Axiom of Subsets Equivalents

From ProofWiki
Jump to navigation Jump to search

Theorem

The Axiom of Specification states that:

$\forall z: \forall \map P y: \exists x: \forall y: \paren {y \in x \iff \paren {y \in z \land \map P y} }$

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

$\forall z: \forall A: \paren {\paren {z \cap A} \in U}$
$\forall z: \forall A: \paren {A \subseteq z \implies A \in U}$

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


Proof of the First Statement

The Axiom of Specification states:

$\forall z: \forall \map P y: \exists x: \forall y: \paren {y \in x \iff \paren {y \in z \land \map P y} }$

$y \in A$ is substituted for the propositional function $\map P y$.


This leads to the statement:

$\forall z: \forall A: \exists x: \forall y: \paren {y \in x \iff \paren {y \in z \land y \in A} }$

By definition of intersection:

$\forall z: \forall A: \exists x: \forall y: \paren {y \in x \iff y \in \paren {z \cap A} }$

By definition of class equality:

$\forall z: \forall A: \exists x = \paren {z \cap A}$

This is equivalent to:

$\forall z: \forall A: \paren {z \cap A} \in U$

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

$\Box$


Re-derivation of the Axiom of Specification

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 Specification 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:

$\forall z: \forall A: \paren {\paren {z \cap A} \in U}$

We will now take the definition of subset:

$A \subseteq B \iff \forall x: \paren {x \in A \implies x \in B}$

From Intersection with Subset is Subset:

$A \subseteq B \iff \paren {A \cap B} = A$

Thus:

$A \subseteq B \implies \paren {\paren {A \cap B} \in U \implies A \in U}$

We will take the result of the first statement:

$\forall z: \forall A: \paren {\paren {z \cap A} \in U}$

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

$\forall z: \forall A: \paren {A \subseteq z \implies A \in U}$

$\Box$


Re-derivation of the Axiom of Specification

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

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

$\forall z: \forall A: \paren {A \cap z} \in U$

$\blacksquare$


Sources