This argument is maybe less confusing than the one I gave in class today.

Suppose $f: A \to P(A)$ is a mapping from $latex A$ to its power set. We want to show that $latex f$ cannot be onto, i.e. there exists a subset $latex B$ of $latex A$ that is not in the range of $latex f$.

We define $B=\{a \in A : a \not\in f(a) \}$. Suppose now there exists $b$ in $A$ with $f(b) = B$. Since $B$ is a subset of $A$, either $b$ is in $B$ or not.

If $b$ is in $B$, then, by definition of $B$, $b$ is not in $f(b) = B$.

If $b$ is not in $latex B$, then, since $latex B = f(b)$, $latex b$ is not in $latex f(b)$ and hence, since $latex f(b) = B$, $latex b$ is in $latex B$.

In both cases we obtain a contradiction, hence such $latex b$ cannot exist, and $latex f$ is not onto.