Unique nearest point property

Question

Consider $\mathbb{R}^2$ with a norm defined by $\|(x,y)\| = |x|+|y|$. Define $\mathrm{dist}(K,p) = \inf_{q \in K} \|q-p\|$. Why are there infinitely many points $q \in K$ that satisfy $\|p-q\| = \mathrm{dist}(K,p)$ when $K$ is the line $y=x$ and $p=\left(\frac{1}{2}, -\frac{1}{2}\right)$?

Answer 1

For a given point $a$ in the plane, the points $z$ located at a constant distance from $a$ are a square whose sides are parallel to the lines $y=x$ and $y=-x$. This is why the point $(1/2,-1/2)$ has infinitely many points located on the line $y=x$ at the same distance to him.

Answer 2

From the definitions, we have: $$ \|p - q\| = \|(\tfrac{1}{2} - x, \tfrac{-1}{2} - x)\| = |\tfrac{1}{2} - x| + |\tfrac{1}{2} + x| = \begin{cases} -2x &\text{if } x < \tfrac{-1}{2} \\ 1 &\text{if } \tfrac{-1}{2} \leq x \leq \tfrac{1}{2} \\ 2x &\text{if } x > \tfrac{1}{2} \\ \end{cases} $$ Thus, we see that any point of the form $q = (x, x) \in K$ where $\tfrac{-1}{2} \leq x \leq \tfrac{1}{2}$ will have minimum distance of $1$ away from $p$.