TITLE: Cantor's Theorem proof seems a bit too convenient
QUESTION [5 upvotes]: I am taking my first analysis class, and I am really enjoying it. I have recently stumbled upon Cantor's Theorem which states that there exists no surjective map $f$ of a set $A$ to its powerset, $P(A)$. The proof is pretty straight forward, that is not the issue, but I was wondering if anyone could speak to the existance of the Cantor Diagonal Set, $B = \{x\in A:x\notin f(x)\}$. It seems a little too convenient to define a set where a mapping is restricted when one is proving that there is a set which cannot be mapped to. 
I am looking for a little intuition. Thanks. 
EDIT: I would like to thank those who provided answers/comments to this question. The Cantor Diagonal Set is now much more clear to me. I withdrawal my statement that the proof to Cantor's Theorem is "convenient." Thanks so much!

REPLY [5 votes]: Better to consider it as $B_f$ - we take a function $f:A\to P(A)$ and define $B$ in terms of it.
In particular, we are stating exactly when $x\in B_f$. That is:
$$x\in B_f\iff x\in A \text{ and } x\notin f(x)$$
That entirely determines $B_f$, given an $f$.
And what the argument shows is that no matter what $f$ you give men, $B_f$ is not in the range of $f$.
So no $f$ is onto.