1
$\begingroup$

On Page 181, Logic, Induction and Sets, Forster(2003),

Show that the collection $\{x: \lnot\exists y( x \in y \in x)\}$ can not be a set.

1 Answers 1

2

Suppose $A=\{x: \lnot\exists y( x \in y \in x)\}$ is a set. Let $B\in{P(A)}$. Suppose, to get a contradiction, that $B\notin{A}$, then there exists $y$ such that $y\in{B}$ and $B\in{y}$, hence $y\in{A}$ but $y\in{B}$ and $B\in{y}$, which implies $y\notin{A}$. Contradiction. Thus we must have that $P(A)\subseteq{A}$, but this contradicts the fact, provable in naive set theory, that $|A|<|P(A)|$. So $A$ is not a set