I need to prove that there is no surjection from $S$ to $P(S)$. I've seen a proof in some posts but I would like to know where mine fails.
My attempt: Since $f$ is surjective, $\{s\}\in f(S)$ for all singletons $\{s\}$, that is, $f(s')=\{s\}$ for some $s'$ (*). Now let $A=\{a_1,a_2\}$ be a subset of $S$. By hypothesis $A=f(a)$ for some $a$. It should happen that $f(a)=\{s\}$. Because $a$ cannot have two different images, $f$ can't be surjective.
Note: I know that $f$ must be an arbitrary function that not necessarily maps elements into singletons so "$f(a)=\{s\}$" is not justified, however I thought of it because surjectivity exhaust all possible pre-images in the (*) step, if $f(a)=A$, then there is a one element set that doesn't have pre-image.
Surjectivity does not exhaust all possible pre-images in the (*) step. Suppose that $S=\Bbb N$. Define part of $f:\Bbb N\to\wp(\Bbb N)$ by setting $f(2n)=\{n\}$ for each $n\in\Bbb N$. We’ve now taken care of all of the singletons and still have infinitely many odd $n\in\Bbb N$ to use for other subsets of $\Bbb N$.
In fact there is a bijection from $\Bbb N$ to the set of all finite subsets of $\Bbb N$, even though we need to use infinitely many $n\in\Bbb N$ to take care of the singletons, infinitely many more to take care of the doubletons, and so on.