TITLE: Intersection of two Jordan curves lying in the rectangle
QUESTION [7 upvotes]: Given a rectangle $ABCD$.
Let a Jordan curve $L_1$ joins the vertexes $A$ and $C$
and all points of $L_1$ belong the rectangle.
Let a Jordan curve $L_2$ joins the vertexes $B$ and $D$
and all points of $L_2$ belong the rectangle.
Then there exists a point of $L_1$ and $L_2$ intersection.
Has this statement a simple proof?
The statement may be connected with the Jordan curve theorem
but I think its proof must be more simple.
Where one can see any proof of the statement?
Thanks.

REPLY [5 votes]: There's a very slick way to demonstrate this using the Hex theorem [which is known to be equivalent to the Brouwer fixed-point theorem]. In a game of Hex, two players compete to connect opposite sides of a rhombus-shaped board built out of regular hexagons. Hex cannot end in a draw: the only way for the blue player to block the red from making a connection is for blue to make a connection.
                                                  
Whether we prove OP's statement for a rectangle, or for a rhombus made by adjoining two equilateral triangles, makes no difference as the two are homeomorphic. So let $A, C$ be the far corners of a rhombus with angles $60°$ and $120°$, respectively, and let $B, D$ be the near corners. The Jordan arcs $L_i$, $i = 1, 2$, are images of continuous injections $\gamma_i : [0, 1] \to ABCD$ where each $L_i$ maps $0, 1$ to opposite corners of the rhombus, $L_1$ (red) connects $A$ to $C$, $L_2$ (blue) connects $B$ to $D$.

As $L_1$ and $L_2$ are continuous images of the compact set $[0, 1]$, they are also compact subsets of the metric space $ABCD$. So to show that they intersect, it suffices to show that the distance $$d(L_1, L_2) = \inf \{ d(x, y): x \in L_1, y \in L_2 \}$$ is $0$.
This is because in any metric space, if $A$ is compact and $B$ is closed, there always exist points $x \in A$, $y \in B$ so that $d(x, y) = d(A, B)$.
But if $d(L_1, L_2) = 0$, that implies $d(x, y) = 0$ for some $x \in L_1, y \in L_2$, and it's an axiom of metric spaces that that's only possible if $x = y$, which means $x$ is also in $L_2$ and $L_1 \cap L_2$ is nonempty.
Now for the argument that $d(L_1, L_2) = 0$. For any $N \geq 2$, we may cover $ABCD$ with an $N \times N$ Hex board whose upper left corner is at $A$ and lower right corner is at $C$. For such a Hex board, the diameter of a single tile is upper bounded by $k/N$ for some absolute constant $k$. If we color all the tiles that intersect $L_1$ red, we get a red-tiled winning Hex path connecting the top to the bottom, while if we color all the tiles that intersect $L_2$ blue, we get a blue-tiled winning Hex path connecting the left to the right:

Now superimpose these two boards. If the winning path for red and the winning path for blue somehow didn't interfere with each other, that would contradict the Hex theorem. So there must be at least one purple tile that belongs to both paths:

But any purple tile must contain points from both the red Jordan arc $L_1$ and the blue Jordan arc $L_2$, which implies that $$d(L_1, L_2) \leq \text{ diameter of a single tile } \leq \frac{k}{N},$$ and since $N$ was arbitrary, that implies $d(L_1, L_2) = 0$ and the two Jordan arcs have at least one point in common. Done.