A description/proof of that no set has itself as member
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
Target Context
- The reader will have a description and a proof of the proposition that no set has itself as a member.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any set, \(S\), \(S \notin S\).
2: Proof
This proposition stems from the regularity axiom.
Suppose that \(S \in S\). By the subset axiom, \(S' := \{s \in S\vert s = S\} = \{S\}\) would be a set. As \(S \in S\), \(S' \cap S = \{S\} \cap S = \{S\} \neq \emptyset\), a contradiction, being against the regularity axiom.
