Axiom of Subsets Equivalents
The flaw in formulation in the first proof puts this whole page without proper foundation. Curse you, Takeuti/Zaring!
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Contents
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 $P \left({y}\right)$.
"$y\in A$ is substituted for $P(y)$". This is definitely NOT valid! Such is the content of Unrestricted Comprehension, the source of Russell...
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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:
- $\forall z: \forall A: \left({A \cap z}\right) \in U$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 5.12$, $\S 5.13$
