Closedness of Voronoi cells.

Question

I have the following problem.

I managed to prove the convexity of $C(x,A)$ but not sure about its closedness. To make things simple I assume $x = \mathbf{0}$ and $C(x,A) =\bigcap H_i^-$, where $H_i$'s are the hyperplanes orthogonal to the segments $[x,y_i]$, $y_i \in A-x$, and divide those segments by half.

Answer 1

Note that the function $$ z\mapsto\|z-x\|-\|z-y\| $$ is continuous. Then for given $y$, the set $\{z: \ \|z-x\|-\|z-y\|\le0\}$ is closed. The set $C(x,A)$ is now the intersection of many such sets, hence closed.