strictly increasing, continuous, and subadditive $ f:\mathbb {R} \to \mathbb {R} _{+}$ such that $d_{2}=f\circ d_{1}$ Let $d_1$ be a metric on $X$. If there exists a strictly increasing, continuous, and subadditive $ f:\mathbb {R} \to \mathbb {R} _{+}$ such that $d_{2}=f\circ d_{1}$. Then $d_1,d_2$ are topologically equivalent.

Note : I found the proposition here:https://en.wikipedia.org/wiki/Equivalence_of_metrics
 A: I came up with a solution, and I will write it below; it seems the conditions of being subadditive is surplus. I think you needed the subadditivity if you wanted to show that $d_2$ is a metric. But when it is given, you won't need it.
Let $B^{1}_{r}(x) = \{ y \in X | d_1(x,y) < r\}$. We will show that $B^{1}_{r}(X)$ is open in the topology induced by the metric $d_2$. let $ y \in B^{1}_{r}(x)$. Then you have that : $d_2(x,y) =f \circ d_1(x,y) < f(r)$. So $y \in B^{2}_{f(r)}(x)$. now let $ z \in B^{2}_{f(r)}(x)$ then $ d_2(x,z) < f(r)$. which means that $f (d_1(x,z)) < f(r)$. As f is strictly increasing, we get $d_1(x,z) < r$. So $z \in B^{1}_{r}(x)$. Until now we have that the topology induced by $d_2$ is finer than the one from $d_1$.
Now let $r^{\prime} \in \Bbb R_+$. and consider the the open ball $B^{2}_{r^{\prime}}(x)=\{z \in X| d_2(x,z) < r^{\prime} \}$. For simplicity, suppose that : $$r^{\prime} < \sup f(d_1(X \times X)).$$ and I do not use the continuity condition in my proof. So it might be that my solution is incorrect.
Now let $z \in B^{2}_{r^{\prime}}(x)$. Thus we have :
$d_2(x,z) < r^{\prime}$. which means $ d_1(x,z) < f^{-1}(r^{\prime})$(we can take inverse from $r^{\prime}$ because $f$ is continuous and strictly increasing. Now let $ d_1(x , y) < f^{-1}(r)$. Then we have $d_2(x,y) < r^{\prime}$.
You should consider that $X, \emptyset$ are open in both of the topologies.
We have shown that the topology of $d_1$ is finer than $d_2$. Thus these metrics are topologically equivalent.
A: It is extremely simple once you read carefully the content in the link you post.
Note that there is an equivalent definition of equivalence of metric:

the open balls "nest": for any point $x \in X$ and any radius $r > 0$, there exist radii $r', r''$ such that
$B_{r'} (x; d_1) \subseteq B_r (x; d_2) \text{ and } B_{r''} (x; d_2)
\subseteq B_r (x; d_1).$

You can try to prove this.
Part of the solution:
Given a point $x$ and a radius $r$, we want to find $r'$ such that $B_{r'} (x; d_1) \subseteq B_r (x; d_2)$. Because the function $f$ is strictly increasing and continuous, it has an inverse function $f^{-1}$ which is also strictly increasing and continuous. Let $r' = f^{-1}(r) $. You can prove $B_{r'} (x; d_1) \subseteq B_r (x; d_2)$. Because $\forall y \in X, d_1(y,x) < r' = f^{-1}(r) \Longrightarrow f \circ d_1(y,x) < f \circ f^{-1}(r) \Longrightarrow d_2(y,x) < r$.
Now Given a point $x$ and a radius $r$, we want to find $r''$ such that $B_{r''} (x; d_2)\subseteq B_r (x; d_1)$. Let $r'' = f(r)$. You can proceed in the similar manner.
So, in fact, the conditions "sub-additivity" are superfluous if you already know $d_2$ is a metric already. However, these conditions are useful when you don't know $d_2$ is a metric and you want to construct a metric $d_2$ from $d_1$. see: this link
