Lipschitz continuity and support of distance function.

Question

I do not see how the following assertion holds

Let $A \subset \mathbb{R}^n$ be a closed subset. define $\psi(x):=\operatorname{dist}(x,A)=\inf_{y \in A}d(x,y)$. Then $\psi$ is Lipschitz with $\operatorname{int}(\operatorname{supp}(\psi)) = A^c$.

Why is $\operatorname{int}(\operatorname{supp}(\psi)) = A^c$?

Thanks a lot in advance!

Answer 1

Let $y,z \in \mathbb{R}^n$.

Then, $\inf_a d(y,a) \leq \inf_a (d(y,z) + d(z,a)) = d(y,z) + \inf_a d(z,a)$ $$\implies \inf_a d(y,a) - \inf_a d(z,a) \leq d(y,z)$$

In the same way:

$$ \inf_a d(z,a) - \inf_a d(y,a)\leq d(y,z)$$

Conclude that $|d(A,y) - d(A,z)| \leq d(y,z)$

Hence, the function is $1-$Lipschitz. Can you handle the other assertion? (I don't have time right now, but if you don't find it I will post a hint)