Question
Given an arbitrary function $f:X → Y$ , consider the new function $f^∗: \mathcal PY → \mathcal PX$ which is defined by the mapping $B → f^{−1}[B]$. Compute $f^{∗∗} : \mathcal{PP}∅ → \mathcal{PPP}∅$ for the unique $f : ∅ → P∅$.
Attempt
We can first make the domains and codomains more explicit. We know the following:
$$ \mathcal P \emptyset = \text{{$\emptyset$}}$$ $$ \mathcal{PP} \emptyset = \text{ { $\emptyset$,{$\emptyset$}}}$$
$$ \mathcal{PPP} \emptyset = \text{ { $\emptyset$ , \{$\emptyset$\}, \{\{$\emptyset$\}\}, \{$\emptyset$,\{$\emptyset$\}\} }} $$
Therefore, we only need to find out $f^{∗∗} ( \emptyset )$ and $f^{∗∗} ( \text{{$\emptyset$}})$ and then we are done.
However, I am struggling to find out what these two values map to. I think what is throwing me off is the fact that the domain of $f$ is the empty set. This is making it a bit difficult for me to intuit exactly how we are going to construct $f^∗$ and $f^{∗∗}$.
I would be grateful for any assistance in solving this problem.
Notation Clarification: I am using $\mathcal P X$ to denote the Power Set of the set $X$
This is an exercise of careful bookkeeping and unwinding of definitions. You correctly wrote down $\mathcal P\emptyset$ and $\mathcal{PP}\emptyset$, so I will get you started on $f^* : \mathcal{PP}\emptyset \to \mathcal P\emptyset$. For this, we need $f^*(\emptyset)$ and $f^*(\{\emptyset\})$. The latter is $f^{-1}[\{\emptyset\}]$, so we need to figure out which elements of the domain of $f$ get mapped to $\emptyset$ (which is the only element of $\{\emptyset\}$). However, the domain of $f$ is $\emptyset$, which has no elements. So there are no elements that even could be mapped to $\emptyset$. Writing the collection of all these (zero) pre-images as a set, we get $\emptyset$. So $f^*(\{\emptyset\}) = \emptyset$.
You should take it from here, first calculating $f^*(\emptyset)$ and then calculating $f^{**}$.