A homeomorphism $h: (X,d_1) \rightarrow (Y,d_2)$ is a bijection between two metric spaces $(X,d_1)$ and $(Y,d_2)$ such that $h$ maps open balls of $X$ to the interior of open balls of $Y$ and $h^{-1}$ maps open balls of $Y$ to the interior of open balls of $X$. It is also true that $h$ maps open sets of $X$ to open sets of $Y$, and $h^{-1}$ maps open sets of $Y$ to open sets of $X$. The open sets of metric spaces are either open balls or unions of open balls.
Is it true that, for a homeomorphism, the image of an open ball of $X$ must be an open ball of $Y$? Does the answer change if both metric spaces are compact?
COMMENT: The convoluted counter-scenario I'm imagining is that $h$ could map an open ball of $X$ into a union of some open balls of $Y$ that isn't itself an open ball, but is contained within some other open ball in $Y$.
Best Answer
No, it is not true. Take, for instance, the identity map from $(\Bbb R^2,d_1)$ onto $(\Bbb R^2,d_2)$, with$$d_1\bigl((x_1,x_2),(y_1,y_2)\bigr)=|x_1-y_1|+|x_2-y_2|$$and$$d_2\bigl((x_1,x_2),(y_1,y_2)\bigr)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}.$$It is a homeomorphism, but the open balls on the left are squares, whereas the open balls on the right are disks. And compacity changes nothing.