# Equivalence of Formulations of Axiom of Powers

## 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 $2$.

$\blacksquare$