Path Metric Topology 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 $d$?
 A: 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.
