I don't get the argumentation of Cantor's theorem proof. I must have gone wrong or misunderstood something somewhere so I will try to explain my reasoning below.
Theorem (Cantor) Let $f$ be a map from set $A$ to its power set $\mathcal{P}(A)$. Then $f : A \to \mathcal{P}(A)$ is not surjective. As a consequence, $\operatorname{card}(A) < \operatorname{card}(\mathcal{P}(A))$ holds for any set $A$
The theorem is straightforward enough, we need to show that $f$ is not surjective, meaning that there is some element $y \in \mathcal{P}(A)$ such that there is no $x \in A$ such that $y = f(x)$. The proof is where I run into trouble.
Proof Consider the set $ B=\{x \in A \mid x \notin f(x)\}$. Suppose to the contrary that $f$ is surjective. Then there exists $\xi\in A$ such that $f(\xi)=B$. But by construction, $\xi \in B \iff \xi \notin f(\xi)= B $. This is a contradiction. Thus, $f$ cannot be surjective. On the other hand, $g : A \to \mathcal{P}(A)$ defined by $x \mapsto \{x\}$ is an injective map. Consequently, we must have $\operatorname{card}(A) < \operatorname{card}(\mathcal{P}(A))$. $\tag*{$\blacksquare$}$
I do not understand the third sentence in the proof. It says that "if $f$ is surjective, then there exists some $\xi \in A$ such that $f(\xi) = B$".
I do not see why such a $\xi$ must exist if $f$ is surjective. because consider the injective/surjective mapping of $f: A \mapsto C$ where $f$ is the identity function and $A$ and $C$ both contain the same single element. There exists no $\xi \in A$ such that $f(\xi)$ = B and $f$ is surjective. Have I misunderstood something? How can I rectify my above statements with the logic of the proof?
