Assume we have an arc \begin{eqnarray*} \varphi:[0,1] &\rightarrow& \mathbb{R}^2, \\ t &\mapsto& (x(t),y(t)), \end{eqnarray*} i.e., $\varphi$ is continuous and injective. Suppose that $x(0)<x(1)$, and we have the following "pairwise monotone" condition: \begin{equation*} \big( (x(t)-x(s)\big)\big( y(t)-y(s)\big) \ge 0,\quad \forall t,s\in [0,1]. \end{equation*}
My question is that: for any $0<t<1$, can we prove that $x(0)\le x(t)\le x(1)$? And hence $y(0)\le y(t)\le y(1)$.
I first started with assuming, contrary to the claim, that there is $t\in (0,1)$ for which $x(t)<x(0)$ (the other case where $x(t)>x(1)$ should be handled similarly). Then the arc would go down from $(x(0),y(0))$ to $(x(t),y(t))$ and then at some time being pulled up to $(x(1),y(1))$ which breaks the pairwise monotone condition around the point $(x(t),y(t)$. I believe the Intermediate Value Theorem should be taken into account, but still cannot "control" simultaneously $x$ and $y$ coordinates.
Thank you for your reading. Any help is very appreciated.
Assume, wlog, $x(0) = y(0) = 0$. All points of the arc will be in the first or third quadrants. It can't pass through the origin again as it is 1-1, so it must stay in one quadrant.