I'm working through Infinitesimal Calculus by Henle and Kleinberg on my own and I'm having some trouble with one of the exercises. Here's what I have currently:
Exercise:
Let $j$ be the hyperreal $\{1, 2, 3,...\}$, prove $j$ > every real number $r$.
My proof:
Let $f$ be a function $\{1, 1/2, 1/3,...\}$, by definition of infinitesimal we know that $\forall{n \in \mathbb N} \{f(n) > 0\}$ and $\forall{n \in \mathbb N}\{f(n) < r\}$.
So $f \cap \mathbb R = \emptyset$ and $f \cap j = \emptyset$.
Consider $c$ the complement of the infinitesimal set ${}^\ast\mathbb R \setminus f$.
...and I'm not sure where to go from here. I'd like to show that $c$ is somehow > $r$, but maybe that's not the right path. I was thinking I could use the fact that $c$ is a superset of $j$ and we know that $j$ is quasi-big.
Thanks for your help!
I would instead take a real number $r$, i.e. the equivalence class $[(r, r, \dots, r, \dots)]$, and show that this is less than $j = [(1, 2, 3, \dots)]$.
This follows from the definition of $<$ in the hyperreals. Indeed, the set of elements $r_n = r$ which are greater than $j_n = n$ is finite, so is small with respect to the ultrafilter of our construction. So $r < j$.