Set Theory and surjective function exercise Let $E$ a set and $f:E\rightarrow P(E)$ any function. If $$A=\{a\in E:a\notin f(a)\}$$ Prove that $A$ has no preimage under $f$.

Suppose $\exists a\in E$, $f(a)=A$. Then, is $a\in f(a)$?
If $a\in f(a)$, that means $a\notin A$, contradicts to $f\in f(a)=A$.
If $a\notin f(a)$, that means $a\in A$.
Thus, such $a$ does not exists i.e. $f(a)$ has no preimage.
