Equivalence of Formulations of Axiom of Powers

From ProofWiki
Jump to navigation Jump to search

Theorem

In the context of class theory, the following formulations of the axiom of powers are equivalent:

Formulation 1

For every set, there exists a set of sets whose elements are all the subsets of the given set.

$\forall x: \exists y: \paren {\forall z: \paren {z \in y \iff \paren {w \in z \implies w \in x} } }$

Formulation 2

Let $x$ be a set.

Then its power set $\powerset x$ is also a set.


Proof

It is assumed throughout that the axiom of extensionality and the axiom of specification both hold.


Formulation $1$ implies Formulation $2$

Let formulation $1$ be axiomatic:

For every set, there exists a set of sets whose elements are all the subsets of the given set.

$\forall x: \exists y: \paren {\forall z: \paren {z \in y \iff \paren {w \in z \implies w \in x} } }$


Thus it is posited that for a given set $x$ the power set of $x$ exists:

$y := \powerset x := \set {z: z \subseteq x}$

and that this is a set.


Formulation $2$ asserts that given the existence of $\powerset x$, it is axiomatic that $\powerset x$ is itself a set.

Hence it follows that the truth of formulation $2$ follows from acceptance of the truth of formulation $1$.

From Power Set Exists and is Unique, which depends on:

the axiom of extensionality
the axiom of specification

$\powerset x$ is unique for a given $x$.

$\Box$


Formulation $2$ implies Formulation $1$

Let formulation $2$ be axiomatic:

Let $x$ be a set.

Then its power set $\powerset x$ is also a set.


Let $x$ be a set.

Let us create the power set $y$ of x:

$\powerset x := \set {z: z \subseteq x}$

From Power Set Exists and is Unique, which depends on:

the axiom of extensionality
the axiom of specification

$\powerset x$ exists and is unique for a given $x$.


We have asserted the truth of formulation $2$.

That is, $\powerset x$ is a set.


As $x$ is arbitrary, it follows that $\powerset x$ exists and is unique for all sets $x$.

That is, the truth of formulation $1$ follows from acceptance of the truth of formulation $2$.

$\blacksquare$


Sources