Let $^\ast\mathbb{R}$ be the set of hyperreals, constructed as a non-principal ultraproduct over the reals, and let $x\in {^\ast\mathbb{R}}$ and $y\in {^\ast\mathbb{R}}$ be two different hyperreal numbers. Is there always a real set $A\subset\mathbb{R}$ such that $x \in {^\ast A}$ but $y\notin {^\ast A}$?
This problem can be stated without mentioning hyperreals: Let $\mathcal U$ be a (non-principal) ultrafilter on $\mathbb N$ and let $(x_n)$ and $(y_n)$ be real sequences such that $\{n: x_n \neq y_n\} \in \mathcal U$. Is there a set $A \subset \mathbb R$ such that $\{n: x_n \in A\} \in \mathcal U$, but $\{n: y_n \in A\} \notin \mathcal U$?
I hardly know anything about the subject, so I only have ruled out two simple approaches:
- If $N:=\{n: x_n \neq y_n\}$, then $A := \{x_n: n\in N\}$ doesn't have to be as demanded: Consider, e.g., $(x_n) = (0,1,0,1,\dots)$ and $(y_n) = (1,0,1,0,\dots)$ with $N = \mathbb N$ and $A = \{0,1\}$. (This approach could be useful if there where some "sufficiently small" $N\in \mathcal U$ such that $x_n \neq y_n$ for all $n\in N$.)
- The sequences $(x_n) = (1/n)$ and $(y_n) = (1/(n+1))$ are not a counterexample: Let $E$ be the set of even natural numbers and assume, w.l.o.g., $E \in \mathcal U$. Let $A := \{1/n: n\in E\}$. Then $\{n: x_n \in A\} = E \in \mathcal U$ and $\{n: y_n \in A\} = \mathbb N\setminus E \notin \mathcal U$ (since $\mathcal U$ is an ultrafilter).
Any corresponding insights about hyperreals or ultrafilters are welcome!
Best Answer
Suppose that whenever $x=\langle x_n:n\in\Bbb N\rangle$ and $y=\langle y_n:n\in\Bbb N\rangle$ are sequences in $\Bbb N$ such that $(\mathscr{U}n)(x_n\ne y_n)$, there is an $A\subseteq\Bbb N$ such that $(\mathscr{U}n)(x_n\in A)$ and $(\mathscr{U}n)(y_n\notin A)$, where $(\mathscr{U}n)\varphi(n)$ abbreviates $\{n\in\Bbb N:\varphi(n)\}\in\mathscr{U}$, and let $x$ and $y$ be such a pair of sequences. Then there is a $U\in\mathscr{U}$ such that
$$\{x_n:n\in U\}\cap\{y_n:n\in U\}=\varnothing\,.\tag{1}$$
Viewing $x$ and $y$ as functions from $\Bbb N$ to $\Bbb N$, we can rewrite $(1)$ as $x[U]\cap y[U]=\varnothing$. Let
$$\mathscr{X}=x(\mathscr{U})=\{X\subseteq\Bbb N:x^{-1}[X]\in\mathscr{U}\}$$
and
$$\mathscr{Y}=y(\mathscr{U})=\{Y\subseteq\Bbb N:y^{-1}[Y]\in\mathscr{U}\}\,;$$
$\mathscr{X}$ and $\mathscr{Y}$ are ultrafilters on $\Bbb N$, $x[U]\in\mathscr{X}$, and $y[U]\in\mathscr{Y}$, so $\mathscr{X}\ne\mathscr{Y}$.
Thus, $x(\mathscr{U})\ne y(\mathscr{U})$ whenever $x\ne y\pmod{\mathscr{U}}$, and $\mathscr{U}$ is therefore a Hausdorff ultrafilter on $\Bbb N$. Unfortunately, in the paper at the link Bartoszynski and Shelah show that it is consistent that there be no Hausdorff ultrafilters on $\Bbb N$, so it is at least consistent that the answer to your question is no.