Is ζ={∅, ζ} a set according to "Axiom of Regularity"?

Question

For the following paragraph, I do not understand the {{∅, ζ}} part.

What about something like ζ = {∅, ζ}? Now that ζ contains the empty set, it’s not in obvious violation of the axiom of regularity. But it is in fact in violation. Here’s the argument: Start with ζ. Pair it with itself to form {ζ}. Now ask, does {ζ} contain any sets that it shares no elements with? It’s easier to see if we expand this using the definition of ζ: {ζ} = {{∅, ζ}}. The problem is that {ζ} contains just one set, ζ, and ζ and {ζ} share an element: namely, ζ itself! So in fact, even adding the empty set won’t help us. The axiom of regularity rules out any sets that contain themselves.

Introduction to Mathematical Logic

So, i think it like this;

1. We could select the element "ζ" from the set {∅, ζ}. And we pair the element "ζ" with set ζ={∅, ζ}. "ζ" is also a set which is {∅, ζ} and when we get intersection with the set ζ={∅, ζ}; Then we get {∅, ζ}. So, intersection is not empty set(not ∅). This is against Axiom of Regularity.

2. We could select the "∅" element(so I select it because it is a set too). And we pair the "∅" element with set ζ={∅, ζ}. And when we get intersection of "∅" with the set ζ={∅, ζ}; Then we get "∅". So, intersection is empty set. Thus, axiom of regularity.

3. According to "1", we have an element of ζ that is against axiom of regularity.

4. But according to "2", we find an element in ζ, namely ∅ which intersection with ζ is ∅. Thus, there is at least one disjoint element as Axiom of Regularity requires. So, this set ζ is valid according to Axiom of Regularity.

Question: So, am I doing anything wrong here?

Side Question: I am selecting an element from a set if it is actually a set i.e. from ζ={∅, ζ} set, I select "∅", "ζ" as these are already sets. If there is some element "a" in ζ which is not a set, then I will not select it. Am I doing right here?

Answer 1

You’ve misunderstood the argument. The set $\zeta$ itself does not directly violate regularity, because $\zeta$ does contain an element whose intersection with $\zeta$ is empty: $\varnothing\in\zeta$, and $\varnothing\cap\zeta=\varnothing$. The problem is that if $\zeta$ is a set, the pairing axiom tells us that $\{\zeta,\zeta\}=\{\zeta\}$ is also a set, and it violates regularity: its only element is $\zeta$, and

$$\zeta\cap\{\zeta\}=\{\varnothing,\zeta\}\cap\{\zeta\}=\{\zeta\}\ne\varnothing\,.$$