Explanation of a "sentence" of Halmos's Naive Set Theory While reading Halmos's Naive Set Theory I found that he has remarked in a place,

If $\mathcal{C}$ be a collection of subsets of a set $E$ (that is, $\mathcal{C}$ is a subcollection of $\mathcal{P}(E)$  [the power set of $E$]), then we write, $$\mathcal{D}=\{X\in\mathcal{P}(E):X'\in \mathcal{C}\}$$(To be certain that the definition of $\mathcal{D}$ is a sentence in the precise technical sense, it must be rewritten in something like the form,$$\exists Y[Y\in\mathcal{C}\land\forall x(x\in X\iff(x\in E\land x\not \in Y))]\tag{1}$$...

My questions is,

What exactly $(1)$ means and how? 

From the reading of earlier chapters it seems to me that what $(1)$ actually means is the "sentence"(to use Halmos's terminology) $$X'\in C\tag{2}$$ but I can't get how one can say that $(1)$ and $(2)$ are equivalent.
Any help will be appreciated.    
 A: You copied the statement wrong. In the book it says
$$
\mathcal D=\{X\in\mathcal P(E):\ X'\in\mathcal C\}.
$$
So its exactly as Henning was saying. The sentence needs to be expanded because "complement" is not in the basic language used by the axiom system he is constructing. 
And what (1) is saying is, 

there exists $Y$ such that $Y\in\mathcal C$ and $Y=X'$.

Edit: as a clarification because there was some confusion about it, in Halmos'book "Not an element of $A$" is written $x\in'A$, while the complement of $X$ is written $X'$. 
A: I agree. What $(1)$ seems to mean is that $X$ is the complement of something in $\mathcal C$, not that $X$ itself is not in $\mathcal C$.
Perhaps there's a hidden condition somewhere that $\mathcal C$ always contains exactly one of $X$ and $E\setminus X$?
A: $\mathcal D$ is the collection of the "complements" of memebers of $\mathcal C$.
We have : $\mathcal C \subseteq \mathcal P(E)$, because $\mathcal C$ is a collection of subsets of $E$.
For the definition of $\mathcal D$, we have that : $X \in \mathcal D$ iff $X \in \mathcal P(E) \land X^c \in \mathcal C$.
Halmos's statement says that :

$X \in \mathcal D$ iff $X \in \mathcal P(E) \land \exists Y [Y \in \mathcal C \land \forall x(x \in X \leftrightarrow (x \in E \land x \notin Y))]$.

We have to compare the "simple" :

$X^c \in \mathcal C$ 

with the "complex" : 

$\exists Y [Y \in \mathcal C \land \forall x(x \in X \leftrightarrow (x \in E \land x \notin Y))]$ 

to check if they are equivalent.
$X^c \in \mathcal C$ can be rewritten as : $\exists Y [Y \in \mathcal C \land X = Y^c]$
and this in turn can be "unwinded", using the def of $=$, as :

$\exists Y [Y \in \mathcal C \land \forall x (x \in X \leftrightarrow x \notin Y)]$.

Thus, "inserting" the caluse $x \in E$, we can see that it works ...
