Existence of Set is Equivalent to Existence of Empty Set
Theorem
Let $V$ be a basic universe.
Let $P$ be the axiom:
- $V$ has at least one element.
Then $P$ is equivalent to the Axiom of the Empty Set:
- The empty class $\O$ is a set.
Proof
Necessary Condition
Let the Axiom of the Empty Set hold.
That is:
- $\O$ is a set.
By definition of a basic universe, $V$ is a universal class.
Hence, by definition, every set is an element of $V$.
We have that $\O$ is a set by hypothesis.
Thus:
- $\O \in V$
and by definition of empty class it is seen that $V$ is not empty.
$\Box$
Sufficient Condition
Let $V$ have at least one element.
Let $x \in V$ be an element of $V$.
Then by the Axiom of Transitivity $x$ is also a class.
By Empty Class is Subclass of All Classes:
- $\O \subseteq x$
By the Axiom of Swelledness, every subclass of $x$ is an element of $V$.
That includes $\O$.
Thus:
- $\O \in V$
By definition of a basic universe, $V$ is a fortiori a universal class.
Hence all of its elements are by definition sets.
Hence it follows that $\O$ is a set.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 3$ Axiom of the empty set