# Does a homeomorphism between metric spaces map open balls to open balls

compactnessgeneral-topologymetric-spacesopen-map

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 $$Y$$.

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.