# Infinite union of internal sets not internal

This is homework problem. I need to give an example of internal sets $A_n \subset \mathbb{R}^*$ for which the union $\bigcup _{n=i}^\infty A_n$ is not internal.

Definition from my lecture material:

Let $B_n \subset \mathbb R, \ \forall n \in \mathbb N$. Then we say that $\Psi[B_n]$ is the set of equivalence classes $[u_n] \in \mathbb R^*$, for which $u_n \in B_n$ for $\mathscr U$-almost-all $n$, given the ultrafilter $\mathscr U$. A set of the form $\Psi[B_n] \subset \mathbb R^*$ is called internal. If for all $n,\ B_n = B$, we call $B$ as non-standard extension of $B$ and use symbol $B^*$ for it.

Any help? Can you explain the internal set any more intuitively, cause it is clearly a very important concept?

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I assume $u_n=x_n$ in your fourth paragraph? – Andrés E. Caicedo Dec 6 '12 at 20:36
Likewise, $B_n=A_n$. – Andrés E. Caicedo Dec 6 '12 at 20:58
Oh dear, I seem to have tried too hard not to use the same notation as I used in the first paragraph and lost concentration. I will fix that too. – Valtteri Dec 6 '12 at 21:01
Anyway, the usual example is to consider $\bigcup_n\{n\}^*$. You ask for intuition on the concept of internal, but I think that intuition is built through problems like this one. The idea is that the non-standard universe is just an "inflated" version of the standard one. You have an elementary embedding relating the two. The embedding is rather "discontinuous". For example, $\mathbb N^*$ should be much larger than simply taking $\{ n^*\mid n\in\mathbb N\}$. How comfortable you are with model theory may help understand this more easily. – Andrés E. Caicedo Dec 6 '12 at 21:03
@AndresCaicedo OK, I see that ${n}^*$ is internal, because $\Psi[{n}] = [n]^*$. But why is the infinite union not internal? Is it because we have stuff like ${1,2,3,4,5,6...}$ where every member is a member of some ${n}^*$, but the whole is not? – Valtteri Dec 6 '12 at 21:13

For $n\in\Bbb N$ let $A_n=\left\{[x^n]\right\}$, where $x^n\in\Bbb R^\omega$ such that $x_k^n=n$ for each $k\in\omega$, and let $A=\bigcup_{n\in\Bbb N}A_n$. Suppose that $B_n\subseteq\Bbb R$ for $n\in\omega$ are such that $A\subseteq\Phi[B_n]$; I’ll show that $A\subsetneqq\Phi[B_n]$.

For $k\in\omega$ let

$$m_k=\begin{cases} \max B_k,&\text{if }B_k\text{ is finite}\\\\ \min\big\{n\in B_k:\forall i<k(n>m_i)\big\},&\text{otherwise}\;, \end{cases}$$

and let $m=\langle m_k:k\in\omega\rangle\in\Bbb N^\omega$. Clearly $[m]\in\Phi[B_n]$. Let $F=\{k\in\omega:B_k\text{ is finite}\}$; clearly $m\upharpoonright(\omega\setminus F)$ is strictly increasing, so if $\omega\setminus F\in\mathscr{U}$, $[m]\notin A$. Suppose, then, that $F\in\mathscr{U}$ and that $[m]\in A$. Then there is some $\ell\in\Bbb N$ such that $\{k\in F:m_k=\ell\}\in\mathscr{U}$, and it follows at once that $[x^{\ell+1}]\in A\setminus\Phi[B_n]$, contradicting the choice of the sets $B_n$.

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OK, first I want to ask what $\omega$ is? It is used as a superscript for real numbers, but also as a set? Then you build the $B_n$ that are subsets of reals and assume that $A$ is subset of the internal set of these $B_n$. Then you go on to show that $A$ is strictly a subset of the int. set $B_n$. Then you construct a hypernatural m. It is in the int. set $B_n$ because for almost all indexes $m_k \in B_n$. If that set omega minus $F$ is in ultrafilter, $m$ is not, for not almost all indexes match. Select $F$ is in filter. Then there is number $l$ with that property, why? – Valtteri Dec 7 '12 at 0:29
@Valtteri: Sorry; I automatically assumed that anyone working with ultrafilters would be familiar with the ordinal $\omega$. You can replace it everywhere by $\Bbb N$ without changing the meaning. (I used both because I wanted to distinguish indices from elements of $\Bbb R$, but that was just an æsthetic preference.) If $F\in\mathscr{U}$ and $[m]\in A$, then $[m]=[x^\ell]$ for some $\ell\in\Bbb N$, so $m_k=\ell$ almost everywhere. – Brian M. Scott Dec 7 '12 at 0:34
No problem, I probably SHOULD know $\omega$... Some more notation questions: Is $[x^n]$ a vector or is the $n$ just an index? Is $[x^1] = (1,1,1,1...)$? Now $B_n$ are sets of reals. If $A$ is a union of classes of infinite vectors, then $\Psi [B_n]$ is a set of classes of infinite vectors too, yes? If $F$ is in $U$ and $[m] \in A$ then sequence $[m]$ must be same as infinite vector $[x^l]$ for some l, correct? Finally $[x^{l+1}]$ is not in $\Psi B_n$ for the elements are too big. $\max$ and all. So..why A is not internal? Is it because it is a strict subset of an internal set? – Valtteri Dec 7 '12 at 0:57
@Valtteri: In $x^n$ the $n$ is just an index; $x^1=\langle 1,1,1,\dots\rangle$, and $[x^1]$ is the corresponding equivalence class mod $\mathscr{U}$. The argument shows that $A$ is not equal to any $\Phi[B_n]$, which by definition means that it’s not internal. – Brian M. Scott Dec 7 '12 at 1:03

OK, thanks to Andres Caicedo and his mention of overspill lemma. We got that today on lecture and I gave it a bit more thought and I got it.

The set $\{ n \}^*$ is internal for all $n \in \Bbb N$. The infinite union of these is $\{ n^* \ | \ n \in \mathbb N \} = \ ^{\sigma} \Bbb N \subset \Bbb N^*$.

Now consider the true sentence $\forall \ n \in \ ^{\sigma} \Bbb N, \ n \in \ ^{\sigma} \Bbb N$. This is true for arbitrarily large finite hypernaturals but it is not true for any infinite natural number. Thus, by the overspill lemma, $^{\sigma} \Bbb N$ cannot be internal.

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