Show that these metrics induces the same topology on X Let $X$ be the set of positive integers. Let $d_1$ be the usual metric space on $X$ and $d_2$ be the discrete metric on $X$. Define $d_3:X\times X \rightarrow R$ by $d_3(m,n)=|\frac{1}{m}-\frac{1}{n}|$ for $m,n\in X$. Prove that $d_3$ is also a metric on $X$ and $d_1,d_2,d_3$ all induce the same topology on $X$.

I've proved that $d_3$ is also metric.. But I could not proved that all of these induces the same topology. I know that, in order to prove that, these metrices induces that the same topology... I have to prove that all are equivalent metrices. I am stuck at this part.
I'll give one equivalence and edit this answer if OP struggles with the others:

Claim: $d_2 \tilde{} \space d_3$ with the notation as in the question.

Proof:

Observe that $d_3(m,n)= \vert \frac{1}{m}-\frac{1}{n} \vert \leq 1$ for all positive integers $m,n$. Therefore we have:

$$d_3(m,n) \leq d_2(m,n) \leq 2d_3(m,n)$$ 
since (w.l.o.g $n &gt;m$):

$$ 2\vert \frac{1}{m}-\frac{1}{n} \vert=\frac{2}{m}-\frac{2}{n}=\frac{2(n-m)}{mn}= 1+\frac{2n-(mn+2m)}{mn}$$ and because

$$1 \geq \frac{m}{2}+ \frac{m}{n} \implies 2n \geq mn+2m$$

we can conclude the equivalence.
All metrics induce the discrete topology (i.e. all sets are open). Forget about any equivalence of metrics, this is a stronger requirement (which does not hold in this case) that is not needed for equivalence of topologies