# Question about the definition of superstructure

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.

## 1 Answer

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.