My main problem is that there's two methods yielding two different results:

1.We can count through $\mathbb{^*R}$ using nothing but infinitesimal steps. For example, we can partition the interval $^*[0,1]$ into $h\in\mathbb{^*N-N}$ steps, creating the sequence $x_i := \frac i h$ which fulfills $x_i\approx x_{i+1}$ for all $i$.

We therefore have the increasing chain $0\approx x_0 \approx x_1 \approx ... \approx x_{h-1} \approx x_h \approx 1$, and as such, for every $x\in{^*[0,1]}$ the assessment $x\approx x_i$ holds for some $i$. Therefore, any $x\in{^*[0,1]}$ can be reached in a series of infinitesimal steps, and so $f$ is monotonic in $^*[0,1]$.
This procedure then can easily be extendend to $\mathbb{^*R}$.

2. Given $x,y\in\mathbb{^*R}$ with $x\approx y , x\le y $ the following is true:
$st(x) = st(y)$, and
monad(x) = monad(st(x)) = monad(y)

I.e. $x$ and $y$ both have the same standard part, and therefore their monads (i.e. the set of points infinitesimally close to them) are identical.

This is in stark contrast to 1. (which should be the correct one), as it says that the premise $(x,y\in\mathbb{^*R},x\approx y , x\le y ) \implies f(x)\le f(y)$ merely says that $f$ is monotonic on any monad, but as each two monads around different standard points are disjunct, we can't extend this argument to conclude that $f$ on all of $^*\mathbb{R}$ is monotonic.

I'm hoping to gain some understanding as to why my argument 2. fails, and how it should be correctly used to form a proof of the question in the title.

  • $\begingroup$ For 2., note that the set of $n\in{}^*\mathbb N$ such that $f(x)\le f(y)$ for all $y\in[0,1/n]$ has a least element. $\endgroup$ – Andrés E. Caicedo Jul 19 '18 at 14:52
  • $\begingroup$ Could you expand your argument? We can't use the overspill principle here (external function), so we can't prove that there's a $n\in\mathbb{N}$ in your set, and therefore we can't prove that there's a least element in the set. $\endgroup$ – Sudix Jul 19 '18 at 15:07
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    $\begingroup$ Since $f$ is external, your 1. does not work (there is a hidden use of induction, which would require $f$ to be internal). $\endgroup$ – Andrés E. Caicedo Jul 19 '18 at 16:13
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    $\begingroup$ No, that is just an instance of induction -- you are proving something of the form "for all $n$, either blah holds at $n$, or $n>h$". $\endgroup$ – Andrés E. Caicedo Jul 19 '18 at 18:27
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    $\begingroup$ If I got your last answer correctly, i.e. $f(\xi)=-sh(\xi)$, that almost works (where $sh$ is the standard part of a finite number). Property 1 is satisfied since if $x \approx y$ then $f(x)=f(y)$, and $f$ is non-increasing. If you want a non-monotonic $f$ you can tweak a bit the definition, e.g. to $f(\xi)=\sin(sh(\xi))$. $\endgroup$ – Emanuele Bottazzi Aug 20 '18 at 20:59

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