Equivalence of Formulations of Axiom of Empty Set for Classes

From ProofWiki
Jump to navigation Jump to search

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 $2$.

$\blacksquare$