Is there an example of a metric space $(X,d)$ whose corresponding [path metric](https://en.wikipedia.org/wiki/Intrinsic_metric), $d^\prime$ generates a strictly finer topology compared to the topology generated by $d$?



Path Metric Topology
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.


