I'll state the Cantor's theorem proof as is it is in my study texts:
Theorem (Cantor): Let $X$ be any set. Then $|X|<|\mathcal{P}(X)|$
Proof: Define map $\varphi:X\rightarrow\mathcal{P}(X)$ by $\varphi:x\mapsto\{x\}$. $\varphi$ is injective, thus $|X|\leq|\mathcal{P}(X)|$. Now suppose there is a bijection (surjection) $\psi:X\rightarrow\mathcal{P}(X)$. Denote the set $A=\{x\in X, x\notin \psi(x)\}$. By assumption, $\psi$ is surjection so we find some $a\in X$ such that $\psi(a)=A$. Then we have two cases, either $a\in A$, but the, by definition of $A$: $a\notin \psi(a)$ which is a contradiction. So $a\notin A$ but then $a\in \psi(a)=A$ which is a contradiction aswell thus such surjection cannot exist.
So now, for my question. The definition of $A$ seems really weird to me. Because, in the first place. Assume $X=\{1,2\}$ then $\mathcal{P}(X)=\{\{1,2\},\{1\},\{2\},\emptyset\}$. In the proof $A$ is supposed to be the set of all members of $X$ (thus numbers) that are not in the range of $\psi$ but, the range of $\psi$ are sets, aren't they? Thus $A=\{1,2\}$ and the case is $\forall a\in X:a\in{A}$ thus the second case of the proof applies. Basically this leads me to the following idea: instead of constructing this set $A$, we can say that:
Estabilish injection and thus $|X|\leq|\mathcal{P}(X)|$ by $\varphi:x\mapsto\{x\}$ thus $\forall{S}\in\varphi(x):|S|=1$, but for any set $X:$ $\emptyset\subset\mathcal{P}(X)$ but $|\emptyset|=0$ so $\emptyset\notin ran(\varphi)$ thus $\varphi$ is not surjective.
Best Answer
No, this is not quite true. The construction is more subtle than that. For any $x \in X$, the value $\psi(x) \in \mathcal P(X)$ is some set -- a subset of $X$ to be precise. Note that this is not the range of $\psi$, it is the value of $\psi(x)$ for a single $x$.
Then we have either $x \in \psi(x)$ or $x \notin \psi(x)$, and we add $x$ to the set $A$ if and only if $x \notin \psi(x)$.