How can one measure distance between point and the line in maximum metric space?

Question

Given metric space $M = (\mathbb{R}^2, d)$ where $d = \operatorname{max}\{|x_1 - y_1|, |x_2 - y_2|\}$, how can one measure distance from some arbitrary point $X$ to the line $y = 3$, let's say?

How it can be done, by use of the identity $$\operatorname{max}\{x, y\} = \frac{1}{2}(x + y+ |x - y|)\text{?}$$

Answer 1

Denote your point by $(x_0,y_0)$ and the line $y = 3$ by $l$. You have

$$ d((x_0,y_0),l) = \inf_{x \in \mathbb{R}} d((x_0,y_0), (x,3)) = \inf_{x \in \mathbb{R}} \max \{ |x_0 - x|, |y_0 - 3| \}. $$

It is clear that $\max \{ |x_0 - x|, |y_0 - 3| \} \geq |y_0 - 3|$ and so $d((x_0,y_0),l) \geq |y_0 - 3|$. By taking $x = x_0$, we see that the infimum is actually attained and $d((x_0,y_0),l) = |y_0 - 3|$. It is worth interpreting this calculation geometrically using the description of the circles in this norm as squares.

Answer 2

Hint: the "circles" of radius $r$ about $X$ in $M$ are squares with edges parallel to the axes that are centred on $X$ and have sides of length $r/2$. See https://en.wikipedia.org/wiki/Uniform_norm or https://en.wikipedia.org/wiki/Norm_(mathematics). To find the distance from $X$ to a line $l$ find the smallest of these squares that has a vertex on $l$.