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I have a question concerning the two implications of the axiom of regularity given in the third edition of Analysis I by Terence Tao. I am aware this question has been raised before but there only one of the implications of the axiom of regularity has been been included. The statement of axiom 3.9 gives:

Axiom 3.9 (Regularity). If A is non-empty set, then there is at least one element x of A which is "either not a set", or is disjoint from A.

I put the first requirement of in quotation marks as the implication of this definition is where I get confused. Exercise 3.2.2 reads:

Exercise 3.2.2. Use the axiom of regularity (and the singleton set axiom) to show that if $A$ is a set, then $A \notin A$ Furthermore show that if A and B are two sets, then either $A\notin B$ or $B\notin A$ (or both).

The first part of the exercise is the part I want clarified. I have already shown that if $A$ contains an element, $x$ disjoint from $A$, then this implies that $A \notin A$ as a consequence of $A \cap \{A\} = \emptyset$, but I am unsure how the presence of primitive objects (objects that aren't sets) in the set, $A$ implies the same result. Every source I have found doesn't include this discussion of primitive objects in the axiom of regularity and only states that there is an object, $x$, disjoint from $A$. I would appreciate any help with understanding this, thank you.

P.S. Apologies if I have got any of the terminology wrong, I am a physics student, not a mathematician.

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    $\begingroup$ Your argument still works, because we are assuming by hypothesis in exercise 3.2.2 that $A$ is a set. To flesh this out; since $A$ is a set, so is $\{A\}$, and so by the axiom of regularity, the set $\{A\}$ contains an element that is either (a) not a set, or (b) disjoint from $\{A\}$. The only element of $\{A\}$ is $A$, which is by hypothesis a set, so (a) cannot apply and hence we must have $A\cap\{A\}=\emptyset$ as desired. $\endgroup$
    – Atticus Stonestrom
    Commented Nov 10, 2020 at 6:21
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    $\begingroup$ This could just be an answer, @AtticusStonestrom :) $\endgroup$
    – SK19
    Commented Nov 10, 2020 at 7:25
  • $\begingroup$ @AtticusStonestrom I think I get this now, the first part of the axiom ensures that primitive objects can exist within set say A = {1,{1,2}} but as the A and {1,2} are not disjoint this needs to be stated separately and shown using the second argument that this does not lead to self contained sets, is that correct? Thank you! $\endgroup$
    – Kristian Stokkereit
    Commented Nov 10, 2020 at 13:49
  • $\begingroup$ Also @SK19 I am unsure what you mean by, "This could just be an answer". $\endgroup$
    – Kristian Stokkereit
    Commented Nov 10, 2020 at 13:50
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    $\begingroup$ Also I agree with @SK19, your answer was very well explained and might be good for anyone else who is interested. Definitely worthy of an upvote. $\endgroup$
    – Kristian Stokkereit
    Commented Nov 12, 2020 at 1:35

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