I have been reading Vector calculus by Peter Baxandall. It defines : A curve $C$ in $\Bbb R^n$ is a simple arc if $C$ has a $1-1~~ C^1$ parametrization of the form $f:[a,b] \subseteq \Bbb R \rightarrow \Bbb R^n$
Now, I want to prove that in $\Bbb R^2 : y+1=(x-2)^2$ is not a simple curve.
Attempt: Let $\Big (g(t),h(t) \Big)$ be a parametrization of the given curve $y+1=(x-2)^2$ so that
$$ f: \Bbb R \rightarrow \Bbb R^2~~|~~f(t)=\Big (g(t),h(t) \Big)$$
represents the curve $C$. $~f$ is a continuously differentiable $C^1$ function. $~g: \Bbb R \rightarrow \Bbb R$ and $h: \Bbb R \rightarrow \Bbb R$ are onto functions. Now, we need to prove that $f$ is not one-one.
As per the given curve : $h(t)=\Big(g(t)-2\Big)^2-1 $
Thus, $f(t)=\Big (g(t),\Big(g(t)-2\Big)^2-1 \Big)$
Now, suppose $f(t_1) = f(t_2) \implies \Big (g(t_1), \Big(g(t_1)-2\Big)^2-1 \Big)=\Big (g(t_2), \Big(g(t_2)-2\Big)^2-1 \Big)$
$\implies g(t_1)=g(t_2)$ and $\Big(g(t_1)-2\Big)^2-1=\Big(g(t_2)-2\Big)^2-1$
$\implies g(t_1)=g(t_2)$ and $ \Big(g(t_1)-2\Big)=\pm \Big(g(t_2)-2\Big)$
But how can we deduce from here that $t_1 \ne t_2 ?$ Did I deduce the meaning of a simple curve correctly? Thanks a lot!
Best Answer
You're forgetting an important part of the definition of a simple curve. Remember that the parametrisation's domain is a closed interval $[a,b]$ ! Therefore the first step of your proof should read
$$ f : [a,b] \rightarrow \mathbb R^2 : t \mapsto (g(t),h(t))$$
instead.
If you don't take this into account then you can clearly find one to one differentiable parammatrisations of $C$ like
$$ x \mapsto \big( x,(x-2)^2 - 1 \big)\in \mathbb R^2$$
However if we consider $ f: [a,b] \rightarrow \mathbb R^2$ then it should be obvious why $C$ is not a simple curve. You need only notice that the set $C$ is unbounded (it's the graph of $p(x) = (x-2)^2 - 1$.) while the image of $[a,b]$ by $f$ is bounded by continuity of $f$.