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In any metric space we can define the length of any continuous map $f:I\to X$ from the unit interval to $X$ as the supermum over all finite partitions $P$ of $I$ of the following sum $$\ell(f)=\sup_P\underset{x_i\in P}{\Sigma}d(f(x_i),f(x_{i+1}))$$

To any metric space $(X,d)$ we can associate a length (or intrinsic) metric $$d_\ell(x,y)=\inf\{\ell(f)|f:I\to X;\ x,y\in\operatorname{Im}(f)\}$$

The metrics $d$ and $d_\ell$ are generally different, and it is not difficult to check (by triangular inequality) that the topology induced by the latter is finer or equal. I can easily produce examples where the two metrics are different, for example any non-planar surface embedded in $\mathbb{R}^3$ with the subspace metric, but the two topologies are still the same.

Question: What are some nice examples where the topologies induced by $d$ and $d_\ell$ are different?

Dinisaur
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    How do you define $\ell(f)$ ? – Surb Jan 14 '21 at 09:57
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    Here is an idea: find a subset of the plane with two "ends" (like an interval) such that in the extrinsic point of view (in the plane), the two "ends" are really close, but in the intrinsic point of view (in the subset with the length metric), the two "ends" are far away from each other (infinitely far away if possible). – Didier Jan 14 '21 at 09:57
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    @Surb I would assume $$\ell(f):=\sup\left{\sum_{k=1}^n d(f(t_{k-1}),f(t_k)),:, n\in\Bbb N\land 0=t_0< t_1<\cdots < t_n=1\right}$$ –  Jan 14 '21 at 10:12
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    Not answerable. Define your terms. What is $l(f),$ what is $I,$ and what restrictions or conditions are there on $f$? – DanielWainfleet Jan 14 '21 at 10:18
  • Daniel Wainfleet is right. I therefore voted to close your question. – Paul Frost Jan 14 '21 at 10:20
  • Thank you for asking clarifications, I added the definitions, which in fact are as Gae suggested – Dinisaur Jan 14 '21 at 10:39
  • What if there's no continuous map $f:I\to X$ connecting $x,y$? E.G., take $X={(x,y)\in\mathbb{R}^2:x\neq 0}$: $(-1,0)$ and $(1,0)$ cannot be joined by a continuous ma $f:I\to X$ –  Jan 14 '21 at 10:53
  • I retracted my close vote. But be aware that if $X$ is not path connected, then you do not get a metric (unless you allow $d(x,y) = \infty$). – Paul Frost Jan 14 '21 at 11:36
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    @PaulFrost Also if the space is path connected and some points aren't joined by rectifiable paths. –  Jan 14 '21 at 12:03

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As others have pointed out (namely, user DIdier_ ), if you have a non-rectifiable topological embedding $\gamma:I\to (X,d)$ (say, a portion of Koch curve) then the image of $\gamma$ is path-connected in the subspace topology, but it is disconnected in the intrinsic metric (say, $\gamma(0)$ and $\gamma(1)$ aren't in the same galaxy). In fact, a portion of Koch curve has discrete intrinsic topology.

A compact example where every two points are joined by a rectifiable path is this: call \begin{align}\Gamma_1&=\left\{(x,y)\in\Bbb R^2\,:\, 0<x\le\frac1\pi \land y=\sin\frac 1x\right\}\\\Gamma_2&=\{0\}\times [-1,2]\\ \Gamma_3&=\left(\left\{\pi^{-1}\right\}\times[0,2]\right)\cup\left(\left[0,\pi^{-1}\right]\times \{2\}\right)\end{align}

and consider $\Gamma=\Gamma_1\cup \Gamma_2\cup \Gamma_3$ with the Euclidean distance. Obviously, $(\Gamma,d_\ell)$ is homeomorphic (in point of fact, isometric) to $[0,\infty)$, so it isn't compact. But $\Gamma$ is.