Curve $\psi(t) = (t \cos\frac{1}{t}, t\sin\frac{1}{t})$ is simple, bounded and not rectifiable Let $\psi(t)$ be a curve defined on $t\in[-1,0)$ by
$\psi(t) = (t \cos\frac{1}{t}, t\sin\frac{1}{t})$
Prove that $\psi(t)$ is simple, bounded and not rectifiable.
I know this curve is simple (if you take $t_1\ne t_2 \in [-1,0)$ then $\psi(t_1)\ne \psi(t_2)$) but I don't know how to prove it properly.
Instead, I really don't know where to start to show that is bounded and not rectifiable. To show that isn't rectifiable I need to prove that the length of the curve is infinitive, but I can't find any inscribed chord-polygons to use. I just need some hint.

Building off of Raffaele's answer, use the fact that
$$
L(\psi|_{[-1, -\epsilon]}) = \int_{-1}^{-\epsilon} \| \psi'(t) \| \, dt
$$
for all $\epsilon > 0$. Then, suppose for contradiction that $L(\psi)$ is finite, i.e. $\psi$ is rectifiable. Observe
$$
L(\psi) \geq L(\psi|_{[-1, -\epsilon]} )
$$
for all $\epsilon > 0$. Since $L(\psi|_{[-1, -\epsilon]})$ gets arbitrarily large as $\epsilon \to 0^+$, we obtain a contradiction. Thus, $\psi$ is not rectifiable.
