Axiom of Subsets Equivalents

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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:

$\forall z: \forall A: \left({\left({z \cap A}\right) \in U}\right)$
$\forall z: \forall A: \left({A \subseteq z \implies A \in U}\right)$

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


Proof of the First Statement

The Axiom of Subsets states:

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

$y \in A$ is substituted for the propositional function $P \left({y}\right)$.


This leads to the statement:

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

By definition of intersection:

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

By definition of class equality:

$\forall z: \forall A: \exists x = \left({z \cap A}\right)$

This is equivalent to:

$\forall z: \forall A: \left({z \cap A}\right) \in U$

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

$\Box$


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 by Biconditional is Commutative.

$\Box$

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:

$\forall z: \forall A: \left({\left({z \cap A}\right) \in U}\right)$

We will now take the definition of the subset:

$A \subseteq B \iff \forall x: \left({x \in A \implies x \in B}\right)$

From Intersection with Subset is Subset:

$A \subseteq B \iff \left({A \cap B}\right) = A$

Thus:

$A \subseteq B \implies \left({\left({A \cap B}\right) \in U \implies A \in U}\right)$

We will take the result of the first statement:

$\forall z: \forall A: \left({\left({z \cap A}\right) \in U}\right)$

Using the above two statements, substituting $z$ for $B$:

$\forall z: \forall A: \left({A \subseteq z \implies A \in U}\right)$

$\Box$


Re-derivation of the Axiom of Subsets

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

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

$\forall z: \forall A: \left({A \cap z}\right) \in U$

$\blacksquare$


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