A useful condition ensuring equivalency of metrics

Question

I obtained a useful condition ensuring when two metrics generate the same topology, but for sake of confidence I post it here to see if this is correct.

“Suppose $d_1,d_2$ are metrics on a set $X$ and suppose there are positive constants $D,c,C$ such that $d_1(x,y)\leq D$ implies $cd_1(x,y)\leq d_2(x,y)\leq Cd_1(x,y)$. Then the topology generated by $d_2$ is contained in the topology generated by $d_1$. By symmetry, reversing the order of $d_1$ and $d_2$ gives a condition implying the reverse inclusion.” Is this claim right? This idea is from the textbook Elementary Topology, Problem Textbook.

If it is right, for example we can readily check that for any metric $d$, the two metrics $d$ and $d/(1+d)$ are equivalent.

Answer 1

Yes that's correct. Can you prove it?

You can also show the following, which is, I think, commonly used to show that two metrics induce the same topology (at least in all the exercises or other situations I met).

Let $d_1$ and $d_2$ be two metrics on a nonempty set $X$.

Suppose that there exists $\alpha,\beta>0$ such that $$\forall x,y\in X,\alpha d_1(x,y)\le d_2(x,y)\le\beta d_1(x,y).$$

Then $d_1$ and $d_2$ induce the same topology on $X$.