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 $2$.
$\blacksquare$