Let $d[(x,y),(x',y')]=max\{d(x,x'),d(y,y')\}$, show that the sphere of center $(a,b)$ and radius $r$ in $M\times N$ is equal to $(B[a,r]\times S(b,r))\cup (S(a,r)\times B[b,r])$.
I know that the metric in $\mathbb{R}^2$, represent a square of sides parallel to axis cordinates, and the closed ball is the disjoint union of open ball with a sphere. This exercises take me much time, any helps ,pls!
Denote by $S((a,b),r)$ the sphere in $M\times N$ whose centre is $(a,b)$ and radius $r$.
Inclusion in one direction:
We want to prove: $$S((a,b),r)\subset (B[a,r]\times S(b,r))\cup(S(a,r)\times B[b,r])\enspace (1)$$
Assume that $(x,y)\in S((a,b),r)$. By definition, $\max\{d(x,a),d(y,b)\}=r$.
Case 1. The maximum is attained with $d(x,a)$. So $d(x,a)=r$, and $d(y,b)\leq r$. In this case $(x,y)\in S(a,r)\times B[b,r]$, and in particular, $(x,y)$ belongs to the right-hand-side of (1).
Case 2. The maximum is attained with $d(y,b)$. So $d(y,b)=r$ and $d(x,a)\leq r$. In this case $(x,y)\in B[a,r]\times S(b,r)$, and in particular, $(x,y)$ belongs to the right-hand-side of (1).
Since either case 1 or case 2 occurs, we have proved the inclusion (1).
Inclusion in the other direction:
We want to prove: $$S(a,b,r)\supset (B[a,r]\times S(b,r))\cup(S(a,r)\times B[b,r])\enspace (2)$$
Assume that $(x,y)\in (B[a,r]\times S(b,r))\cup(S(a,r)\times B[b,r])$.
Then either
$(x,y)\in B[a,r]\times S(b,r)$,
$\Longrightarrow$ $d(x,a)\leq r$ and $d(y,b)=r$,
$\Longrightarrow$ $\max\{d(x,a),d(y,b)\}=r$,
$\Longrightarrow$ $(x,y)\in S((a,b),r)$;
or
$(x,y)\in S(a,r)\times B[b,r]$,
$\Longrightarrow$ $d(x,a)=r$ and $d(y,b)\leq r$,
$\Longrightarrow$ $\max\{d(x,a),d(y,b)\}=r$,
$\Longrightarrow$ $(x,y)\in S((a,b),r)$.
Since either one or the other case occurs, we have proved the inclusion (2).
From the two inclusions (1) and (2), we obtain the set-equality:
$$S(a,b,r)= (B[a,r]\times S(b,r))\cup(S(a,r)\times B[b,r])$$
Q.E.D
