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Is there a sequence of the following operation that change a closed curve with finite number of self-intersections to a simple closed curve?

Also, every self-intersection differs at least $\epsilon$ in distance. The curve never pass though the same point 3 times.

If there is a intersection that locally looks like

a b
c d

we can change it to one of the following

a b
c d

a b
c d

If the answer is different on different spaces, I'm interested in $\mathbb{R}^2$.

From some example I tried, it seems one of the move can create 2 closed curves, but not both.

Edit: I don't know how to capture the notion of the curves. I'm wondering about this because I draw some closed curve on a notebook, and figured I can erase a intersection and connect in the above way, and eventually it become a simple closed curve.

Just assume this curve is well behaved enough that I can draw on a notebook.

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What's your definition of curve a $C^0$ path, $C^1$ or $C^\infty$? I ask because I'm skeptical something like this would be able to smooth out a closed Peano Curve. – JSchlather Oct 8 '11 at 20:04
Are you implicitly assuming that there are only finitely many self-intersection points, or at least that every self-intersection has a finite minimal distance from all other self-intersections? – Henning Makholm Oct 8 '11 at 20:10
@Jacob You can pick the one that behave the best. $C^\infty$? – Chao Xu Oct 8 '11 at 20:17
@Henning yes, there are only finite many self-intersections, I will fix my question. – Chao Xu Oct 8 '11 at 20:18
@Jacob, I don't think even $C^\infty$ is enough to make this well-defined. Consider a curve that behaves as $t\mapsto(t,e^{-1/t^2}\sin(2\pi/t))$ near $t=0$ (which ought to be smooth) and later passes through the origin along the $x$ axis. The self-intersection at (0,0) does not look like the OP's assumption locally. – Henning Makholm Oct 8 '11 at 20:19
up vote -1 down vote accepted

I think I got this.

Parametrize the curve as f(t) = (x,y). Pick any intersection, mark 4 local points a,b,c,d, such that it looks like

a c
b d

and $f(t_a)=a, f(t_b)=b, f(t_c)=c, f(t_d)=d$. such that $t_a<t_b<t_c<t_d$

I claim the following would not make two closed curves.

a c
b d

The a,b,c,d divide the curve into 4 pieces. $ab, bc, cd, da$. Where $xy$ is the piece of curve $\{f(t)|t\in[t_x,t_y]\}$, where $f(t_x) = x$ and $f(t_y)=y$.

This movement only change two pieces. $ab$ to $ac$, $cd$ to $bd$. $bc$, $ad$ are not changes. Therefore if we travel though a, we will get $acbd$,

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Expressed in simpler terms: If $P$ is a transversal self-intersection of a closed curve $\gamma\subset{\mathbb R}^2$ then you can always apply one of the proposed moves, which leaves a closed curve $\gamma'$ with one self-intersection less. – Christian Blatter Oct 9 '11 at 9:42

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