$A=\{A,\emptyset\}$ and axiom of regularity

Question

The axiom of regularity says:

(R) $\forall x[x\not=\emptyset\to\exists y(y\in x\land x\cap y=\emptyset)]$.

From (R) it follows that there is no infinite membership chain (imc).

Consider this set: $A=\{A,\emptyset\}$.

I am confused because it seems to me that A violates and does not violate (R).

It seems to me that A does not violate (R) because $\emptyset\in A$ and $\emptyset\cap A=\emptyset$.

It seem to me that A violates (R) because we can define the imc $A\in A\in A\in ...$

I checked some sources, but I am still confused. Can anyone help me? Thanks.

Answer 1

Suppose there were a set $A$ with the property $A=\{A,\emptyset\}$. Clearly $A \not = \emptyset$ since $A$ would have at least one element and possibly two.

Now let $x=\{A\}$. Clearly $x \not = \emptyset$ since $x$ would have exactly one element. So the axiom of regularity would imply $\exists y\,(y\in x\land x\cap y=\emptyset)$.

Since $x=\{A\}$ would only have one element, this would mean $y\in x$ would imply $y=A$ and then $x\cap y = \{A\} \cap \{A,\emptyset\} = \{A\} = x \not = \emptyset$.

So the property $A=\{A,\emptyset\}$ is inconsistent with the axiom of regularity for any set $A$.

Answer 2

Let's suppose that there is a set $A$ such that $$A=\{A,\emptyset\}.$$ Now, consider the set $$B:=\{A,A\}.$$ This set exists by Pairing (if $A$ exists), and we can show by Extensionality that $$B=\{A\}.$$ Now we apply Regularity to get our contradiction.

You have correctly observed that such a set $A$ does not violate Regularity. However, Regularity is not our only axiom.

Answer 3

The other answers and the various comments clarified this important point. There are two senses of the expression

(a) 'A violates (R)'.

First, (a) may mean: if you let x=A in (R), then you get a false sentence. This is not true: A does not violate (R) in this sense.

Second, (a) may mean: if you assume that A exists (plus the axioms of ZFC minus (R)), then the denial of (R) holds. This is true (as people showed). A violates (R) in this sense. Many thanks.

Answer 4

The set $A=\{A, \emptyset\}$ itself does not violate Regularity, since indeed there exists an element inside $A$ whose intersection with $A$ is empty. $$\emptyset\cap A = \emptyset$$ (Incidentally this property holds true for any set $A$.)

However the set $\{A\}$ does violate Regularity: every element (there is only one) inside $\{A\}$ intersects $\{A\}$ in a nonempty way. $$A \cap \{A\} = A \ne \emptyset$$ because $A \in \{A\}$ and $A \in A$.

Therefore although it may seem that $A$ doesn’t violate regularity, it has a consequence that does.