Thus we have:

- $\bigcap B = \bigcap A = V$

and neither $\bigcap B$ nor $\bigcap A$ are in fact sets.

So while in this case $\bigcap B \subseteq \bigcap A$, $\bigcap B$ is a subclass of $\bigcap A$ and not actually a subset.

$\Box$

Next we cover the case where $A = \O$ is the empty class.

It is noted that Empty Class is Subclass of All Classes, and so $A \subseteq B$.

By Intersection of Empty Class we have that:

- $\bigcap A = V$

and so by definition of basic universe $\bigcap B \subseteq V$.

By Intersection of Non-Empty Class is Set, $\bigcap B$ is a set.

By definition of basic universe:

- $\bigcap B \in V$

By the Axiom of Transitivity, $V$ is transitive.

Hence as $\bigcap B \in V$ it follows that:

- $\bigcap B \subseteq V$

and the result holds.

$\Box$

By Intersection of Non-Empty Class is Set, both $\bigcap A$ and $\bigcap B$ are sets.

Let $x \in \bigcap B$.

Then:

- $\forall y \in B: x \in y$

That is:

- $y \in B \implies x \in y$

But as $A \subseteq B$, we have that:

- $y \in A \implies y \in B$

and therefore:

- $y \in A \implies x \in y$

That is:

- $\forall y \in A: x \in y$

and so by definition of intersection:

- $x \in \bigcap A$

Hence the result by definition of subset.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.2. \ \text {(b)}$