The domain for the function $A, B ⊆ X$, $A f B \iff A ∪ B=X$ and $A \cap B = ∅$

I have this function $f$:
$$A, B ⊆ X,\ A f B \iff A \cup B = X\quad\text{ and }\quad A \cap B = \emptyset$$

My solution: Let sets $A, B \subseteq 2^X$
${\rm Domain} = \{ A \cup B : A \cap B = \emptyset\}$. Is this true? I feel like this might work but on the other side it does not define a precise set. It is like on-demand set, if I can say!

Any help on what would be improved on my solution or what would be the domain for the function $f$ is appreciated!

Let $\mathcal P(X)$ denote the power set of $X$.

Let $A\in\mathcal P(X)$. Then $A$ is in the domain of $f$ iff there exists $B\in\mathcal P(X)$ such that $A\cup B=X $ and $A\cap B=\emptyset$. Set $B=X\setminus A$. Then $A\cup B=X$ and $A\cap B=\emptyset$. Hence, $A$ is in the domain of $f$. Since $A$ was arbitraty, the domain of $f$ is $\mathcal P(X)$.