Direct proof for: If $|a|<\epsilon$ for all $\epsilon>0$, then $a=0$ I am asked to prove: if $|a|<\epsilon$ for all $\epsilon>0$, then $a=0$
I can prove this as follows.
Assume $a \not= 0$
I want to show then that $|a| \geq \epsilon$ for some $\epsilon$
We let $\epsilon = \frac{|a|}{2}$ and we are done.
However, I am curious if you can use the idea of infinitesimals to prove this directly.
Can I let $\epsilon$=an (or maybe the?) infinitesimal value and then just show that:
$|a|<\epsilon$ $\implies$ $-\epsilon<a<\epsilon$ $\implies$ $|a|=a=0$
I don't know much about infinitesimals but wanted to try and prove this directly. This was the first thing that occurred to me for some reason. Can anyone shed some light on infinitesimals and whether or not I can use them this way? this might be completely off base because I really have almost no knowledge about how infinitesimals fit into mathematics and mathematical thinking.
 A: You can do so if you do work with infinitesimals (say Robinson's hyperreal numbers) and, to really make the proof this short, make sure that $a$ is a standard number. One needs to know that application of the $\operatorname{st}()$ part to both sides of a strict inequality produces a corresponding  un-stricted inequality:
$$\begin{align} \forall\epsilon>0\colon |a|<\epsilon&\implies \forall\epsilon>0\colon \operatorname{st}(|a|)\le\operatorname{st}(\epsilon)\\&\implies 0\le |a|=\operatorname{st}(|a|)\le \operatorname{st}(\epsilon)=0\text{ for any positive $\epsilon\approx 0$}\end{align}$$
A: Infinitesimals aren't really going to help. What you are trying to prove is true in any ordered field. A proof along the lines you suggest requires the extra assumption that $a$ is "standard" (see Hagen von Eltzen's answer) and so proves something weaker than you want. If you accept a proof by contradiction as "direct", then how about: assume $a \not= 0$, then $|a| > 0$ and hence, by assumption, with $\epsilon = |a|$, $|a| < |a|$, a contradiction.
