¿Can you help me with axiom of regularity?

Question

I understand that if we define a set such that $A=\{A\}$ we have then a contradiction because of regularity axiom, thus, a set can not be a member of itself. My question is; how can it be so, if I define a set $A=\{A,b\}$ ($b$ disjoint from $A$)?

Thanks.

Answer 1

In general, if we suppose that $A$ is any set such that $A\in A,$ then let $C=\{A\}$. That will give you your contradiction from the Axiom of Regularity.

Answer 2

Given $A\in A$, regardless or not $A=\{A\}$, we can define $\{A\}$ in one of many ways:

1. By the axiom of pairing, if one takes it as an axiom of $\sf ZF$, the pair $\{A,A\}$ exists, and using extensionality we know this is really just $\{A\}$.
2. By the axiom of power set we know that $\mathcal P(A)$ exists, and by the axiom of separation, if one takes it - or as a consequence of replacement instead, one can define $\{A\}$ as the set of maximal elements of $\mathcal P(A)$ when ordered by $\subseteq$.
3. One can make a direct appeal to replacement, and define the function from the singleton $\{\varnothing\}$ (which exists due to power set and empty set axioms), and make it so the image is exactly $\{A\}$.
4. One can use a Russell-like trick and another appeal to separation, and require $\{x\in A\mid A\in x\}$, so $A$ is a member of this set, and it is sufficient to contradict the axiom of regularity.