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 $d$?


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.

  • 2
    $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.