# 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