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Let $u : \mathbb{R} \to \mathbb{R}$ be the right continuous version of the Heaviside step function. What does the natural extension $u^*$ of $u$ to the set $\mathbb{R}^*$ of the hyperreals look like? Specifically, what does it look like on an infinitely narrow symmetric interval around $0$?

This question arose out of the example I used in Why it is absolutely mistaken to cancel out differentials?. Specifically, if

$$u_n(x)=\begin{cases} 0 & x \in (-\infty,-1) \\ (x+1)^n & x \in [-1,0] \\ 1 & x \in (0,\infty) \end{cases}$$

then $u_n \to u$ pointwise but (as far as I can tell) $u_n^* \not \to u^*$ pointwise.

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Ian
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  • what does $x \in (−\infty,−1)$ mean in $\mathbb{R}^\star$ ? I think the closest you can get is $x \in [− N,−1+ \mu]$ where $N$ is infinite and $\mu$ is infinitesimal. – user48672 Jul 06 '16 at 16:53
  • @user48672 If you prefer, $u_n(x)=\begin{cases} 0 & x<-1 \ (x+1)^n & -1 \leq x \leq 0 \ 1 & x>0 \end{cases}$. Now everything in the definition of $u_n$ itself should translate smoothly. (Also note that $u_n$ is continuous.) – Ian Jul 06 '16 at 16:57
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    There is a theorem in Robinson's 1966 book that $f_n$ converges uniformly iff the natural extensions converge pointwise in the sense of the epsilon-N defintiion to the natural extension of $f$. – Mikhail Katz Jul 07 '16 at 07:34

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As you note, every real function $f$ has a natural hyperreal extension $f^\ast$. This is true regardless of whether $f$ is continuous or not. Consider for example the Heaviside function $h$. This is defined by $h(x)=0$ if $x\leq 0$ and $h(x)=1$ if $x>0$. By the transfer principle, the same conditions will be satisfied for $h^\ast$ when $x$ is hyperreal. Thus for all negative or zero hyperreal $x$ one will have $h(x)=0$ whereas for all positive hyperreal $x$ one will have $h(x)=1$. In particular at all negative infinitesimals $h$ will be zero, whereas at all positive infinitesimals $h$ will be equal to one.

Mikhail Katz
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