Theorem: Paths Crossing Rectangle Necessarily Meet

Let $R = \closedint a b \times \closedint c d$ be a closed rectangle.
Let $p, q : \closedint {-1} 1 \to R$ be paths in $R$.
Suppose:
* $\map {p_1} {-1} = a$
* $\map {p_1} 1 = b$
* $\map {q_2} {-1} = c$
* $\map {q_2} 1 = d$
where:
:$\map p t = \tuple {\map {p_1} t, \map {p_2} t}$
:$\map q t = \tuple {\map {q_1} t, \map {q_2} t}$
Intuitively, that is to say:
:The path $p$ begins on the left edge of $R$, and ends on the right edge.
:The path $q$ begins on the bottom edge of $R$, and ends on the top edge.
Then, there exist $t_p, t_q \in \closedint {-1} 1$ such that:
:$\map p {t_p} = \map q {t_q}$
which is to say, the two paths meet at some point in $R$.

Proof:
{{AimForCont}}, suppose there are no such $t_p, t_q$.
Define $N : \closedint {-1} 1^2 \to \R_{\ge 0}$ as:
:$\map N {s, t} = \map \max {\size {\map {p_1} s - \map {q_1} t}, \size {\map {p_2} s - \map {q_2} t}}$
By our assumption:
:$\map p s \ne \map q t$
for all $s, t \in \closedint {-1} 1$.
Therefore, for each $s, t$, at least one of $\map {p_1} s \ne \map {q_1} t$ or $\map {p_2} s \ne \map {q_2} t$ holds.
Thus, $\map N {s, t} > 0$ for all $s, t \in \closedint {-1} 1$.
Additionally, by Combination Theorem for Continuous Mappings on Metric Space:
:$N$ is continuous on $\closedint {-1} 1^2$
Now, we define $F : \closedint {-1} 1^2 \to \closedint {-1} 1^2$ as:
:$\map F {s, t} = \tuple {\dfrac {\map {q_1} t - \map {p_1} s} {\map N {s, t}}, \dfrac {\map {p_2} s - \map {q_2} t} {\map N {s, t}}}$
First, we must show that $F$ is well-defined.
Since:
:$\map N {s, t} \ge \size {\map {q_1} t - \map {p_1} s}$
:$\map N {s, t} \ge \size {\map {p_2} s - \map {q_2} t}$
it follows that the image of $F$ is indeed contained in $\closedint {-1} 1^2$.
Next, we want to prove that:
:$\Img F \subseteq \closedint {-1} 1 \times \set {-1, 1} \cup \set {-1, 1} \times \closedint {-1} 1$
or in other words, for all $s, t \in \closedint {-1} 1$:
:$1 \in \set {\size {\map {F_1} {s, t}}, \size {\map {F_2} {s, t}}}$
Observe that, by definition of $N$:
:$\map N {s, t} = \size {\map {p_i} s - \map {q_i} t}$
for at least one of $i \in \set {1, 2}$.
Then:
{{begin-eqn}}
{{eqn | l =  \size {\map {F_i} {s, t} }
      | r = \size {\frac {\map {p_i} s - \map {q_i} t} {\map N {s, t} } }
      | c = Definition of $F$
}}
{{eqn | r = \frac 1 {\map N {s, t} } \size {\map {p_i} s - \map {q_i} t}
      | c = Absolute Value Function is Completely Multiplicative
}}
{{eqn | r = \frac 1 {\size {\map {p_i} s - \map {q_i} t} } \size {\map {p_i} s - \map {q_i} t}
      | c = Above
}}
{{eqn | r = 1
}}
{{end-eqn}}
By:
* Brouwer's Fixed Point Theorem
* Compact Convex Sets with Nonempty Interior are Homeomorphic
* Fixed-Point Property is Topological
it follows that $F$ must have a fixed point $\tuple {s_0, t_0}$.
That is:
:$\map F {s_0, t_0} = \tuple {s_0, t_0}$
By the above argument, at least one of the following holds:
* $s_0 = -1$
* $s_0 = 1$
* $t_0 = -1$
* $t_0 = 1$
Suppose, for instance, $s_0 = -1$.
Then, {{hypothesis}}, $\map {p_1} {s_0} = a$.
Since $\map {q_1} {t_0} \in \closedint a b$, it follows that:
:$\map {q_1} {t_0} \ge \map {p_1} {s_0}$
Therefore:
{{begin-eqn}}
{{eqn | l = \map {F_1} {s_0, t_0}
      | r = \dfrac {\map {q_1} {t_0} - \map {p_1} {s_0} } {\map N {s_0, t_0} }
      | c = Definition of $F$
}}
{{eqn | o = \ge
      | r = 0
      | c = $\map {q_1} {t_0} \ge \map {p_1} {s_0}$
}}
{{eqn | o = >
      | r = -1
}}
{{eqn | r = s_0
}}
{{end-eqn}}
which contradicts our supposition that $\map F {s_0, t_0} = \tuple {s_0, t_0}$
In the same manner, we can show:
* $s_0 = 1 \implies \map {F_1} {s_0, t_0} \le 0$
* $t_0 = -1 \implies \map {F_2} {s_0, t_0} \ge 0$
* $t_0 = 1 \implies \map {F_2} {s_0, t_0} \le 0$
all of which are clearly contradictory.
Since every case leads to a contradiction, we conclude that there must exist $t_p, t_q$ such that:
:$\map p {t_p} = \map q {t_q}$
{{qed}}