Does a homeomorphism between two compact metric spaces preserve open balls? More specifically, if $X$ and $Y$ are compact metric spaces, and there is a $\phi: X \to Y$ a homeomorphism, then is it true that $\phi^{-1}(B(\phi(x),r))= B(x,r)$ ? If so, how? Thanks in advance!
 A: Not necessarily. For that, you would need an isometry.
Consider $X=Y=[0,1]$ with the usual metric. Then for any integer $n>1,$ the map $x\mapsto x^n$ is a homeomorphism, but doesn't preserve open balls.
A: No: Consider $(x,y,z) \mapsto (3x,y,z).$ That is a homeomorphism from $\mathbb R^3$ to itself, and it changes spheres to ellipsoids.
Alright so one wants an example in which the space is compact.
Let the space be $\{z \in \mathbb C : |z| = 1\}.$
Let distances be arc lengths along this circle.
Let $f(z) = \dfrac{2z+1}{z+2}.$
I'll leave it as an exercise to see that this is a homeomorphism of this space.
It maps $1$ to $1,$ $-1$ to $-1,$ and $i$ to $(4+3i)/5.$ The distance between $1$ and $i$ is more than the distance between $f(1)$ and $f(i).$
A: Let $X$ be the set of vertices of an isosceles right triangle, and let $Y$ be the set of vertices of an equilateral triangle. Then $X$ and $Y$ are homeomorphic compact metric spaces, but $X$ has $6$ open balls, while $Y$ has only $4.$
A: The metrics $d_1((x,y),(x',y'))=((x-x')^2+(y-y')^2)^{1/2}$ and $d_2((x,y),(x',y'))=\max (|x-x'|,|y-y'|)$ generate the same topology on $\mathbb R^2.$ 
So $id_{\mathbb R^2}$ is a homeomorphism from $(\mathbb R^2,d_1) $ to $\mathbb R^2,d_2).$ But $d_1$-balls are round and $d_2$-balls are square.
