Let $h:\mathbb{R}^2 \to \mathbb{R}^2$ be a surjective function such that $||h(x)-h(y)||\ge 3||x-y||$ for all $x,y\in \mathbb{R}^2$.Here $||.||$ denote the Eucleadian norm on $\mathbb{R}^2$.Show that the image of every open set(in $\mathbb{R}^2$) under the map $h$ is an open set(in $\mathbb{R}^2$)
Well the map h is surjective as well as injective and if we prove that h is lipschitz then continuous bijective mapping maps open sets to open sets but proving the continuous part Not being able to prove the continuity part
A function that maps open sets to open sets need not be continuous. In fact, this is in a sense opposite to the definition of continuity (where the inverse image of an open set is open). As you have observed, $h$ is a bijection, and so it has an inverse. What does your inequality imply about this inverse function?