In what way does the nonstandard definition of microcontinuity differ from that of epsilon-delta continuity, and related quibbles During a very cursory glance over the Wikipedia articles on non-standard calculus, I spotted the following definitions of continuity, uniform continuity and "microcontinuity", and I wonder what the actual difference is between them and their standard analytic counterparts!
First definition:
A real function $f$ is continuous if for any real number $x$, the hyperreals $x'$ infinitely close to $x$ have the property $f^*(x')\approx f^*(x)$, where $f^*$ is the extension of $f$ to the hyperreals, and the approximation sign means "infinitely close to". Now, to me this is consistent with the intuition of epsilon-delta statement; for all $x'$ such that $0\lt\|x'-x\|\lt\delta$, we have $0\lt\|f^*(x')-f^*(x)\|\lt\epsilon$ where $\epsilon$ and $\delta$ can be "infinitely close" to $0$ - I'm used to "arbitrarily close", but with a stretch of the imagination I can see how the two definitions could be rigorously shown to be the same. However, I am told that if $f$ satisfies this definition for all real and all hyperreal $x$, then $f$ is not only continuous but also microcontinuous. I am unsure of how different this statement is from normal continuity - I see no reason why we could not slap an $\epsilon-\delta$ argument onto this as well! If $\forall x',x'\approx x\implies f^*(x')\approx f^*(x)$, then $\forall x', \|x'-x\|\lt\delta\implies\|f^*(x')-f^*(x)\|\lt\epsilon$, which is the same statement as before. The only difference is that in the canonical continuity definition, $\delta,\epsilon$ are strictly real numbers greater than zero, whereas in the statement I just made up for the purposes of "microcontinuity", I allowed $\delta,\epsilon$ to be smaller than every real number yet greater than zero - i.e. they were allowed to be hyperreal. To me, as a naive student who's learnt next to nothing about nonstandard analysis, I cannot imagine a counterexample of a real function which is classically continuous but not microcontinuous! The difference between the two definitions seems microscopic to me...
Second definition:
A function $f$ is uniformly continuous over some interval $I$, where $I^*$ is $I$'s hyperreal extension, if for every pair of hyperreals $x',y'\in I^*$, the following holds: $x'\approx y'\implies f^*(x')\approx f^*(y')$. This can be restated as: $f$ is uniformly continuous over $I$ if and only if $f^*$ is microcontinuous at every point in the domain of $I^*$. This seems to align itself fairly well with the standard uniform continuity definition: $\forall\epsilon\gt0,\,\exists\delta\gt0:\forall x,y\in I,\,\|x-y\|\lt\delta\implies\|f(x)-f(y)\|\lt\epsilon$, which is loosely speaking saying that $f$ is continuous with the same $\delta$ at every point in the domain $I$. So, since someone has presumably shown these two definitions to be equivalent, this suggests that my elusive counterexample of a continuous-but-not-microcontinuous function is a function that is not uniformly continuous anywhere - but what does that even mean, I ask?
Additionally, what would the nonstandard definitions of absolute continuity be, or of Holder continuity? Could absolute continuity be something like: $\sum_{x'\in N^*}\|x'-x\|\approx0\implies\sum_{x'\in N^*}\|f^*(x')-f^*(x)\|\approx0$, where $N^*$ is some collection of hyperreal points in the neighbourhood of $x$?
Can anyone enlighten me on:

*

*Whether the classical and non-standard definitions of continuity and uniform continuity are the same, rigorously speaking, or just extremely similar?

*How can a function ever be continuous everywhere but not microcontinuous anywhere?

*Whether or not there exists analogous definitions of most standard analysis ideas of convergence, continuity and limits (with all the typical extra statements of uniform, weak, strong, absolute,...) in the nonstandard world?

Many thanks. The article I took the definitions from is this one.
 A: 
1. Whether the classical and non-standard definitions of continuity and uniform continuity are the same, rigorously speaking, or just extremely similar?

The classical and nonstandard definitions of continuity are the same, for real functions, in the sense that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the usual standard continuity condition precisely if its $\:^\star$-extension $\:^\star\! f : \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{R}$ satisfies the nonstandard continuity condition presented in the linked article.
Similarly, $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the usual standard uniform continuity condition precisely if its $\:^\star$-extension $\:^\star\! f : \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{R}$ satisfies the nonstandard uniform continuity condition presented in the linked article.
However, the usual definitions of continuity and microcontinuity do not coincide for arbitrary hyperreal functions $g : \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{R}$. This happens because most such functions are not $\:^\star$-extensions of any real function. Assuming that your usual standard definition of continuity and your definition of microcontinuity are both phrased in such a way that they make sense for hyperreal functions, you get examples of hyperreal functions that are continuous in the usual standard sense but not in the nonstandard sense (the indicator function of the non-real hyperreal numbers) and vice versa (the indicator function of the rationals multiplied by an infinitesimal).

I cannot imagine a counterexample of a real function which is classically continuous but not microcontinuous!

This one's easy to settle. Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by the equation $f(x) = x^2$. Take your favorite infinite hypernatural $\omega$. This hypernatural is positive, and larger than any natural number. Since if $x \leq y$ then $\frac{1}{y} \leq \frac{1}{x}$ for all $x$, $y$, we get that $\frac{1}{\omega}$ is a positive hyperreal that is smaller than any positive natural number, and is therefore infinitesimal.
So we have that $\omega$ is infinitesimally close to $\omega + \frac{1}{\omega}$. Is $f(\omega)$ infinitesimally close to $f\left(\omega + \frac{1}{\omega}\right)$? Well, $\left(\omega + \frac{1}{\omega}\right)^2 - \omega^2 = \omega^2 + 2\omega\frac{1}{\omega} + \frac{1}{\omega^2} - \omega^2 = 2 + \frac{1}{\omega^2} > 2$, which is therefore not infinitesimal. So $f(\omega)$ and $f\left(\omega + \frac{1}{\omega}\right)$ are not infinitesimally close, and consequently $f$ is not microcontinuous, since if you take $x=\omega$ and $x' = \omega + \frac{1}{\omega}$, then the premise of the implication is satisfied, but the conclusion is not. And this same argument shows that the real function $f$ is not uniformly continuous on the real line.
Which brings us to uniform continuity. You write:

This seems to align itself fairly well with the standard uniform continuity definition: $\forall x,y\in I,\,\forall\epsilon\gt0,\,\exists\delta\gt0:\|x-y\|\lt\delta\implies\|f(x)-f(y)\|\lt\epsilon$

This is not the definition of uniform continuity at all. You need $$\forall \varepsilon > 0. \exists \delta > 0. \forall x, y \in I. |x - y|<\delta \rightarrow |f(x)-f(y)| <\varepsilon$$
instead. Your reasoning, that "the counterexample of a continuous-but-not-microcontinuous function is a function that is not uniformly continuous anywhere" does not work. In fact, uniform continuity is not a local property: a continuous real function is uniformly continuous on every closed and bounded real interval, so it doesn't make much sense to talk about a continuous function that is not uniformly continuous anywhere. This has nothing to do with nonstandard analysis: it should be familiar from introductory courses on standard real analysis.

2. How can a function ever be continuous everywhere but not microcontinuous anywhere?

There is no such real function. If a real function is continuous, then it is microcontinuous around every real point. But possibly not around non-real hyperreal points (see the squaring function above).
There are such hyperreal functions, though (provided that you define microcontinuity appropriately for such functions). Consider e.g. the function $f : \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{R}$ that takes the value $0$ on every real number, but takes the value $1$ on every non-real hyperreal number. For any $x \in \mathbb{R}, \varepsilon \in \mathbb{R}^+$, there is clearly a $\delta \mathbb{R}^+$ so that for all $y \in \mathbb{R}$, if $|x - y| < \delta$ then $|f(x) - f(y)| = |0 - 0| = 0 < \delta$. So $f$ is continuous. But if you take $x = 0$ and any non-zero infinitesimal $x' \approx 0$, then $1 = f(x') \not\approx f(0) = 0$. So $f$ is not microcontinuous.

Whether or not there exists analogous definitions of most standard analysis ideas of convergence, continuity and limits (with all the typical extra statements of uniform, weak, strong, absolute,...) in the nonstandard world?

There are analogous definitions of all these. You can find them in any nonstandard analysis textbook, e.g. Goldblatt's Lectures on the Hyperreals.
