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Note: I do not know for sure that it's wrong, but have a strong suspicion, as the authors implicitly mentioned it but in the end chose another (more elaborate) proof.

Let $\hat S$ be a superstructure.
Let $^*$ denote the transfer of an element of $\hat{S}$ via the transfer principle.
Let $\varphi[x_1,...,x_n]$ be a statement in the variables $x_1,...,x_n$ and $A$ a set, and $A, \varphi[x_1,...,x_n]\in\hat S$.

To show:
$^*\{(a_1,...,a_n)\in A : \varphi[a_1,...,a_n] \text{ is valid} \} = \{(b_1,...,b_n)\in {^*A} : {^*\varphi[b_1,...,b_n] }\text{ is valid} \} $

The proof:
Let $A_0 := \{(a_1,...,a_n)\in A : \varphi[a_1,...,a_n] $ is valid $\}$.
Since $A_0 \subset A$ it follows that $A_0\in \hat S$. Therefore the statement
$\forall x_1,...,x_n \in A: \bigg((x_1,...x_n)\in A_0 \Leftrightarrow \varphi[x_1,...,x_n] \bigg)$
is a valid formula of $\hat S$. By transferring it, we get: $\forall x_1,...,x_n \in {^*A}: \bigg((x_1,...x_n)\in {^*A_0} \Leftrightarrow {^*\varphi[x_1,...,x_n] } \bigg)$

And since $^*A_0 \subset {^*A}$ (follows from $A_0 \subset A$ ), the transferred formula tells us which elements are in $^* A_0$ :

$$ \{(b_1,...,b_n)\in {^*A} : {^*\varphi[a_1,...,a_n] }\text{ is valid} \} = {^*A_0} = {^*\{(a_1,...,a_n)\in A : \varphi[a_1,...,a_n] \text{ is valid} \}} $$

The book it's from is "Nichtstandardanalysis by Landers, Rogge" (a German book).

The transfer principle so far is (let $\hat S$ and $\hat W$ be two superstructures): $$ \begin{align} &(1) & {^* \hat S }&= \hat{W} \\ &(2) & {^*s} &= s, \text{ for } s\in S \\ &(3) & \varphi\in\hat S \text{ valid} &\Leftrightarrow {^*\varphi}\in\hat W \text{ valid} \end{align} $$

(where in $(3)$, $\varphi$ is a propositional formula)

Sudix
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  • 1
    I corrected a typo. There are some additional (similar) typos to correct. Otherwise the proof is OK. Which book are you referring to? Perhaps they are only working with a restricted form of the transfer principle. – Mikhail Katz Apr 24 '18 at 09:37
  • In both the statement you transferred and the result after you tranferred it, all the $a_i$'s should be $x_i$'s. Otherwise, the proof looks good. – Andreas Blass Apr 24 '18 at 11:51

1 Answers1

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I found the mistake:

My proof assumes that for every tuple $(a_1,...,a_n) \in A$, the elements in the tuple $a_1, ..., a_n$ are in $A$ as well.

Hmm... yes, you're right. If we go by the definition $\hat S = \bigcup_{v\in \mathbb{N}} S_v$, we get $A\in \hat S \Rightarrow A\in S_v \text{ for some }v\in\mathbb{N}$, therefore $A\subset S_v$, and hence $^*A \subset {^* S_v}$.
So if we correct the statement
$\forall x_1,...,x_n \in A: \bigg((x_1,...x_n)\in A_0 \Leftrightarrow \varphi[x_1,...,x_n] \bigg)$ into $\forall x_1,...,x_n \in S_v: \bigg((x_1,...x_n)\in A_0 \Leftrightarrow \varphi[x_1,...,x_n] \bigg)$, my argument is void.

Sudix
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