Intersection of Class is Subset of Intersection of Subclass

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


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

$A \subseteq B$


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

$\ds \bigcap B \subseteq \bigcap A$


Proof

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

By Intersection of Empty Class we have that:

$\ds \bigcap \O = V$

Thus we have:

$\ds \bigcap B = \bigcap A = V$

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

So while in this case $\ds \bigcap B \subseteq \bigcap A$, $\ds \bigcap B$ is a subclass of $\ds \bigcap A$ 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:

$\ds \bigcap A = V$

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


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

By definition of basic universe:

$\ds \bigcap B \in V$

By the axiom of transitivity, $V$ is transitive.

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

$\ds \bigcap B \subseteq V$

and the result holds.

$\Box$


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

Let $x \in \ds \bigcap B$.

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:

$x \in \ds \bigcap A$

Hence the result by definition of subset.

$\blacksquare$


Sources