Let $X$ be a set and $d_1, d_2$ be metrics on
$X$ so that for constants $m,M > 0$ and any $x,y \in X$ we have
$md_1(x,y) \leq d_2(x,y) \leq Md_1(x,y)$
Prove that $f: X \rightarrow \mathbb{R}$ is continuous with respect to the metric $d_1$ on $X$ iff it is continuous with respect to the metric $d_2$ on $X$.
My attempt:
$\Rightarrow f$ is continuous w.r.t. to the metric $d_1$
Then $\exists > 0$ and $\delta > 0$ such that $d_1(x,y) < \frac{\delta}{M}$ implies $d_1(f(x),f(y)) < \epsilon$.
Then $d_2(x,y) \leq Md_1(x,y) < \delta$.
And so $d_2(x,y) < \delta$.
From here, I don't know how to show that $d_2(x,y)$ implies $d_2(f(x),f(y))$.
Best Answer
Take $x\in X$ and suppose that $f$ is continuous at $x$ with repect to the metric $d_1$. Take $\varepsilon>0$. You know that there is a $\delta>0$ such that$$d_1(x,y)<\frac\delta m\implies\bigl\lvert f(y)-f(x)\bigr\rvert<\varepsilon.$$But then$$d_2(x,y)<\delta\implies md_1(x,y)<\delta\iff d_1(x,y)<\frac\delta m\implies\bigl\lvert f(y)-f(x)\bigr\rvert<\varepsilon.$$So, $f$ is continuous at $x$ with respect to the metric $d_2$.
By a similar argument, if $f$ is continuous at $x$ with repect to the metric $d_2$, then it is continuous with respect to the metric $d_1$.