EDIT: This question was in-part made unclear by my misconception of the hyperreals not being dense, since I thought there were no numbers between zero and consecutive infinitesimals
($0, 1/\infty, 2/\infty, ...$). Furthermore, it was a question based on an assumption begging more questions. My question has now deferred to the question of that assumption.
I believe the following is an account of all non-fractal, continuous curves on the Cartesian plane between $a$ and $b$. Such curves can be though of as the limit, as $n \rightarrow \infty$, of linearly interpolating more and more points along the curve, where $n$ is the number of sampled points. This shrinks the line segments between the points down to points, eventually leaving us with a continuous string of points.
Now, in a curve $C$ made by linearly interpolating between some set of points, the following property holds:
$$\forall p \in C, \exists q \in C, \ \text{s.t}. \overline{pq} \in C \tag 1$$
I might be abusing notation here, but in English, I am simply saying the for all points on such a curve, there must exist another point, such that the curve $C$ between $p$ and $q$ is a straight line. This is obviously true for a curve made by linearly interpolating between some set of points. However, does this property extend to continuous, non-fractal curves, given the above account?
You might say that: $$\lim_{q \rightarrow p} p-q = 0$$
But if that were applicable in this context, it would mean that one would be sampling all the points in the curve in order to replicate it. However, as we approach the limit as $n \rightarrow \infty$, the number of points between the sampled points shrinks to one, not zero. I mean, you cannot linearly interpolate between a point and itself? You don't need to at least. If you agree with the account given at the beginning of this post, then you agree as we approach the limit, the line segments between the points shrinks towards single points, not zero points. Perhaps for the real numbers, this is the same. For the hyperreals and above, it would not be, since among them you have the infinitesimals.
So, that is my question. Is the property $(1)$ true for continuous, non-fractal curves in $\Bbb R^2$? And if not, is it perhaps true for such curves in $^*\Bbb R^2$?
What can be shown is that every continuous curve can be approximated (to within an infinitesimal) by an internal curve consisting of infinitesimal line segments. But the curve itself cannot be such a infinite-sided polygon if you work in the traditional setting of classical logic. Something like that can be achieved in Synthetic Differential Geometry where you need to work in a category-theoretic framework and use intuitionistic logic; see Bell's book.