Empty Set is Unique

Theorem

The empty set is unique.

Proof 1

Let $\O$ and $\O'$ both be empty sets.

From Empty Set is Subset of All Sets, $\O \subseteq \O'$, because $\O$ is empty.

Likewise, we have $\O' \subseteq \O$, since $\O'$ is empty.

Together, by the definition of set equality, this implies that $\O = \O'$.

Thus there is only one empty set.

$\blacksquare$

Proof 2

Let $A$ and $B$ both be empty sets.

Thus:

$\forall x: \neg \paren {x \in A}$

and:

$\forall x: \neg \paren {x \in B}$

Hence:

$x \notin A \iff x \notin B$

and so:

$x \in A \iff x \in B$

vacuously.

From Axiom of Extension:

$\forall x: \paren {x \in A \iff x \in B} \iff A = B$

Hence the result.

$\blacksquare$

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.1: \ 1$