$A\notin B$ set thoery

Question

Let $A$ and $B$ be sets. Is true or false that $A\notin B$? I think It's false.

Proof

Suppouse that $A\notin B$. Then for any set $A$ and $B$ is the case that $A\notin B$, this implies that $\emptyset\notin\{ \emptyset, \emptyset\}$, but this is false.

Is this proof correct?

Thank you so much.

Answer 1

While you've removed the term "arbitrary set", you have the same error.

Suppouse [sic] that [$A,B$ are sets and] $A \not\in B$.

That's fine.

Then for any set $A$ and $B$ is the case that $A \not\in B$,

No. This is a faulty generalization. Your use of the same names just confuses the issue, but there's absolutely no reason that your assumption of $A \not\in B$ should be able to, and in fact it does not, imply that "For any sets $x,y$, it is the case that $x \not\in y$."

Answer 2

It's not correct, hence it's not a proof.

As an example let's take the set of all real numbers as A and a set of points in a unit circle as B. Certainly $A\notin B$ because elements of B are points in a circle, not sets of numbers.

On the other hand a set $\mathbb P$ of all prime numbers is a proper subset of $\mathbb N$, the set of all natural numbers $\mathbb P\subset \mathbb N$.
As such, it is an element of a powerset of natural numbers: $\mathbb P \in \mathcal P(\mathbb N)$.

Clearly, $A \notin B$ does not imply $\mathbb P \notin \mathcal P(\mathbb N)$.