Conditon for $f(x)$ such that $f(d(x, y))$ induces the same topology with $d(x, y)$ On Munkres' Topology, Sec 20, Q 11, for $f(x)=\frac{x}{1+x}$, $f(d(x, y))$ induces the same topology with $d(x, y)$.
And I have some question here : what conditions for $f(x)$ could make same result?
My attempt : If $f(x)$ satisfies

*

*$f(a+b) \leq f(a)+f(b)$,

*$f(0)=0$, and $0< f(x) \leq 1$ for $0 < x$
edit) Maybe, latter of condition 2) can be replaced by "$f$ is bounded"!
But, I can't prove that it induces the same topology. Does it need some specified condition?
 A: This can be seen as part of a more general problem. We say that two metrics $d_1$ and $d_2$ are equivalent if they induce the same underlying topology. The basic theorem characterizing equivalent metrics is the following.
Theorem: Let $d_1$ and $d_2$ be metrics on some space $X$. Then $d_1$ and $d_2$ are equivalent if and only if for all $\epsilon > 0$ and $x\in X$, there exists $\delta_1 > 0$ and $\delta_2 > 0$ such that
$$B_1(x,\delta_1) \subseteq B_2(x,\epsilon),\qquad\text{and}\qquad B_2(x,\delta_2) \subseteq B_1(x,\epsilon),$$
where $B_i(x,r)$ is the open ball of radius $r$ centered on $x$, as determined by the metric $d_i$.
I won't prove this theorem here. It's not difficult, and you should be able to find a proof rather easily when you search for "topologically equivalent metrics". Roughly speaking, the theorem says that two metrics are equivalent if and only if you can find an open ball of one inside any open ball of the other.
What does this mean for the functions $f$? First, let us require that $f(d(x,y))$ remain a metric. This means that we should have $f(r) \ge 0$ for all $r\in \mathbb{R}^+$, with $f(r) = 0$ if and only if $r=0$. We should also require that $f$ be a non-decreasing function such that $f(r+s) \le f(r) + f(s)$ for all $r,s\in \mathbb{R}^+$. This will ensure that the triangle inequality is satisfied.
Fix some $\epsilon > 0$ and $x \in X$. Let $d_1 = d$ and $d_2 = f\circ d$. Then the condition $B_1(x,\delta_1) \subseteq B_2(x,\epsilon)$ is equivalent to requiring the existence of some $\delta_1$ such that $d(x,y) < \delta_1$ implies $f(d(x,y)) < \epsilon$. This is simply continuity of $f$ at $0$.
The condition $B_2(x,\delta_2) \subseteq B_1(x,\epsilon)$ is equivalent to requiring the existence of some $\delta_2$ such that $f(d(x,y)) < \delta_2$ implies $d(x,y) < \epsilon$. This latter condition is automatically satisfied for any non-decreasing function. Indeed, pick $\delta_2 = f(\epsilon)$. Then $f(d(x,y)) < \delta_2 = f(\epsilon)$ implies $d(x,y) < \epsilon$.
Therefore the following requirements are sufficient for $f:[0,\infty)\rightarrow [0,\infty)$ to determine an equivalent metric:

*

*$f$ is non-decreasing.

*$f(r) = 0$ if and only if $r=0$.

*$f$ is sub-additive, i.e., $f(r+s) \le f(r) + f(s)$.

*$f$ is continuous at $0$.

Notice that the first 3 conditions are just generic conditions for $f\circ d$ to be a metric again. Any metric determined by such a function will define a topology which is at least as fine as the original topology. The new topology will be equivalent if and only if $f$ is continuous at $0$. In particular, note that all four of the conditions above are satisfied by the function $f(r) = r/(1+r)$, so that it determines an equivalent topology.
