I want to prove "There is no set to which every function belongs." Can I approach it as follows?
Attempt No. 1: Let $F: A\rightarrow B$ Since $F\subset A\times B,$ it follows that $F\in \mathcal{P}(A\times B).$ Now, let $\mathcal{P}(A\times B)$ be the set of all functions from A into B. Since $\mathcal{P}(A\times B)\subseteq \mathcal{PP}(A\times B),$ it follows that $\mathcal{PP}(A\times B)$ is also a set of all functions from A into B. Therefore $\mathcal{P}(A\times B)=\mathcal{PP}(A\times B).$
Attempt No. 2 is to approach it by the concept of Russel Paradox. Then I must build a set of all functions that are not in itself, but, honestly, I have not a clue of what constitute a function that is not in itself.