Axiom of Subsets Equivalents

Theorem

The Axiom of Subsets states that:

$\forall z: \forall P \left({y}\right): \exists x: \forall y: \left({y \in x \iff \left({y \in z \land P \left({y}\right)}\right)}\right)$

We will prove that this statement is equivalent to the following statements:

$\forall z: \forall A: ( ( z \cap A ) \in U )$

$\forall z: \forall A: ( A \subseteq z \implies A \in U )$

In the above statements, the universe is $U$.

Proof of the First Statement

The Axiom of Subsets states:

$\forall z: \forall P \left({y}\right): \exists x: \forall y: \left({y \in x \iff \left({y \in z \land P \left({y}\right)}\right)}\right)$

We will substitute $y \in A$ for the propositional function $P(y)$. We arrive at the statement:

$\forall z: \forall A: \exists x: \forall y: ( y \in x \iff ( y \in z \land y \in A ) )$

We apply the definition for intersection:

$\forall z: \forall A: \exists x: \forall y: ( y \in x \iff y \in ( z \cap A ) )$

We now apply the definition for set equality:

$\forall z: \forall A: \exists x = ( z \cap A ) )$

This is equivalent to:

$\forall z: \forall A: ( z \cap A ) \in U$

because $A \in U \iff \exists x = A$. $\blacksquare$

Re-derivation of the Axiom of Subsets

Only bi-conditional ($\iff$) statements were used to prove the first result, so it is possible to reverse the step order and arrive at the original Axiom of Subsets. $\blacksquare$

Although this statement is shorter, it uses defined terms, and is thus unsuitable as an axiom.

Proof of the Second Statement

We will take the result of the first statement:

$\forall z: \forall A: ( ( z \cap A ) \in U )$

We will now take the definition of the subset:

$A \subseteq B \iff \forall x: ( x \in A \implies x \in B )$

From Intersection with Subset is Subset we have:

$A \subseteq B \iff ( A \cap B ) = A$

Thus,

$A \subseteq B \implies ( ( A \cap B ) \in U \implies A \in U )$

We will take the result of the first statement:

$\forall z: \forall A: ( ( z \cap A ) \in U )$

Using the above two statements, substituting $z$ for $B$, we arrive at the desired statement:

$\forall z: \forall A: ( A \subseteq z \implies A \in U )$

Re-derivation of the Axiom of Subsets

Because $( A \cap z ) \subseteq z$, the antecedent of $\forall z: \forall A: ( A \subseteq z \implies A \in U )$ is satisfied.

We now arrive at the first statement (above), which in turn can prove the Axiom of Subsets:

$\forall z: \forall A: ( A \cap z ) \in U$

$\blacksquare$