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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 $d_2(f(x),f(y))$.

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  • $\begingroup$ You can show the following: Given any open set in $(X,d_1)$, it is also open in the metric $(X,d_2)$ and vice versa. So this means that if $f:(X,d_1)\rightarrow \mathbb{R}$ is continuous, then for any open $E\subset \mathbb{R}$, the set $f^{-1}(E)$ is open in $(X,d_1)$ and hence open in $(X,d_2)$. So $f:(X,d_2)\rightarrow \mathbb{R}$ is continuous. $\endgroup$ – thedilated May 20 at 6:52
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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 $d_1$.

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In context:

Suffices to show that

1) $O \subset X$, open in $(X,d_1)$, then open in $(X,d_2)$;

2) $O \subset X$, open in $(X,d_2)$, then open in $(X,d_1)$.

1): Let $O$ be open in $(X,d_1)$.

There is a $\delta /m >0$ s.t.

$B_1(x): =$ { $y|d_1(x,y) <\delta/m$} $\subset O$.

Let $d_2(x,y) < \delta$.

Then

$m d_1(x,y) \le d_2(x,y) < \delta$ implies

$d_1(x,y) <\delta/m$, i.e. $y \in O$, and

$B_2(x) = $ { $y|d_2(x,y)<\delta $} $\subset O$.

Similarly for 2).

To sum up:

Every open set in $(X,d_1)$ is open in $(X,d_2)$, and vice versa.

Hence:

If $f$ continuos on $(X,d_1)$ then $f$ continuos on $(X,d_2)$, and vice versa.

Cf. Comment of user thedilated.

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