The text book in my course has an exercise about finding a metric space whose (usual) length metric is not a metric. It wants me to find a metric space $(X,d)$ satisfying $d'(x,y)=0 \ \ $for some distinct $x,y\in X$ ; $d'$ is a length metric.
In my conclusion, such a metric space doesn't exist.
The followings are some definitions in the problem set (and my 'wrong' solution).
Thanks in advance!
Let $(X, d)$ be a metric space.
(1) For a continuous map $ α:I=[0,1] → X$, we define the $length$ of $α$ as $$l(α) = \sup{\sum_{i=1}^{k} d(α(t_{i−1})), α(t_i))},$$ where the supremum is taken for all finite sequences $\{t_i\}$ satisfying $0 = t_0 < t_1 < t_2 < · · · < t_k = 1$. If the supremum is finite, we say $α$ is $rectifiable$.
(2) Define for $x, y ∈ X$, $$d' (x, y) = \inf \ l (α)$$ where the infimum is taken over all rectifiable curves joining $x$ and $y$. We call $ d' $ as the $length\ metric$ on $X$.
;;;Suppose such a metric space exists
$\exists \ x,y \in X \ $ s.t. $\ 0=d'(x,y) < d(x,y)$
By the definition of $d'(\cdot,\cdot)$,
$\exists \ a \ rectifiable \ curve \ \gamma \ joining \ x \ and\ y \ $s.t.$ \ d'(x,y) \leq l(\gamma)<d(x,y) $
However, $$l(\gamma) \geq \sum_{i=1}^{k} d(\gamma(t_{i−1})), \gamma(t_i)) \geq d(\gamma(t_0)), \gamma(t_k)) =d(x,y)$$ holds for all finite sequences $\{t_i\}$ satisfying $0 = t_0 < t_1 < t_2 < · · · < t_k = 1$ (for some positive integer $k$) ;Which leads to a contradiction.