Microcontinuous vs Continuous

Question

I've been studying infinitesimals and came upon the idea of uniformly microcontinuous functions. My question is: if a function $f^*: \mathbb{R}^* \to \mathbb{R}^*$ the natural extension of $f: \mathbb{R} \to \mathbb{R}$ is microcontinuous on $A\subset\mathbb{R}$ does that imply that it is continuous with the $\epsilon-\delta$ definition on $A$? If this is not true does the (I think)weaker statement hold that if the function is uniformly microcontinuous on all of $\mathbb{R^*}$ does that imply that it is continuous with the $\epsilon-\delta$ definition on $\mathbb{R}$?

Just a note that the definition I'm using for microcontinuous at $x$ is $\forall x' (x \approx x') \implies (f(x) \approx f(x'))$ where $a \approx b$ means that the standard part of $a-b$ is $0$. From this a function is micrcontinuous on some domain $S$ means that $\forall x\in S\, \forall x' (x \approx x') \implies (f(x) \approx f(x'))$

Answer 1

Yes, it does. To see that, choose any (standard) real $x_0\in A$. Now for any $x$ such that $$\models \{x_0-1/n<x<x_0+1/n\mid n\in {\bf N}\}$$ we have $$\models \{ f(x_0)-1/m<f(x)<f(x_0)+1/m\mid m\in {\bf N}\}$$ This means that in particular, for each $m\in {\bf N}$, we have $$\{x_0-1/n<x<x_0+1/n\mid n\in {\bf N}\}\vdash f(x_0)-1/m<f(x)<f(x_0)+1/m$$ and by compactness there is some $n\in {\bf N}$ such that $$x_0-1/n<x<x_0+1/n\vdash f(x_0)-1/m<f(x)<f(x_0)+1/m$$ but then $f$ is clearly ($\varepsilon$-$\delta$) continuous at $x_0$ when considered as a function on real numbers, so it's also just continuous, since $x_0$ was arbitrary. Since $\varepsilon$-$\delta$ continuity is first order, it is also $\varepsilon$-$\delta$ continuous on any given nonstandard model.

Note that it only applies if $f$ is a standard real function (a symbol of the language) to begin with. Otherwise the part about restricting to reals does not make sense and neither does the compactness argument, and indeed we can find a counterexample: choose an infinitesimal $\varepsilon$ and let $f(x)=0$ for $\lvert x\rvert >\varepsilon$, $f(0)=0$, and $f(x)=\varepsilon$ for $x\in [-\varepsilon,0)\cup (0,\varepsilon]$. Then $f$ is microcontinuous but not continuous.

Answer 2

It is not clear what you mean exactly by "uniformly microcontinuous". If you simply mean "microcontinuous", then microcontinuity of $f^*$ at every point of a real set $A$ implies continuity of $f$ on $A$. If you assume that $f^*$ is microcontinuous on all of $\mathbb{R}^*$ then it follows that $f$ is not merely continuous on $\mathbb{R}$ but is actually uniformly continuous on $\mathbb{R}$.