# Axiom of regularity from Terence Tao

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.

• 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. Commented Nov 10, 2020 at 6:21
• This could just be an answer, @AtticusStonestrom :)
– SK19
Commented Nov 10, 2020 at 7:25
• @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! Commented Nov 10, 2020 at 13:49
• Also @SK19 I am unsure what you mean by, "This could just be an answer". Commented Nov 10, 2020 at 13:50
• 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. Commented Nov 12, 2020 at 1:35