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I have the following exersice which I have no idea how to approach.

Let γ=$φ[0, L]$ to $R^2$ be a closed simple curve parametrized by arc length.
Now we set $si$ = $iL/n$ for $i = 0, 1, . . . , n$,
let $ Pn (γ) = (φ (s0), φ (s1), . . . , φ (sn))$ be the $n$-th polygonal approximation of γ.
Show that for $n$ sufficiently large, $Pn(γ)$ has no self-intersections, that is, the interiors of the straight line segments joining $φ(si)$ and $φ(si+1)$ for $0 ≤ i ≤ n$ are disjoint
The result is intuitive however I have no idea how to begin proving this.
Does anyone have any idea?
Thanks in advance

TheGeometer
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1 Answers1

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I don't know how to use Latex as such I will post just a sketch of my proof to avoid ugliness.(its not very rigorous but is all I could make, please let me know if you think the proof is wrong or anything I could add up to make it more rigorous.)
We suppose for a contradiction that for all n we have a point of intersection, then the points of intersection will have a convergent subsequence say at p . Now we note that as n tends to infinity the lines that are supposed to intersect become points therefore it must be that the lines of intersection converge to p. Now we may assume WLOG that x'(p)>0 and we consider a neigbourhood U, p in U so that inside U any point q has x'(q)>0(such a neigborhood U exist by continuity). Now we consider a ball around p so small so that the intersection of the ball with the curve is still a curve and this curve is in U.(a sort of subcurve of out initial curve inside U). Such a ball can be found since our initial curve is simple. We now note that in this subcurve the polygonal approximations will never intersect no matter how small or big n is.(This because x'(q)>0 for any q in our subcurve) And now we arrive to our contradiction by taking n big enough so that the point of intersection lies in the ball around p we just mentioned and the vertices of lines of intersection lie in in the subcurve. So we constructed a point of intersection in our subcurve, contradiction

TheGeometer
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  • You have assumed $\phi\in C^1$ (which was not mentioned in the question). Consider two congruent logarithmic spirals spiraling towards the origin and concatenated there, and the outer ends joined somehow. (Maybe this is not a counterexample; it just shows that things can become tricky without the $C^1$ assumption.) – Christian Blatter Mar 03 '15 at 14:14