Ultrapower Construction of Hyperreal System 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


*

*$a_n=n\ (\text{mod}\ 3)$

*$a_n=\sin(n)$

*Let $U^*=\bigcap_{U\in\mathcal{U}}U\neq\emptyset$, construct a sequence as follows


  
*
  
*Set $i=1, n=0$;
  
*If $n\in U^*$, then $a_n=(-1)^i$ and set $i=i+1$, else $a_n=0$;
  
*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


*

*From ultrapower construction of hyperreal, if it is not infinite, why is it infinitesimally close to some real? 

*Is there a way to determine the standard part of a hyperreal?

 A: *

*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 $\mathbf{k}=\langle k,k,k,\ldots\rangle$.

*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.

*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.
A: 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 $x$.
