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I think I understand the ultrapower construction of hyperreals. Given a free ultrafilter $\mathcal{U}$ (take $\mathcal{U}\subset\mathcal{P}(\mathbb{N})$) for simplicity, then the hyperreal system is the ultrapower of $\mathbb{R}$ associated with $\mathcal{U}$.

I also understand that either $\chi_E$ or $\chi_O$ (sequence generated by indicator function of even or odd numbers) shall be 1 depending on $\mathcal{U}$. But what about some weired sequences like

  1. $a_n=n\ (\text{mod}\ 3)$
  2. $a_n=\sin(n)$
  3. Let $U^*=\bigcap_{U\in\mathcal{U}}U\neq\emptyset$, construct a sequence as follows
    1. Set $i=1, n=0$;
    2. If $n\in U^*$, then $a_n=(-1)^i$ and set $i=i+1$, else $a_n=0$;
    3. Set $n=n+1$ and turn to step 2.

Each sequence represents a hyperreal, buy why is it always infinitesimally close to some real? More precisely, how to determine the standard part of a hyperreal?

Thanks.

Edit version 1

  1. From ultrapower construction of hyperreal, if it is not infinite, why is it infinitesimally close to some real?
  2. Is there a way to determine the standard part of a hyperreal?
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  1. Suppose that $a_n=n\bmod3$ for each $n\in\omega$. For $k=0,1,2$ let $A_k=\{n\in\omega:n\bmod 3=k\}$; then $\{A_0,A_1,A_2\}$ is a partition of $\omega$, so exactly one of the sets $A_k$ belongs to $\mathscr{U}$. If $A_k\in\mathscr{U}$, then $\langle a_n:n\in\omega\rangle_{\mathscr{U}}=\mathbf{k}_\mathscr{U}$, where $\mathbf{k}=\langle k,k,k,\ldots\rangle$.

  2. Now suppose that $a_n=\sin n$ for each $n\in\omega$. The sequence $a=\langle a_n:n\in\omega\rangle$ is bounded, so this answer shows that there is a real number $x\in[-1,1]$ such that $x=\mathscr{U}$-$\lim a$, i.e., such that $\{n\in\omega:|a_n-x|<\epsilon\}\in\mathscr{U}$ for each $n\in\omega$. It follows that $a_{\mathscr{U}}$ is infinitesimally close to $\mathbf{x}_{\mathscr{U}}$. Note that this argument works for every bounded sequence of reals.

  3. Since $\mathscr{U}$ is a free (non-principal) ultrafilter, $\bigcap\mathscr{U}=\varnothing$, so your construction is impossible.

The answer to your final question is that it’s not true that every hyperreal is infinitely close to some real: the infinite hyperreals are not. The simplest example of such a hyperreal is $\alpha=\langle n:n\in\omega\rangle_{\mathscr{U}}$. If $\mathbf{x}_{\mathscr{U}}$ is any real in ${^*\Bbb R}$, let $m$ be any integer larger than $|x|+1$, say; then $|n-x|\ge 1$ for $n\ge m$, so $\alpha\ge\mathbf{x}_{\mathscr{U}}+1$, which certainly implies that $\alpha$ is not infinitely close to $\mathbf{x}_{\mathscr{U}}$.

Added: More generally, given any sequence $a=\langle a_n:n\in\omega\rangle$ of reals, if there is a $U\in\mathscr{U}$ such that $\langle a_n:n\in U\rangle$ is bounded, then we can apply the argument of $(2)$ above to see that $a_{\mathscr{U}}$ is infinitely close to some standard real: any two sequences that agree on a member of $\mathscr{U}$ give rise to the same element of ${^*\Bbb R}$.

A hyperreal $\alpha=\langle a_n:n\in\omega\rangle_{\mathscr{U}}$ is infinite if $\mathbf{x}<|\alpha|$ for each $x\in\Bbb R$. Thus, if $\alpha$ is not infinite, there is an $x\in\Bbb R$ such that $|\alpha|\le\mathbf{x}$. Let $U=\{n\in\omega:|a_n|\le x\}$; then $U\in\mathscr{U}$ so $\langle a_n:n\in\omega\rangle$ is bounded on $U$, and $\alpha$ is therefore infinitely close to some standard real.

Conversely, suppose that $x\in\Bbb R$, and $a_{\mathscr{U}}$ is infinitely close to $\mathbf{x}$. Let $U=\{n\in\omega:|a_n-x|<1\}$; then $U\in\mathscr{U}$, and $a$ is bounded on $U$. Thus, $a_{\mathscr{U}}$ is infinitely close to some standard real if and only if the sequence $a$ is bounded on some element of $\mathscr{U}$, and infinite if and only if $a$ is unbounded on every element of $\mathscr{U}$.

In general there is no way to determine the standard part of the hyperreal defined by a given sequence of reals, because there’s no way to pin down exactly which subsets of $\omega$ belong to $\mathscr{U}$. About all that you know for sure (in general) is that $\mathscr{U}$ contains all of the cofinite sets.

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  • $\begingroup$ Oh thanks. I made a few mistakes. Could you please elaborate more about the last two questions? 1. From ultrapower construction of hyperreal, if it is not infinite, why is it infinitesimally close to some real? 2. Is there a way to determine the standard part of a hyperreal? I'd appreciate very much. $\endgroup$ – Shuchang Sep 17 '15 at 8:55
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Since the hyperreals are totally ordered, each finite hyperreal $x$ defines a Dedekind cut on the standard rationals (which are included as a subset). The real number corresponding to that cut is the standard part of $x$.

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