prove a function $f(x,y) = \max(x,y)$ is continuous Consider the function $f : R^2 →R$ given by $f(x,y) = \max(x,y)$. (That is, $f(x,y)$ is the larger of $x$ and $y$, so $f(−3,2) = 2$, $f(1,4) = 4$, and $f(−3,−2) = −2$.) (assume that $R^2$ has sup metric)
prove that $f$ is continuous.
I got $d(x,y)-d(p,q) = \max \{|x-p|,|y-q|\} $  and $d(f(x,y)-f(p,q)) = \max(x,y)-\max(p,q)$

Hint: $\max\{x,y\} = \dfrac{x + y + |x - y|}{2}$
