1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.