Timeline for Two arcs in the complement of a disc must intersect?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2023 at 21:06 | answer | added | Wlod AA | timeline score: 1 | |
| Nov 28, 2023 at 8:24 | comment | added | Jochen Wengenroth | Olivier Bégassat's suggestion using the inversion reduces the situation to a classical application of Brouwer's fixed point theorem (which, of course can be proved using degrees but there are many other approaches like in some of the answers). | |
| Nov 28, 2023 at 7:46 | comment | added | Sam Nead | The answer that springs to mind is "apply some version of the Jordan curve theorem". The fact that the arcs are outside of the disk (instead of inside) is artificial (either add a point or invert). So I am voting to close. | |
| Nov 28, 2023 at 7:17 | answer | added | Iosif Pinelis | timeline score: 4 | |
| Nov 28, 2023 at 4:48 | history | became hot network question | |||
| Nov 28, 2023 at 2:37 | comment | added | Ryan Budney | Linked $0$-spheres cobound intersecting discs. This is true in higher dimensions, as well. For example, a linked $S^0$ and $S^1$ in the boundary of a $D^3$ must cobound intersecting embedded copies of $D^1$ and $D^2$ respectively. This is basic Poincare/Alexander duality. | |
| Nov 28, 2023 at 0:41 | comment | added | Daniel Asimov | It occurs to me to add a point at infinity, to get a topological closed disk (S^2 - U) that we now identify boundary points at 180º from one another ,,, resulting in the projective plane. Thus we are talking about the intersection of a homology cycle representing the nonzero class of H_1(P^2; Z/2Z) with itself ... which is known to be nonzero. That proves there must be a geometric intersection point, since each of the arcs — now closed curves — represents that same class. | |
| Nov 28, 2023 at 0:07 | answer | added | Karl Fabian | timeline score: 9 | |
| Nov 27, 2023 at 23:40 | answer | added | Kostya_I | timeline score: 7 | |
| Nov 27, 2023 at 22:42 | comment | added | D.S. Lipham | These both seem like good answers. | |
| Nov 27, 2023 at 22:41 | review | Close votes | |||
| Nov 28, 2023 at 16:29 | |||||
| Nov 27, 2023 at 22:37 | comment | added | Olivier Bégassat | There should be a standard homotopy argument. Something in the spirit of "By applying an inversion at 0 you may assume that the two arcs are drawn in $D \setminus 0$ and thus in $D$. If these two arcs never intersected you could consider the map $c:I\times I \to \mathbb{S}^1$ where $I = [0, 1]$ as the quotient $c(s,t) = \frac{a(t)-b(s)}{|a(t)-b(s)|}$. This map, when considered along the boundary $\partial (I \times I)$ induces a homotopy equivalence (it has degree $\pm1$) but by construction it extends to the disk thus is nullhomotopic and should have degree $0$. Contradiction." | |
| Nov 27, 2023 at 22:03 | answer | added | D.S. Lipham | timeline score: 1 | |
| Nov 27, 2023 at 21:25 | comment | added | The Amplitwist | Extend the arc $A$ to a closed curve by joining its endpoints with a straight line segment, and do the same with the arc $B$. Tait proves that any two closed curves (without "double points") in the plane intersect an even number of times, in this paper: Some elementary properties of closed plane curves. Messenger (2) 6 (1877), 132–133. JFM 09.0393.01. | |
| Nov 27, 2023 at 20:44 | history | asked | D.S. Lipham | CC BY-SA 4.0 |