Talk:Joining Arcs makes Another Arc

We are looking for a general attack on two arcs to eliminate any crossings. It is easy if they intersect finitely many times.

However, pathological examples intersecting infinitely many times do exist. I think that this is an example (in $\R^2$):

$\displaystyle f: t\mapsto (t,0)$

$\displaystyle g: t\mapsto \frac1{1+k+(2^{k+1}t+2(1-2^k)}(\cos(2\pi(2^{k+1}t+2(1-2^k)),\sin(2\pi(2^{k+1}t+2(1-2^k)))$ where $k \in \N$ is such that $t \in [1-2^{-k}, 1-2^{-k-1}]$, and $g(1) = (0,0)$.

$g$ does spiral to (0,0), intersecting the positive $x$-axis at every $(\frac1n,0)$

I think however, that the only possibility such pathologies can occur is when in fact $f(0) = g(1)$, but I haven't been able to prove this. --Lord_Farin 12:06, 27 October 2011 (CDT)
 * An easier example is probably $g: t\mapsto t \sin(\frac1t)$. --Lord_Farin 12:10, 27 October 2011 (CDT)
 * It would be delightful if we could come up with a counterexample, to prove it false ... --prime mover 12:31, 27 October 2011 (CDT)

I think the theorem is not true if the space is not Hausdorff; for then we put a second point on top of $f(0)$ and take $g(1)$ there. One has to be cautious with the topology though... I will try to formulate it nicely here. --Lord_Farin 12:39, 27 October 2011 (CDT)
 * Cf. http://en.wikipedia.org/wiki/Connected_space#Arc_connectedness for an example similar to what I thought of. --Lord_Farin 17:25, 27 October 2011 (CDT)
 * In such scenarios one needs to make quite certain that what you're starting with is still arc-connected. Just sayin'. --prime mover 00:27, 28 October 2011 (CDT)