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.
Also, this whole internal set business has a tad eluded me. 
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?
 A: 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 $B_n$.
A: 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. 
