Cantor's Theorem proof seems a bit too convenient 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!
 A: 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.
A: Sets are the mathematical attempt to formalize the idea of an object which is itself a collection of other mathematical objects.
As such we expect collections to have certain behaviors.
For example, if I have a collection of marbles, I expect that the collection of all the white marbles also to be a collection. If I have a collection of bottles of liquor, I expect the collection of all bottles of whisky to be a collection as well (not to be confused with the collection of all the bottles of whiskey!).
This is formalized in the idea that if $A$ is a set, and $\varphi$ is a first-order property in the language of set theory, then all those elements of $A$ which satisfy the property $\varphi$ also forms a set.
Since properties might refer to other existing objects, this might allow parameters. And so if $f$ is a function we can write the property $\varphi(x)$ to be $x\notin f(x)$. So if $A$ is a set, then $\{a\in A\mid a\notin f(a)\}$ should also be a set.
This is known as comprehension, or rather bounded comprehension (also known as separation and subset). And this is the crux that allows us to ensure that $B$ is a set.
There are set theories like Quine's New Foundations, in which not every property defines a set. In that set theory there is, in fact, a universal set, and Cantor's theorem is not quite provable in its easy formulation because the restrictions on which properties define sets disallow the diagonal set to exist.
