Let $f: \mathcal{P}(A) \to \mathcal{U}$ be the function given by $f(x) = \mathcal{P}(x)$ for every $x \in \mathcal{P}(A)$. Here, $\mathcal{U}$ is the universe, and $A$ might be a set or it might be a proper class (we can assume it is non-empty).
Note: In the case that $A$ is a proper class, $P(A)$ is the usual definition, except $A \not \in P(A)$. That is, when $A$ is a proper class, $P(A)$ is the set of all proper subsets of $A$.
It is not difficult to show that $\text{range}(f) \subseteq \mathcal{P(P}(A))$. However, I need to show that there exists some non-empty $z \in \mathcal{P(P}(A)) \setminus \text{range}(f)$, i.e. $\text{range}(f) \subsetneq \mathcal{P(P}(A))$. The formal definition of the range of a function f is: $\text{range}(f) = \{f(x) : x \in \text{domain}(f)\}$.
I am struggling to prove this (and I have yet to even convince myself that it is true as it seems quite counterintuitive to me). Any advice to point me in the correct direction is appreciated.
Best Answer
I'm going to use 'set' throughout this answer, but substitute 'class' at your leisure.
Think about what $\mathrm{range}(f)$ and $\mathcal{P}(\mathcal{P}(A))$ mean:
So suppose $A \ne \varnothing$ and fix some $a \in A$. Then $\{ a \} \in \mathcal{P}(A)$, so $\{ \{ a \} \} \in \mathcal{P}(\mathcal{P}(A))$, however $\{ \{ a \} \} \ne \mathcal{P}(x)$ for any $x \in \mathcal{P}(A)$ since $\varnothing \not\in \{ \{ a \} \}$.
Hence $\{ \{ a \} \} \in \mathcal{P}(\mathcal{P}(A)) \setminus \mathrm{range}(f)$.