# Prove that $f: X \rightarrow R$ is continuous with respect to the metric $d_1$ on $X$ iff it is continuous with respect to the metric $d_2$ on $X$.

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

• 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. – thedilated May 20 at 6:52

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 $$d_1$$.

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.