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)

Hm... From the definition of arc, it seems that an arc, joined with its reversal (that is, eg. $f(t)$ and $f(1-t)$), do not satisfy this hypothesis. I think some thought is needed to adapt the definition of arc. I suggest to replace (See the definition of arc):

That is, an arc from $a$ to $b$ is a continuous injection $f: I \to T$ such that $f \left({0}\right) = a$ and $f \left({1}\right) = b$.

By:

That is, an arc from $a$ to $b$ is a continuous $f: I \to T$ such that:
 * $f \left({0}\right) = a$;
 * $f \left({1}\right) = b$;
 * whenever $r \le s \le t$ and $f(r) = f(t)$, we have $f(r) = f(t)$.

If this is not done, we have the theorem is false, because: In the above example case, one must have that the restriction $\tilde h$ of $h$ has $\tilde h(t) = f(0)$ for all $t$, in order to avoid actual crossings (this is obvious). But then $h$ is not injective, so not an arc. Therefore, there is no restriction $\tilde h$ that is injective. --Lord_Farin 17:26, 28 October 2011 (CDT)


 * I'm not a fan of "just" redefining the objects so as to get the result we intuitively want, unless we can find something out there that backs it up. I'm not going to be up for looking at this till at least tomorrow though. --prime mover 17:48, 28 October 2011 (CDT)