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
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.