Set Consisting of Empty Set is not Empty

Theorem

Let $S$ be the set defined as:

$S = \set \O$

Then $S$ is not the empty set.

That is:

$\O \ne \set \O$

Proof

We have:

$\O \in \set \O$

and so:

$\neg \paren {\forall x: x \notin \O}$

The result follows by definition of the empty set.

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 2$. Sets of sets
• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions