# Can any Neutrix be written as a countable intersection or union of internal sets? (IST)

I need help understanding a counter-example in non-standard analysis. First a few preliminaries:

Def.: A Neutrix is a convex additive subgroup of the hyperreals ${}^*\mathbb R$.

The two simplest examples of non-trivial neutrices are the set of infinitesimals $\mathbb I$ and the limited numbers $\mathbb L$. Moreover if $N$ is a neutrix, then $\mathbb L N = N$ and $C N$ is also a neutrix for any convex set $C$. Note that besides $\{0\}$ and ${}^*\mathbb R$ itself, neutrices are always external sets.

Question: Given a neutrix $N$, does there always exist a sequence $a_n$ of hyperreals such that $N$ is equal to the countable union or intersection

$$N = \bigcup_{n\in\mathbb N} (-a_n,+a_n) \qquad\textbf{or}\qquad N = \bigcap_{n\in\mathbb N} (-a_n,+a_n)$$

For example $\mathbb L = \bigcup_{n\in\mathbb N} (-n,+n)$ and $\mathbb I = \bigcap_{n\in\mathbb N} (-\frac{1}{n},+\frac{1}{n})$.

In his book Nonstandard Asymptotic Analysis, Imme van den Berg provides a counter-example that I don't understand; my biggest problem is that he works with Nelson's Internal Set Theory (IST) and I don't know how to properly translate it into Robinson's framework.

Let $w\in\mathbb R^+$ be unlimited and let $F$ be the set of all increasing real functions. Then $$G_w = \{x\in\mathbb R | \exists f\in F^\text{st} : 0\le x\le f(w)\}$$ where $F^\text{st} = \{f\in F| f\text{ standard}\}$, is a strictly generalized galaxy. I.e. $G_w$ is external set of the form $G_w = \bigcup_{x\in X^\text{st}} A_x$ for a internal family $\{A_x\}_{x\in X}$ of internal sets, where $X$ must be uncountable (= strict).

Proof: (1): $G_w$ is external because on one side it is obviously a convex additive semigroup of $\mathbb R^+$, but then again $G_w\neq \mathbb R^+$ since by the idealization principle applied on the formula $$B(f,y) \iff f\in F \wedge f(w)<y$$ it follows that there exists $y\in\mathbb R^+$ with $y>G_w$.

(2): $G_w$ is strict: Assume this would not be the case. Then, w.l.o.g., there exists a sequence of increasing standard functions $(f_n)_{n\in\mathbb N^\text{st}}$ such that $f_{n+1}$ dominates $f_n$ with $$G_w = \bigcup_{n\in\mathbb N^\text{st}} [0,f_n(w)]$$ Then by the lemma of du Bois-Reymond there exists an increasing standard function $g$ that dominates all $f_n$ for large inputs, i.e. $\forall n \exists a_n \forall x\ge a_n : g(x)>f_n(x)$. But then both $g(w)\in G_w$ and $g(w)>G_w$, contradiction!

So as far as I understand it - and correct me if I'm wrong - the set $F$ corresponds to the set of internal increasing functions ${}^*\mathbb R \to {}^*\mathbb R$ and a standard function $f\colon\mathbb R\to\mathbb R$ in IST corresponds to a hyperreal function ${}^*f\colon {}^*\mathbb R \to {}^*\mathbb R$ where ${}^*f$ is the canonical extension of a real function $f\colon\mathbb R\to\mathbb R$ to the hyperreals, hence $$G_w = \{x\in {}^*\mathbb R | \exists f\colon \mathbb R \to \mathbb R \text{ increasing} : 0\le x\le {}^*f(w)\}$$

Now the idealization principle says that if for any finite collection $\{f_1,\ldots,f_k\}\subset F^\text{st}$ there exists $y$ such that $B(f_1,y),\ldots,B(f_k,y)$ is true, then there exists $\hat y$ such that $B(f,\hat y)$ is true for all $f\in F^\text{st}$.

However on the other side it is trivial to construct, for any given $y\in {}^*\mathbb R^+$, an increasing function $f\colon\mathbb R\to\mathbb R$ such that ${}^*f(w)>y$: if $y = [y_k]$ and $w =[(1,2,3,\ldots)]$, simply pick $f$ such that $f(k) > y_k$ for all $k$. So $G_w = {}^*\mathbb R^+$

What am I missing?