Metric Spaces Topology

Question

Let $X_1,X_2,...,X_n$ be metric spaces, and $d_1,d_2,...,d_n$ be metrics for these spaces. Show that $$d\left[(x_1,....,x_n),(y_1,...,y_n)\right]:= \max\left\{d_i(x_i,y_i)|i=1,...n \right\}$$ is a metric for the space $\displaystyle X=\prod_{j=1}^{n}{X_j}$

Answer 1

for $d$ to be a metric on $X$ it has to have the following properties:

• $d(x,y)\geq 0$
• $d(x,y)=0 \Leftrightarrow x=y$
• $d(x,y)=d(y,x)$
• $\forall x,y,z\in X$, $d(x,y)\leq d(x,z)+ d(z,y)$

Now, we shall prove this properties. Let $x,y\in X$; $d(x,y)=d_i(x_i,y_i)\geq 0$ for some $i\in \{1,...,n \}$, if $x=y$, then $x_i=y_i$ for all $i\in \{1,...,n \}$, thus $d_i(x_i,y_i)=0$ for all $i=1,...,n$ and therefore $d(x,y)=0$, the converse is obvious by the same argument, as it is for the 3rd property; now, let $x,y,z\in X$, for $i=1,...,n$ it holds that $d_i(x_i,y_i) \leq d_i(x_i,z_i)+ d_i(z_i,y_i) $, let $j\in \{ 1,...,n\}$ be such that $d_j(x_j,y_j)=d(x,y)$, since $d(x,z) + d(z,y)\geq d_j(x_j,z_j) + d(z_j,y_j) $, we have that $d(x,y)\leq d(x,z)+ d(z,y)$ as desired, therefore $d$ is metric in $X$. $\square$