Does a curve which is homeomorphic to interval always have a finite length independent of the parametrization in general metric spaces?

Question

Let $(X,\rho)$ be a metric space and $C\subseteq X$ be a subset which is homeomorphic to interval $[a,b]$ for some $a,b\in\mathbb{R},a<b$. Each homeomorphism $\gamma:[a,b]\rightarrow C$ is called a parametrization. Is it true that each parametrization has finite length, which is the same for all parametrizations?

Answer 1

Given such a curve $\gamma:\>[a,b]\to X$ and a partition $${\cal P}:\quad a=t_0<t_1<\ldots<t_N=b$$ of the interval $[a,b]$ let $$\ell_{\cal P}(\gamma):=\sum_{k=1}^N d\bigl(\gamma(t_{k-1}),\gamma(t_k)\bigr)\ .\tag{1}$$ The length $L(\gamma)$ is then defined as $$L(\gamma):=\sup_{\cal P}\ell_{\cal P}(\gamma)\leq\infty\ .$$ The value $L(\gamma)$ remains unchanged under a homeomorphism of $[a,b]$ (or some other reparametrization), because the collection of sums $(1)$ remains the same. For an example with $L(\gamma)=\infty$ consider $$\gamma:\quad[0,1]\to{\mathbb R}^2,\qquad t\mapsto\left(t,\ t\,\sin{1\over t}\right)\qquad(t>0),\qquad \gamma(0):=(0,0)\ .$$

Answer 2

This is not true. If $\gamma$ has finite length, then it is called rectifiable. See https://www.encyclopediaofmath.org/index.php/Rectifiable_curve. A non-rectifiable example is

$$c : [0,1] \to \mathbb{R}^2, c(t) = \begin{cases} (t,t\cos(\pi/t) & t > 0 \\ (0,0) & t = 0 \end{cases}$$