2

I understand that the axiom of Regularity excludes $A=\{A\}$ but what about $A=\{A, \emptyset \}$?

Mario Pucci
  • 21
  • 2
    It excludes $A={\emptyset, {A}}$ or $A={\emptyset, {\emptyset, {\emptyset,{\emptyset, A}}}}$, or anything like that where $A$ is a member of some element of (an element of an element of...) $A.$ – spaceisdarkgreen Jan 07 '24 at 19:37

2 Answers2

5

By pairing+comprehension there exists a set $B=\{A,A\}=\{A\}$ and then $B\cap A=\{A\}\cap\{A, \emptyset\}=\{A\}\ne\emptyset$, and since $A$ is the only member of $B$, $B$ violates the axiom of regularity.

Chad K
  • 4,873
1

If $A = \{A,\emptyset\}$ then $A \in A$ which is impossible with Regularity.

C. Dubussy
  • 9,120