TITLE: Pair of curves joining opposite corners of a square must intersect---proof?
QUESTION [23 upvotes]: Reposting something I posted a while back to Google Groups.
In his 'Ordinary Differential Equations' (sec. 1.2) V.I. Arnold says
"... every pair of curves in the square joining different pairs of
opposite corners must intersect".
This is obvious geometrically but I was wondering how one could go
about proving this rigorously. I have thought of a proof using
Brouwer's Fixed Point Theorem which I describe below. I would greatly
appreciate the group's
comments on whether this proof is right and if a simpler proof is
possible.
We take a square with side of length 1. Let the two curves be
$(x_1(t),y_1(t))$ and $(x_2(t),y_2(t))$ where the $x_i$ and $y_i$ are
continuous functions from $[0,1]$ to $[0,1]$. The condition that the
curves join different pairs of opposite corners implies,
$$(x_1(0),y_1(0)) = (0,0)$$
$$(x_2(0),y_2(0)) = (1,0)$$
$$(x_1(1),y_1(1)) = (1,1)$$
$$(x_2(1),y_2(1)) = (0,1)$$
The two curves will intersect if there are numbers $a$ and $b$ in $[0,1]$
such that
$$p(a,b) = x_2(b) - x_1(a) = 0$$
$$q(a,b) = y_1(a) - y_2(b) = 0$$
We define the two functions
$$f(a,b) = a + p(a,b)/2 + |p(a,b)| (1/2 - a)$$
$$g(a,b) = b + q(a,b)/2 + |q(a,b)| (1/2 - b)$$
Then $(f,g)$ is a continuous function from $[0,1]\times [0,1]$ into itself and
hence must have a fixed point by Brouwer's Fixed Point Theorem. But at
a fixed point of $(f,g)$ it must be the case that $p(a,b)=0$ and $q(a,b)=0$
so the two curves intersect.
Figuring out what $f$ and $g$ to use and checking the conditions in the
last para is a tedious. Can there be a simpler proof?

REPLY [4 votes]: How about the following, using the  Nested Intervals Theorem  (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in $\mathbb{R}$ is non-empty. Here goes the proof:
We construct recursively a nested sequence $I_j := [a_j, b_j]$ of closed intervals for $j \geq 0$. Let $I_0 := [0,1]$. For every $j \geq 0$, construct $I_{j+1}$ as follows: let $m_j$ be the midpoint of $I_j$. If the curves intersect at $t = m_j$, then we are done, so stop the sequence. Otherwise set $I_{j+1}$ to be $[a_j, m_j]$ or $[m_j, b_j]$ depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that $c_1$ is to the "left" of $c_2$ at $t = a_j$ and to the "right" of $c_2$ at $t = b_j$). 
If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that $\cap_{j=0}^\infty I_j = \lbrace t\rbrace$ for some $t \in [0,1]$. Then we must have $c_1(t) = c_2(t)$.