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From book Axioms and Set Theory - A first course in Set Theory by Robert Andr´ Axiom A9 (Axiom of regularity): Every non-empty set A contains an element x whose intersection with A is empty.

From Wikipedia In mathematics, the axiom of regularity (also known as the axiom of foundation) is an axiom of Zermelo–Fraenkel set theory that states that every non-empty set A contains an element that is disjoint from A. In first-order logic, the axiom reads:

{\displaystyle \forall x\,(x\neq \varnothing \rightarrow \exists y\in x\,(y\cap x=\varnothing )).}

Now I totally understand that a class A = {A} is not a set by axiom of regularity as A intersection {A} is not null.

But What about class A = {A,B}, now as there exists B belong to A such that B intersection A is null so as per axiom of regularity such a set A = {A, B} and A is an element of A can exist.

But then in Wikipedia it is mentioned that - The axiom of regularity together with the axiom of pairing implies that no set is an element of itself

So how to prove that class A = {A,B} is not a set

(simple understandable answer if possible is very much appreciated.)

Shreyas Bhatt
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    To use LaTeX, simply put things between single dollar signs "$$$" for inline math, and double dollar signs "$$$$" for display math – Vsotvep Feb 15 '20 at 13:17
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    If ${A,B}$ is indeed a set that satisfies $A={A,B}$ then consequently $A$ is a set that satisfies $A\in A$. Then ${A}$ is a non-empty set that does not contain an element that has empty intersection with $A$. This contradicts the axiom of regularity. Our conclusion is then that no set ${A,B}$ exists that satisfies $A={A,B}$. – drhab Feb 15 '20 at 14:32
  • ok thanks now I get it, A={A,B} can not be a set, the proof is elegant. just a follow up question, can it be a class or not even a class? – Shreyas Bhatt Feb 15 '20 at 14:59

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