Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I assume $u_n=x_n$ in your fourth paragraph? – Andrés Caicedo Dec 6 '12 at 20:36
Likewise, $B_n=A_n$. – Andrés 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 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
up vote 1 down vote accepted

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 $B_n$.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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