Embedding $\omega_1$ in the hyperreals I've been trying to show that there exists a subset of the hyperreals which has order type $\omega_1$, that is, that there exists a subset $A \subseteq {}^*\mathbb{R}$ for which there is an order isomorphism from $\omega_1$ onto $A$. 
Here the hyperreals are constructed using a free ultrafilter, $\mathscr{U}$, over $\omega$ - much like construction presented on the wikipedia page. 
I know that one cannot find such a subset of $\mathbb{R}$, but I really just don't see how the situation is much better for the hyperreals. Initially I believed that one ought to be able to find a subset of the hyperreals simply because there were ''more'' hyperreals, but even that isn't (entirely) true because $\mathbb{R}$ and ${}^*\mathbb{R}$ have the same cardinality. I haven't really been able to even get off the ground and I was wondering if anyone could shed some light on how to construct such an embedding/subset of ${}^*\mathbb{R}$? Any feedback is much appreciated.  
 A: What you want is the usual argument that the bounding number $\mathfrak{b}$ is uncountable. Given a countable family $\mathscr{F}=\{f_n:n\in\omega\}\subseteq{}^\omega\omega$, define
$$f:\omega\to\omega:n\mapsto 1+\max_{k\le n}f_k(n)\;;$$
then for each $n\in\omega$ we have $f(k)>f_n(k)$ for all $k\ge n$. Using this, it’s easy to construct recursively a family $\mathscr{F}=\{f_\xi:\xi<\omega_1\}\subseteq{}^\omega\omega$ such that $f_\xi<^*f_\eta$ whenever $\xi<\eta<\omega_1$, where for $f,g\in{}^\omega\omega$ we write $f<^*g$ iff there is an $n\in\omega$ such that $f(k)<g(k)$ for all $k\ge n$. The map 
$$\varphi:\omega_1\to{}^*\Bbb R:\xi\mapsto [f_\xi]_{\mathscr{U}}$$
is then an order isomorphism (though there’s no reason to think that it’s continuous at limits).
A: Let $\alpha$ be some ordinal such that $f : \alpha \to {}^*\mathbb{R}$ is a bijection (i.e. a well-ordering of the hyperreal numbers).  Let $I$ be the subset of $\alpha$ defined by
$$
I = \{i \in \alpha : f(i) \ge f(j) \text{ for all } j \le i\} \quad (f(i) \ge f(j) \text{ as hyperreals.})
$$
It suffices to show that $I$ is uncountable, and we will have $\omega_1 \subset I$.
Suppose towards contradiction that $I$ is only countable.  Then there is a countable set of hyperreal numbers $h_1, h_2, h_3, \ldots$ such that ALL hyperreal numbers $h$ are bounded by some $h_n$.  Pick representative sequences
\begin{align*}
h_1 &= (r_{11}, r_{12} , r_{13}, \ldots) \\
h_2 &= (r_{21}, r_{22} , r_{23}, \ldots) \\
h_3 &= (r_{31}, r_{32} , r_{33}, \ldots) \\
\vdots
\end{align*}
And consider the hyperreal number
$$
q := (1 + r_{11}, 1 + \max\{r_{12}, r_{22}\}, 1 + \max\{r_{13}, r_{23}, r_{33}\}, \ldots)
$$
where $q_i = 1 + \max\limits_{1 \le j \le n} r_{ji}$. 
$q_i$ is real, so $q$ is hyperreal.
But now for any $h_n$, the $i$th coordinate of of $q$ is strictly bigger than the $i$th coordinate of $h_n$ for all $i$ sufficiently large.  So $q > h_n$ for all $n$, contradiction.
