Convergence of a point and that of its projection

Question

Let $K$ be a compact in $\mathbb{R}^{n}$ with $n>1$, and $x$ a point outside $K$. Let $ z_{0} $ be a fixed point on the boundary of $K$ and $ z $ the point on $K$ where the distance of $x$ and $K$ is attained. Intuitively it is clear that as $x\rightarrow z_{0}$ from outside $K$, the point $ z\rightarrow z_{0} $ from inside $K$, but how would you write the details of it?

Answer 1

We define $d_x = \|x-\cdot\| : K\to \mathbb{R}$ and note that the preimage of $$ d(x, K) := \inf\{\|x-z\| : z\in K\} = \inf d_x(K)$$ under $d_x$ is nonempty, because $d_x$ is continuous and therefore $d_x(K)$ is compact and contains its infimum. We therefore let $z(x)\in d_x^{-1}(d(x, K))$ for all $x$. As $d(x, K)$ is continuous (proof here), we have that $d(x, K)\to 0$ as $x\to z_0$. Then, $$\|z(x)-z_0\|\leq \|x-z(x)\|+\|x-z_0\| = d(x, K)+\|x-z_0\|\to 0$$ as $x\to z_0$ by the triangle inequality, so $z(x)\to z_0$.