# Equivalence of Formulations of Axiom of Empty Set for Classes

## Theorem

In the context of class theory, the following formulations of the **Axiom of the Empty Set** are equivalent:

### Formulation 1

- $\exists x: \forall y: \paren {\neg \paren {y \in x} }$

### Formulation 2

The empty class $\O$ is a set, that is:

- $\O \in V$

## 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:

There exists a set that has no elements:

- $\exists x: \forall y: \paren {\neg \paren {y \in x} }$

From the Axiom of Specification it follows that the $x$ so created is a class, and we label it $\O$.

But by hypothesis this $\O$ is a set.

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

$\Box$

### Formulation $2$ implies Formulation $1$

Let formulation $2$ be axiomatic:

The empty class $\O$ is a set, that is:

- $\O \in V$

By the Axiom of Specification, we can define the class $\O$ as:

- $\O := \set {y: \lnot {y \ne y} }$

which is a class with no elements.

We have that $\O$ is a set by hypothesis.

By the Axiom of Extensionality we have that $\O$ is unique.

From Equivalence of Formulations of Axiom of Empty Set, we can express $\O$ as:

- $\O := \set {y: \lnot {y \in \O} }$

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

$\blacksquare$