I may just be misunderstanding something here, but according to ZFC a set can not be a member of itself or this could lead to Russell's paradox.
despite this, I seemed to have found a contradiction. I'm not saying that I think I'm right, but I would like someone to point out my misunderstanding.
the axiom of regularity is:
$(\exists A)(A\ne \emptyset \implies(\forall x)(x\in A\implies (\forall y)(y\in x\implies y\notin A))$
now if we define a set y to be
$y=\{x_1,x_2,x_3,…,x_n,y\}$ where all $x_i$ are individuals and not sets.
the set should not exist so we try to find a contradiction.
the axiom of regularity implies
$(\exists x)(x\in y\implies x\cap y=\emptyset)$
so if we have that $x\in y$ then that must mean
$x=x_1\lor x_2\lor x_3\lor …\lor x_n\lor y$
now if we have x=y we have our contradiction as $x\cap y=y\cap y=y$
but if instead, we have for any $x_i$, $x=x_i$ then no contradiction arises as the intersection of a set and a member is the empty set. or in the case, we say that this is not a well-defined concept than, at least in the book I am reading (Patrick Suppes Axiomatic Set Theory), an impossible set is defined to be the empty set.
Could someone explain the problem in my argument?
Best Answer
If $y$ were a set, then by the Axiom of Separation you have that $$ z= \{x\in y\mid x=y\} = \{y\}$$ is also a set.
But now $z$ does not satisfy the Axiom of Regularity: the only element of $z$ is $y$, but $y\cap z = \{y\}$, because $y\in y$ and $y\in z$. So $z$ does not have any element that is disjoint from $z$.
So while $y$ does not contradict Regularity directly, if $y$ were a set it would allow the construction of a set that does. That contradiction arises from the assumption that $y$ is a set, so we conclude that $y$ is not a set.
By the way, you’ve got the quantifications for Regularity wrong. It should be $$\forall A(A\neq\varnothing\implies (\exists x(x\in A\wedge \forall y(y\in x\implies y\notin A)))).$$
Added.
Finally: the Axiom of Regularity is not what prevents Russell’s Paradox! In fact, there are set theories that specifically deny the Axiom of Regularity, and are as fine as ZFC is. Look up Aczel’s Anti-Regularity Axiom or “Quine atom”.
What allows for Russell’s Paradox is unbounded comprehension: that is, arbitrary set formation. That given any property $\varphi$, the collection of all $x$ with $\varphi(x)$ true is to be a set. Setting $\varphi(x)$ to be “$x\notin x$” is what yields Russell’s Paradox. But ZFC does not allow unbounded comprehension, it only allows separation. Other theories, such as Bernays-Goedel-von Neumann, allow arbitrary class formation, but not arbitrary set formation. The collection of all sets that are not elements of themselves turns out not be a set, but only a proper class, so it does not lead to a contradiction.