Let $\pi$ and $\lambda$ be two distinct permutations of $1, 2, . . . , n$, and consider the points $p := (\pi(1),\pi(2), ... , \pi(n))$ and $r:= (\lambda(1), \lambda(2), ... , \lambda(n))$ in $\mathbb{R}^n$.
Because $\pi$ and $\lambda$ are distinct, there are indices $i$ and $j$ with $1\leq i \leq n$ and $1 \leq j \leq n$, such that $\pi(i) < \pi(j )$ and $\lambda(i) > \lambda(j )$. (if not both permutations are the same)
It has been claimed that:
The points $p$ and $r$ are on different sides of the hyperplane $x_i$ = $x_j$ . Any continuous curve in $\mathbb{R}^n$ between $p$ and $r$ must pass through this hyperplane
I would like to know why is this claim true?
Thanks in advance.
Let the continuous curve $C:[0,1]\to \mathbb R^n$ start at $p$ and end at $q$. So $C(0)=p$ and $C(1)=q$ Consider the continuous function $f:\mathbb R^n\to \mathbb R$ given by $f(x_1\cdots x_n)=x_i-x_j$. Composition of these functions is still continuous, $f\circ C: [0,1]\to\mathbb R$. What beautiful theorem of calculus should we now apply? :-).
P.S: let me know if you want me to finish the argument. I think it would be helpful for you finish it.
$f \circ C(0)=f(p)=\pi (i)-\pi(j)<0 $ and $f \circ C(1)=f(q)=\lambda(i)-\lambda(j)>0$. So what does the intermediate value theorem say about this function somewhere in the middle?