See power set.

"There exists no surjection from $A$ onto $\mathcal{P}(A)$."

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.

Proof

Let $B=\{x\in A:x\notin f(x)\}$
$B\in \mathcal{P}(A)$ because $B\subseteq A$.
If $f$ is surjective, then there exists some $\xi \in A$ such that $f(\xi )=B$.
By the definition of $B$, $\xi \in B⟺\xi \notin f(\xi )$.
This is a contradiction, since $f(\xi )=B$ if $f$ is surjective.
Therefore, $f$ is not surjective.

QED