What role does $\mathfrak D$ play in the definition of the union and intersection of the complements of a collection of sets? I'm puzzled by something that might be complete silly. Halmos writes in his "Naive Set Theory":

If $\mathcal C$ is a collection of subsets of a set $E$ (that is, $\mathcal C$ is a subcollection of $\mathcal P(E)$), then write 
$$\mathfrak D = \{X \in \mathcal P(E):X^\prime \in \mathcal C\}$$
It is customary to denote the union and intersection of $\mathfrak D$ by the symbols:
$$\bigcup_{X \in \mathcal C}X' \;\;\text{  and  
 } \;\;\bigcap_{X \in \mathcal C}X'$$ 

He is defining the union and intersection of the complements of the sets $X \in \mathcal C$, which are subsets of the set $E$. $E$ is the "everything" set, which contains all sets considered in the current section. 
I'm a little confused by why he is considering the set $\mathfrak D$. Why not just consdier the set of complements? $\mathfrak D$ is apparently, as I understand, the set of all subsets of $E$ such that the complement of $X$, namely $X'$ is in the collection of subsets of $E$, $\mathcal C$. By definition of the powerset of $E$, 
$$\mathcal C \subset \mathcal P(E)$$
so $\mathfrak D$ is the set of subsets of $E$, $X$ such that $X'$ is also a subset of $E$.
I still can't see how $\mathfrak D$ enters in the picture here. 
 A: $\mathfrak{D}=\{X\in\mathcal{P}(E):X'\in\mathcal{C}\}$ is the collection of subsets of $E$ whose complements are in $\mathcal{C}$. For instance, if $\mathcal{C}$ is the collection of all finite subsets of $E$, then $\mathfrak{D}$ is the collection of all cofinite subsets of $E$, i.e., those subsets of $E$ whose complements are finite.
Note that for subsets $X$ of $E$ we have $X\in\mathfrak{D}$ iff $X'\in\mathcal{C}$. Take complements: this is equivalent to saying that $X'\in\mathfrak{D}$ iff $X''\in\mathcal{C}$. But $X''=X$, so $X'\in\mathfrak{D}$ iff $X\in\mathcal{C}$, and $\mathfrak{D}=\{C':C\in\mathcal{C}\}$. Thus, we can just as well describe $\mathfrak{D}$ as the set of complements (relative to $E$) of members of $\mathcal{C}$. And this is what justifies writing 
$$\bigcup\mathfrak{D}=\bigcup_{C\in\mathcal{C}}C'\quad\text{ and }\quad\bigcap\mathfrak{D}=\bigcap_{C\in\mathcal{C}}C'\;.\tag{1}$$
If it weren’t the case that $\mathfrak{D}=\{C':C\in\mathcal{C}\}$, $(1)$ would not be justifiable.
