The set $\{x| x\in \mathbb N\land x<1000\}$ is finite and it can easily be presented in the set builder notation...
All integers greater than $0$ but less than $6$:
$$\{n| n\in\mathbb N\land n>0 \land n<6\}$$
All positive even integers less than $11$:
$$\{n| n\in\mathbb N\land n<11\land (\exists k\in\mathbb N: n=2k)\}$$
or, alternatively,
$$\{2k| k\in\mathbb N\land 2k<11\}$$
All perfect squares less than $101$:
$$\{n| n\in\mathbb N\land n<101\land (\exists k\in\mathbb N: n=k^2)\}$$
or, again, alternatively as
$$\{k^2| k\in\mathbb N\land k^2<101\}$$
"all elements of $\mathcal P(A)$ are sets".
Yes; but they are the subsets of $A$. If $A$ has $n$ elements, then $\mathcal P(A)$ has $2^n$ elements, "starting" from $\emptyset$ up to $A$ itself.
Consider the simple example with $n=3$, i.e. $A= \{ 1,2,3 \}$.
We have that $\mathcal P(A)$ has $2^3=8$ elements:
$\mathcal P(A) = \{ \emptyset,\{ 1 \}, \{ 2 \}, \{ 3 \}, \{ 1,2 \}, \{ 1,3 \}, \{ 2,3 \}, \{ 1,2,3 \} \}$.
Thus, for $a=1$ we have that:
$\{ (1,S): S ∈ \mathcal P(A) \} = \{ (1,\emptyset),(1,\{ 1 \}),(1,\{ 2 \}),(1,\{ 3 \}),(1,\{ 1,2 \}),(1,\{ 1,3 \}),(1, \{ 2,3 \}),(1,\{ 1,2,3 \}) \}$
i.e. again $8$ elements.
The same for $a=2$ and $a=3$. Conclusion: $24=3 \times 8= n \times 2^n$.
But the definition is slightly more complicated:
$\{ (a,S) : a∈S, S∈ \mathcal P(A) \}$.
We have to pick-up pairs $(a,S)$ such that $a \in S$: this excludes of course the possibility $S=\emptyset$. But other cases must be excluded as well.
As we can verify in the simplified example, with $a=1$ we have only four pairs with $1 \in S, S \in \mathcal P(A)$.
And we have $4=2^2=2^{n-1}$ for our example with $n=3$.
We have again $3$ possible choices for $a$, and thus - in this case - we have $n \times 2^{n−1}$ pairs.
Some more calculations ...
Why in general, for fixed $a$, we have $2^{n-1}$ elements $S \in \mathcal P(A)$ such that $a \in S$ ?
With $n$ elements we have that $\mathcal P(A)$ has $2^n$ elements.
If we remove $a$, the set left $A'=A \setminus \{ a \}$ has $n-1$ elements; thus $\mathcal P(A')$ has $2^{n-1}$ elements.
The subsets of $A$ are all the subsets of $A'$ "plus" the subsets obtained from the previous ones "adding" $a$.
This is shown by the fact that: $2^n=2 \times 2^{n-1}=2^{n-1}+2^{n-1}$.
In conclusion, the subsets of $A$, i.e. the elements of $\mathcal P(A)$ are $2^{n-1}$ subsets with $a$ inside plus $2^{n-1}$ subsets without $a$ [see again the simplified example].
This gives us the fact that, for fixed $a$, the number of pairs $(a,S)$ such that $a \in S$ are $2^{n-1}$.
Best Answer
That is okay, although the use of words should be discouraged, and it can be compacted a bit more. Any of the following should be acceptable: $${\{(i,j)\mid i\in S, j\in S, i=2j\}\\\{(2j,j)\mid j\in S, 2j\in S\}\\\{(i,j) \in S^2 \mid i= 2 j \} \\\{(2j,j)\in S^2\}}$$
Sometimes there is a trade off between compactness and comprehensibilty. Choose the version that you feel most clearly conveys the intended message.