Measurable selection for argmin to distance

Question

Let $Y$ be a Banach space and equip $Y$ with the weak topology. Now, let $X$ be a closed, bounded, and convex subset of $Y$ and suppose that the relative (weak) topology on $X$ is metrizable with metric $d_X$. Let $x_1,\dots,x_n \in X$ for some natural $n>0$.

Let $\emptyset\neq Z\subseteq X$ be a compact subset of $X$. Then, does the map $$ x\mapsto \min_{1\leq i\leq n}\, d_X(x_i,x) $$ admit a measurable selection? I.e.: Does there exist a measurable function $$ S\mapsto \{1,\dots,n\} \mbox{ s.t. } d_{X}(x_{S(x)},x)=\min_{i=1,\dots,n} d_{X}(x_i,x) $$ for all $x\in Z$ and $S$ is measurable as a function from $Z$ to $\{1,\dots,n\}$ (where the latter has the $\sigma$-algebra $2^{\{1,\dots,n\}}$?)

Answer 1

Here is a simple direct argument. For $i=1,\ldots,n$, let $$C_i=\{z\in Z\mid d(z,x_i)\leq d(z,x_j), j=1,\ldots,n\}.$$ Clearly, each $C_i$ is closed and hence measurable. Let $M_i=C_i\setminus\bigcup_{l=1}^{i-1}C_l$. The nonempty sets of the form $M_i$ form a finite measurable partition of $Z$. Now let $S$ map each $M_i$ to $x_i$.