TITLE: How can I show these two metrics give the same topology?
QUESTION [5 upvotes]: This question came up while showing the composition of a metric with a certain other 
function gives another metric. Suppose I have some metric space $(X,d)$ and a 
continuous, non-decreasing function $f$ on the nonnegative reals. Moreover, suppose 
that $f(x)=0$ iff $x=0$, and $f$ also satisfies the triangle inequality in that $f
(x+y)\leq f(x)+f(y)$. 
Using these properties, it is not difficult to show that $f\circ d$ is yet another 
metric on $X$, so $(X,f\circ d)$ is also a metric space. 
I notice the (open) balls given by the metrics are of form 
$$
B_d(x,r)=\{y\in X\mid d(x,y)\lt r\}
$$ and 
$$
B_{f\circ d}(x,r)=\{y\in X\mid (f\circ d)(x,y)\lt r\},
$$ 
so it seems that the topologies generated by the base of open balls in each case 
would probably be the same. I would like to see how one would go about showing the topologies given by these two metrics are indeed the same. Thanks for any insight.

REPLY [2 votes]: Here is a Theorem from my book 
Topology, Sts. Cyril and Methodius University, Skopje, 2002 (in Macedonian) . 
The proof consists of five rows.  
Theorem.   X is a set with two topologies T1  and T2 , B1   is a basis for topology T1  and B2  is a basis for topology T2 . If for every B1 in B1 and every point x in B1 there exists B2 in B2 , such that 
x is in B2 and B2 is a subset of B1,
then T1 is a subset of T2 . 
Nikita Shekutkovski