Let $S_n=S_0\cup \mathcal{P}(S_{n-1})$ and $V_n=V_{n-1}\cup \mathcal{P}(V_{n-1})$ with $S_0=V_0$. Defining $\hat{S}:=\bigcup_{n\in\mathbb{N}}S_n$ and $\hat{V}:=\bigcup_{n\in\mathbb{N}}V_n$, can we conclude that $\hat{S}=\hat{V}$?
The only thing I could do (which is very simple) is to prove that $\hat{S}\subset\hat{V}$.
I'm asking this question because in the book "Nonstandard Analysis", written by Martin Väth, the author defines the superstructure of a set $S$ as $\hat{S}:=\bigcup _{n\in\mathbb{N}}S_n$ in which $S_n=\boxed{S_0}\cup \mathcal{P}(S_{n-1})$ with $S_0=S$. But looking at the internet I noticed that the superstructure of a set $S$ is actually the set $\hat{S}:=\bigcup _{n\in\mathbb{N}}S_n$ in which $S_n=\boxed{S_{n-1}}\cup\mathcal{P}(S_{n-1})$ with $S_0=S$.
If $\hat{S}\neq\hat{V}$, then I ask: will this new definition of superstructure bring future difficulties in nonstandard analysis?
The author of this book also states that it is possible to build a superstructure $\hat{S}$ of a set $S$ such that $s\in a$ is always false to any $a\in S$ and $s\in\hat {S}$. Is that really possible? For me this is totally counter-intuitive since if we chose $S=\mathbb{R}$ we would not be able to tell which elements the number $1\in\mathbb{R}$ contains.
Yes, in fact $S_n = V_n$ for all $n\in \mathbb{N}$. The difference between the definitions is this: it is clear from the definition that $S_0\subseteq S_n$ for all $n$, but it is not clear that $S_{n-1}\subseteq S_n$. On the other hand, it is clear from the definition that $V_{n-1}\subseteq V_n$ for all $n$, but it is not clear that $V_0\subseteq V_n$. But in fact both assertions are true about both heirarchies.
Claim: $V_0\subseteq V_n$ for all $n$.
In the base case, $V_0\subseteq V_0$. In the inductive step, we have $V_0\subseteq V_{n-1}\subseteq V_n$.
Claim: $S_{n-1}\subseteq S_n$ for all $n\geq 1$.
In the base case, we have $S_0\subseteq S_1$ by definition. In the inductive step, we assume $S_{n-2}\subseteq S_{n-1}$, and suppose $X\in S_{n-1}$. Then either $X\in S_0$, or $X\in \mathcal{P}(S_{n-2}) \subseteq \mathcal{P}(S_{n-1})$ by the inductive hypothesis. In either case, $X\in S_n$.
Having proven the claims, we can see that $$V_n = V_0\cup V_n = V_0 \cup V_{n-1} \cup \mathcal{P}(V_{n-1})$$ and $$S_n = S_{n-1}\cup S_n = S_0 \cup S_{n-1}\cup \mathcal{P}(S_{n-1})$$ from which it follows immediately that if $V_0 = S_0$, then $V_n = S_n$ for all $n\in \mathbb{N}$.
Regarding your second question, it's not clear what "build a superstructure" means, because according to your definition, any set $S$ has just one superstructure $\widehat{S}$. Probably the claim means that this is true up to isomorphism.
So one interpretation is this: For any set $S$, there is a set $S'$ of the same cardinality such that $s\notin a$ for all $a\in S'$ and $s\in \widehat{S'}$. The existence of a bijection between $S$ and $S'$ means that we can transport any structure we like between $S$ and $S'$: from the point of view of structuralist mathematics, we might as well work with $S'$. This is certainly possible, but the details are a little bit annoying; I think they've been spelled out in an answer to your other question.