Proving that every proper elementary extension of the real numbers has an infinitesimal element.

Question

I'm not sure how to show that every proper elementary extension of the real numbers has an infinitesimal element. I've been trying to somehow apply the property

$$\forall x \forall y (((\forall \varepsilon > 0)(\lvert x - y \rvert < \varepsilon)) \leftrightarrow x = y)$$

because it seemed useful, although I have no better reason to believe this. If a nonstandard model has no infinitesimal element, then $(\forall \varepsilon > 0)(\lvert x - y \rvert < \varepsilon)$ is equivalent to $(\forall \varepsilon \in \mathbb{R}^+) (\lvert x - y \rvert < \varepsilon)$ where $\mathbb{R}$ is the natural embedding of the standard model, so perhaps this is somehow useful. Unsure how to proceed.

Answer 1

Let $R$ be a proper elementary extension of $\mathbb{R}$, and $r\in R\setminus \mathbb{R}$.

• Case 1: $r>q$ for all $q\in \mathbb{Q}$. Then $1/r$ is a positive infinitesimal.
• Case 2: $r<q$ for all $q\in \mathbb{Q}$. Then $-1/r$ is a positive infinitesimal.
• Case 3: The sets $A = \{q\in \mathbb{Q}\mid q < r\}$ and $B = \{q\in \mathbb{Q}\mid r < q\}$ are both non-empty. If $A$ has a greatest element $a$, then $(r-a)$ is a positive infinitesimal. If $B$ has a least element $b$, then $(b-r)$ is a positive infinitesimal. If neither, then $(A,B)$ is a Dedekind cut corresponding to some irrational real number $s$. Depending on whether $r<s$ or $s<r$, either $(s-r)$ or $(r-s)$ is a positive infinitesimal.