I was wondering about the following (self-posed) question:
If $f\colon (X, d)\to (X, d')$ is a bi-Lipschitz bijection, then, is $f$ continuous on $(X, d)$ (i.e., is $f\colon (X, d)\to (X, d)$ continuous)?
This popped up in an attempt to analyze the following (again, self-posed) question:
If $f\colon (X, d)\to (X, d')$ is a bi-Lipschitz bijection, then, are the metric topologies on $X$ induced by $d$ and $d'$ the same?
Any insights into any of the above two questions?
If you want, I'd be happy to reproduce the path that led me to the former question from the former.
Partial answer: I could show that if $f$ is the identity function, then the answer to both the above posed questions are true (the first one being trivial).
So I have got an answer to both of my questions now.
The answer to the first and the second questions are both negative in general:
Let $f\colon (X, d)\to (X, d)$ be a discontinuous bijection. Now, we define a metric $d'$ on $X$ by ``transporting'' $d$ so that $$ d'(x, y) := d(f^{-1}(x), f^{-1}(y))\text. $$ Now, $f\colon (X, d)\to (X, d')$ is not just bi-Lipschitz, but also an isometry. Now, if $\mathcal T_d = \mathcal {T_{d'}}$, then the continuity of $f\colon (X, d)\to (X, d')$ would have implied the continuity of $f\colon (X, d)\to (X, d)$ (continuity being a topological property). Hence $\mathcal T_d \ne \mathcal T_{d'}$ even when $(X, d)$ and $(X, d')$ are isometric.
However, if $f$ is the identity function, then answer to the both the questions are affirmative:
The first question is trivially true. For the second question, consider a ball $B_\epsilon(x)$ in $(X, d)$. We show that it is open in $\mathcal T_{d'}$. Let $y\in B_\epsilon(x)$ and let $\delta > 0$ such that $B_\delta(y)\subseteq B_\epsilon(x)$ (we could take $\delta$ to be, for instance, $\epsilon - d(x, y)$). Now, we find a $\delta' > 0$ such that the ball $B'_{\delta'}(y)$ in $(X, d')$ is contained in $B_\delta(y)$. We can simply take $\delta < \epsilon c$ where $c$ is the "left Lipschitz constant": $c d(x, y) \le d'(x, y)$.