Let $D=\{z\in \mathbb C:|z|\leq 1\}$ be the unit disc in the complex plane, with interior $U=\{z\in \mathbb C:|z|<1\}$.

Let $A\subset \mathbb C\setminus U$ be an arc intersecting $D$ only at its two endpoints $\pm i$.

Let $B\subset \mathbb C\setminus U$ be an arc intersecting $D$ only at its two endpoints $\pm 1$.

Must $A$ intersect $B$? This seems intuitively obvious, but I am looking for proofs. What are some different ways of proving this? Might it be equivalent to a problem about planarity of a graph?

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    $\begingroup$ 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. $\endgroup$ Nov 27, 2023 at 21:25
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    $\begingroup$ 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." $\endgroup$ Nov 27, 2023 at 22:37
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    $\begingroup$ These both seem like good answers. $\endgroup$ Nov 27, 2023 at 22:42
  • $\begingroup$ 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. $\endgroup$ Nov 28, 2023 at 0:41
  • $\begingroup$ 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. $\endgroup$ Nov 28, 2023 at 2:37

5 Answers 5


Amplitwist's solution just with Jordan: Consider the curves on the Riemann sphere $S$ and connect $\pm 1$ along the real interval $R$ inside the disk such that $R \cup B$ is a closed curve on $S$. Then $A$ is a curve connecting inside and outside of $R \cup B$ and thus intersects $R \cup B$ by Jordan curve theorem, but not inside the disk on $R$. Therefore $A$ and $B$ must intersect.

Update: $i$ and $-i$ are on different sides of $R\cup B$ because the straight line $[-i,i] \subset D$ intersects $R$ exactly once and does not intersect $B$.

  • $\begingroup$ How do you know that $A$ connects the inside and outside? $\endgroup$ Nov 28, 2023 at 0:45
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    $\begingroup$ @D.S. Lipham: I added an update to answer your comment. $\endgroup$ Nov 29, 2023 at 6:50

First, apply inversion so that the arcs are inside the unit circle. Assume that the arcs don't intersect. Then, since they are compact, they stay at least $10\delta$ apart, for some $\delta>0$. Consider a tiling by of the plane by regular hexagons of diameter $\delta$, and color all hexagons within distance $\delta$ from the first curve white, and all hexagons within distance $\delta$ from the second curve black. Also, divide the boundary of the hexagonal tiling of the disc into four arcs, containing $i,1,-i,-1$ repsectively, and color them white/black/white/black. Color all other hexagons arbitrarily. You've got a game of hex where both players won, so we need to show that this is impossible.

Consider the "exploration path" running along edges of hexagons, starting on the boundary between black and white boundary arc, and defined inductively so that it always has a black hexagon on its left and white hexagon on its right. This path cannot intersect itself, and so it has to end somewhere. It can only end at the other end of either black or white boundary arc where it started; suppose it's black. Complete the path to a simple closed loop by tracing the outside of the black boundary arc.

Now apply the easy Jordan theorem for polygons: the black hexagons on one black boundary arc are inside the loop, and the black hexagons the other boundary arc are outside (all you need here is that the properly counted number of intersections of the polygon with a ray from any point not on the polygon has a well defined parity). Hence, any path on black hexagons connecting the two must cross the loop, at which point it will contain a white hexagon, a contradiction.


$\newcommand{\R}{\mathbb R}$Here is an elementary (albeit somewhat longish) solution, without using the Jordan curve theorem.

Let us borrow the inversion idea from Olivier Bégassat, so that the two arcs be in $D$, with endpoints $-1,1$ and $-i,i$. Inscribing the disk into the closed square with horizontal and vertical sides, we see that it is enough to prove the following:

Claim 1: Let $R$ be a closed rectangle with horizontal and vertical sides; in the sequel, we will only consider such rectangles. Let $A$ be an arc in $R$ connecting the two vertical sides of $R$. Let $B$ be an arc in $R$ connecting the two horizontal sides of $R$. Then $A$ and $B$ intersect.

Proof: We have $A=a([0,1])$ and $B=b([0,1])$ for some continuous functions $a$ and $b$ from $[0,1]$ to $\R^2$ such that $a(0)\in\{0\}\times[0,1]$, $a(1)\in\{1\}\times[0,1]$, $b(0)\in[0,1]\times\{0\}$, $b(1)\in[0,1]\times\{1\}$.

Let us say that a sub-rectangle $Q=[x_1,x_2]\times[y_1,y_2]$ of $R$ is good if for some subintervals $[s_1,s_2]$ and $[t_1,t_2]$ of $[0,1]$ we have $a([s_1,s_2])\subseteq Q$, $b([t_1,t_2])\subseteq Q$, $a(s_1)\in\{x_1\}\times[y_1,y_2]$, $a(s_2)\in\{x_2\}\times[y_1,y_2]$, $a(s_1)\in\{x_1\}\times[y_1,y_2]$, $a(s_2)\in\{x_2\}\times[y_1,y_2]$. That is, the goodness of $Q$ is witnessed by the subarc $a([s_1,s_2])\subseteq Q$ (of the arc $A=a([0,1])$) connecting the left and right sides of $Q$, and by the subarc $b([t_1,t_2])\subseteq Q$ (of the arc $B=b([0,1])$) connecting the bottom and top sides of $Q$.

By compactness and continuity, there is a good sub-rectangle $Q$ of $R$ with the smallest perimeter; let us call such a good sub-rectangle $Q$ minimal.

Without loss of generality (wlog) $R$ is a minimal good sub-rectangle of itself. The case when the area of $R$ is $0$ is easy. So, wlog the area of $R$ is $>0$.

Moreover, by approximation and compactness, the connecting arcs $A$ and $B$ in $R$ are wlog polygonal (chains), say $A_0\cdots A_m$ and $B_0\cdots B_n$ for some natural $m$ and $n$ and some points $A_0,\ldots,A_m,B_0,\ldots,B_n$ in $R$ such that the points $A_0,A_m,B_0,B_n$ are respectively on the left, right, bottom, and top sides of $R$.

Let us prove the so-reduced Claim 1 (with a minimal $R$ and polygonal $A$ and $B$) by induction on $m+n$. The smallest value of $m+n$ is $2$, and then $m=n=1$, in which case Claim 1 is a simple exercise.

If for some $j\in\{1,\dots,m-1\}$ the point $A_j$ is on the left side of $R$, then we can replace the arc $A_0\cdots A_m$ by $A_j\cdots A_m$ and then use the induction on $m+n$. So, wlog the points $A_1,\dots,A_{m-1}$ are not on the left side of $R$; similarly, these points are not on the right side of $R$. Similarly, the points $B_1,\dots,B_{n-1}$ are not on the bottom or top side of $R$.

So, we can move the left side of $R$ slightly to the right and thereby get a contradiction with the minimality of $R$ unless a point $B_l$ is on the left side of $R$ for some $l\in\{0,\dots,n\}$. So, we do have a point $B_l$ is on the left side of $R$ for some $l\in\{0,\dots,n\}$. Similarly, we have a point $B_r$ is on the right side of $R$ for some $r\in\{0,\dots,n\}$. Similarly, we have a point $A_b$ on the bottom side of $R$ for some $b\in\{0,\dots,m\}$ and a point $A_t$ on the top side of $R$ for some $t\in\{0,\dots,m\}$.

The arc $B_l\cdots B_r$ (possibly with $r<l$) connects the left and right sides of $R$, and the arc $A_b\cdots A_t$ (possibly with $t<b$) connects the bottom and top sides of $R$. So, if $l\notin\{0,n\}$, then the arcs $B_l\cdots B_r$ and $A_b\cdots A_t$ intersect by the induction on $m+n$, and we are done.

So, wlog $l\in\{0,n\}$. Similarly, wlog $r\in\{0,n\}\setminus\{l\}$, $b\in\{0,m\}$, $t\in\{0,m\}\setminus\{b\}$. Also, by shifting and rescaling, wlog the rectangle $R$ of area $>0$ is the unit square $[0,1]^2$. So, if the arcs $A$ and $B$ do not intersect, then wlog $B_0=(0,0)$, $B_n=(1,1)$, $A_0=(0,1)$, $A_m=(1,0)$. Let $k,-u,v,-w$ stand for the slope coefficients of the lines $B_0B_1,A_0A_1,B_nB_{n-1},A_m A_{m-1}$, respectively. Note that $B_{n-1}$ is in $R$ but not on the top side of $R$ and is not on the right side of $R$. So, $v\in(0,\infty)$. Similarly, $k,u,w$ are each in $(0,\infty)$.
Wlog $uw\ge kv$. Then for all small enough real $x>0$ the rectangle $R'$ with vertices $B'_0:=(x,kx)$, $B'_n=(1-\frac uv\,x,1-ux)$, $A'_0:=(x,1-ux)$, $A''_m:=(1-\frac uv\,x,kx)$ is good -- witnessed by the subarc $A'_0A_1\cdots A_{m-1}A'_m$ with $A'_m:=(1-\frac uv\,x,w\frac uv\,x)$ (of the arc $A=A_0A_1\cdots A_{m-1}A_m$) connecting the left and right sides of $R'$ and by the subarc $B'_0B_1\cdots B_{n-1}B'_n$ (of the arc $B=B_0B_1\cdots B_{n-1}B_n$) connecting the bottom and top sides of $R'$.

Thus, we get a good rectangle $R'$ whose perimeter is smaller than that of the minimal rectangle $R$, the final contradiction. $\quad\Box$

Here is a picture showing the rectangle $R$ (which was wlog the unit square), the rectangle $R'$ (dotted), and the points $A_0,A_m,B_0,B_m,A'_0,A''_m,A'_m,B'_0,B'_m$ as in the penultimate paragraph of the proof with $(k,u,v,w,x)=(1, 2, 1, 3,1/20)$:

enter image description here


Here's my attempt at solving this with graph theory (probably overkill). Assuming $A$ and $B$ do not intersect, we get a planar embedding of $K_{3,3}$ (which is known to not be planar).

Let $I_1,I_2,I_3,I_4$ be the 4 "quadrants" of the unit circle ($I_1$ is the arc along the circle from $1$ to $i$, $I_2$ from $i$ to $-1$, etc.).

Suppose $A$ and $B$ do not intersect. Then $S=A\cup B\cup I_2\cup I_4$ is a simple closed curve. By the Jordan Curve Theorem, $\mathbb C\setminus S$ has two components $G$ and $H$. Clearly the interiors of $I_1$ and $I_3$ are contained the same component because they are connected by a straight arc through $D$ which misses $S$. Say they are both in the component $G$.

Let $a\in A\setminus D$ and $b\in B\setminus D$. Then $a$ and $b$ are connected with an arc $J$ in $\overline H$ which touches $S$ only at $a$ and $b$. (By the Theorem: The closure of any complementary component of a simple closed curve is locally (path)-connected.)

The graph on 6 vertices $\pm 1, \pm i, a,b$ with edges $$I_1,I_2,I_3,I_4, (-i\leftrightarrow a), (a \leftrightarrow i), (-1 \leftrightarrow b), (b \leftrightarrow 1),J$$ is easily seen to a plane embedding of $K_{3,3}$, as the edges above are pairwise disjoint (except at their ends). The independent vertex sets are $X=\{\pm i,a\}$ and $Y=\{\pm 1,b\}$.

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    $\begingroup$ You were already given an answer by The Amplitwist in the comments. Also, this is basically circular. The proof that $K_{3,3}$ is not planar uses the same kinds of things that go into proving the Jordan curve theorem type results that are needed to prove what you're trying to prove. $\endgroup$ Nov 27, 2023 at 22:29
  • $\begingroup$ @AndyPutman The one in the comment above is probably the simplest and I would accept it if written as an answer here. But I suspect there are many others that are less elementary but use established (well-known) results. That is the nature of my proof. $\endgroup$ Nov 27, 2023 at 22:39

In the complex plane $\ \mathbb C,\ $ consider six vertices  (they belong to the unit circle $\ S:=D\setminus U\ $ that it the boundary of $D$):

$$ a_1:= 1\qquad\text{and}\qquad A_1=-a_1; $$ $$ a_2:= \frac{-1-i}{\sqrt 2}\qquad\text{and}\qquad A_2=-a_2; $$ $$ a_3:= i\qquad\text{and}\qquad A_3=-a_3; $$

Then consider the nine (curved or straight) edges that connect vertices $\ a_k\ $ to vertices $\ A_m,\ $ where six of those edges are the just obtained arcs of $\ S,\ $ and two of them are the given arcs that connect $\ a_1\ $ to $\ A_1,\ $ and $\ a_3\ $ to $\ A_3,\ $ and the ninth edge is the straight diameter interval that connects $\ a_2\ $ and $\ A_2.\ $

If the mentioned two arcs are disjoint then every pair consisting of the different arcs of the said nine arcs have disjoint interiors. Thus our graph of six vertices and nine edges would be isomorphic to the second Kuratowski's graph $\ K_{3,3}.\ $ This however would mean that our graph is not planar -- the two special edges have to intersect.


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