Epsilon-Delta Continuity in Hyperreals

Question

I am wondering what's exactly the reason why we do not use the $\epsilon-\delta$ definition of continuity in Hyperreals? I mean, I know that one of the purposes of building this set is that we replace that $\epsilon-\delta$ concept with infinitesimals. But there must be some other reasons, right? Will that $\epsilon-\delta$ definition still work in Hyperreals? Cheers!

Answer 1

Yes, the epsilon-delta definition does work in the hyperreals. This is because the epsilon-delta definition is a first-order formula, meaning that quantification is only over elements (not over sets, as in the completeness property of the real field). Thus, a real function $f$ is continuous at a real point $c$ if $$(\forall \epsilon>0)(\exists\delta>0)(\forall x) \big[|x-c|<\delta \implies |f(x)-f(c)|<\epsilon\big].$$ By the transfer principle applied to this first-order formula, it is still valid over the hyperreals: $$(\forall \epsilon\in{}^\ast\mathbb R^+)(\exists\delta\in{}^\ast\mathbb R^+)(\forall x\in{}^\ast\mathbb R) \big[|x-c|<\delta \implies |f(x)-f(c)|<\epsilon\big].$$ This is now true for all positive values of $\epsilon$, for instance if $\epsilon$ is a positive infinitesimal. In the second formula $f$ denotes the natural extension of the real function $f$ (sometimes this is denoted ${}^\ast\! f$ but the asterisk is usually deleted when dealing with functions when the meaning is unambiguous).

As to your question "But there must be some other reasons, right?" you may be interested in our work where we developed a framework for differential geometry via infinitesimal displacements:

Nowik, Tahl; Katz, Mikhail G. Differential geometry via infinitesimal displacements. J. Log. Anal. 7 (2015), Paper 5, 44 pp.

There is actually a subtle difference between the two definitions of continuity: (A) "continuity" at $x$ via epsilon-delta, and (B) "microcontinuity" at $x$ via Cauchy's idea of requiring that for each infinitesimal $\Delta x=\alpha$, the change in function $\Delta y=f(x+\alpha)-f(x)$ is also infinitesimal.

The two definitions (A) and (B) are equivalent when applied at a real point $x=c$, namely $c\in\mathbb R$. However, in general they may not be equivalent. For example, if $f(x)=x^2$ and $x=H$ is infinite, then $f$ will be continuous at $x=H$ but not microcontinuous at $x=H$.

In fact, uniform continuity of a real function $f$ on, say, $\mathbb R$ is equivalent to microcontinuity of ${}^\ast\! f$ at all points of ${}^\ast\mathbb R$. Meanwhile, continuity of $f$ on $\mathbb R$ is equivalent to microcontinuity at all real points $x\in \mathbb R$.