A question about a continuous curve in $\mathbb{R}^2$ Let $f:[0,1]\longrightarrow\mathbb{R}^2$ be a continuous function such that 
$f(1)-f(0)=(1,0).$ Is it right that we can find $0\leq t_1<t_2\leq1$ such that $f(t_2)-f(t_1)=(\pm1,0)$ and for any $t_1\leq t<t'< t_2$ or $t_1< t<t'\leq t_2$, $f(t')-f(t)=(a,0)$ for some $a\in \mathbb{R}$ gives $|a|<1$?
 A: Yes, this is true, and it is an application of compactness.
For the proof, first consider the continuous $\mathbb{R}^2$-valued function
$$F(t,t') = f(t')-f(t)
$$
defined on the compact set 
$$D = \{(t,t') \in \mathbb{R}^2 \,\bigm|\, 0 \le t \le t' \le 1\}
$$
Consider also the set
$$A = D \cap F^{-1} \bigl(\{(a,0) \,\bigm|\, a \in \mathbb{R}, |a| \ge 1\}\bigr)
$$
This is a closed subset of $D$, and since $D$ is compact it follows that $A$ is compact. By assumption we have $(0,1) \in A$, hence $A$ is not empty. Also, $A$ is disjoint from the diagonal where $t=t'$ because $F(t,t) = f(t)-f(t)=(0,0)$.
Next, consider the following continuous real valued function defined on $A$:
$$g(t,t') = t'-t
$$
Since $A$ is compact and nonempty, $g$ attains a minimum value at some pair $(t_1,t_2) \in A$. This minimum value $g(t_1,t_2)=t_2-t_1$ cannot equal $0$ because $A$ is disjoint from the diagonal where $t=t'$. Thus, $0 \le t_1 < t_2 \le 1$. 
For any $t,t'$ such that $t_1 \le t < t' \le t_2$, we have $t'-t < t_2-t_1$, and so $(t,t') \not\in A$; it follows that if $F(t,t') = f(t')-f(t) = (a,0)$ then $|a|<1$.
