TITLE: Distances inducing the same topology but with different balls
QUESTION [8 upvotes]: I'm looking for two different metrics $d,d'$ on the same underlying space $X$, such that 
1) $d$ and $d'$ induce the same topology on $X$;
2) there exists $x \in X$ such that small balls for $(X,d)$ and for $(X,d')$ with center $x$ are not homeomorphic.
This might be a trivial question, but I can't find an easy example.
EDIT: I'm mostly interested in the case $X = \mathbb{R}^n$ and $d$ is the Euclidean metric.

REPLY [2 votes]: Given any injective function $f:\mathbb R\to\mathbb R$ you can define a metric $d_f$ by
$$d_f(x,y):=|f(x)-f(y)|.$$
Consider the function $f$ qualitatively pictured below:

A possible analytic expression could be
$$f(x)=x+\sum_{n\in\mathbb Z} 4^{-n}\chi_{(4^{-n},2\cdot4^{-n})}+\sum_{n\in\mathbb Z} 2\cdot 4^{-n}\chi_{[2\cdot 4^{-n},4^{-(n-1)}]}$$
where the important thing is the alternation between open and closed intervals. 
Consider then
$$d:=d_f\qquad \text{and} \qquad d':=d_{2f}=2d_f.$$
They induce the same topologies because the collection of all balls is exactly the same, since $B^d(x,r)=B^{d'}(x,2r)$.
However if you choose any $r$ not in $\overline{f(\mathbb{R})}$ (which you can do for arbitrarily small values of $r$), the ball of center zero and radius $r$ for one metric will be a right-open interval, while for the other metric it will be a right-closed interval, therefore they are not homeomorphic.