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Is there an example of a metric space $(X,d)$ whose corresponding path metric, $d^\prime$ generates a strictly finer topology compared to the topology generated by $d$?

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Consider "the topologist's sine"

$$X = \left\{ \left(x, \sin \frac 1 x \right) \mid x>0 \right\} \cup \Big( \{0\} \times [-1,1] \Big) \subset \mathbb R^2$$

endowed with the distance induced by its natural embedding in $\mathbb R^2$. Clearly the sequence given by $s_n = (\frac 1 {2n \pi}, 0)$ converges to $(0,0)$ in the induced topology.

Endow now $X$ with the Riemannian structure induced by the Euclidean one on $\mathbb R^2$. A moment of reflection will convince you that in the path metric associated to it, the distance between $s_m$ and $s_n$ (with $m<n$) is given by

$$\int \limits _{\frac 1 {2n \pi}} ^{\frac 1 {2m \pi}} \sqrt {1 + \frac 1 {x^4} \cos^2 \frac 1 x} \ \mathrm d x = \int \limits _{2m \pi} ^{2n \pi} \frac 1 {t^2} \sqrt{1 + t^4 \cos^2 t} \ \mathrm d t \ge \int \limits _{2m \pi} ^{2n \pi} |\cos t| \ \mathrm d t = 4 (n-m) ,$$

which shows that in the path metric $(s_n)_{n \ge 1}$ is not Cauchy. We conclude that the path metric has strictly fewer convergent sequences, therefore it is strictly finer.

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    $\begingroup$ Rigorously speaking, $X$ is not a smooth submanifold of $\mathbb R^2$, therefore one cannot pull the Riemannian structure from $\mathbb R^2$ onto it. On the other hand, $\left\{ \left(x, \sin \frac 1 x \right) \mid x>0 \right\}$ is a smooth submanifold, and this is what matters in the above example. Intuitively, the closer you get to $(0,0)$ in the induced distance, the more the intrinsic path distance grows. In particular, $X$ is bounded in the induced distance, but unbounded (to the left) in the path distance. $\endgroup$ – Alex M. Jul 26 '18 at 9:32
  • $\begingroup$ I am glad you took time to explain the intuition behind the example as I was trying to construct similar ones. Thank you! $\endgroup$ – Temari Jul 27 '18 at 4:10

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