Let $\mathbb{B}(X; M \times N)$ be the set of bounded functions $f: X \rightarrow M\times N$ where $X$ is a set and $M,N$ are metric spaces. Find an isometry between $\mathbb{B}(X;M \times N)$ and $\mathbb{B}(X;M) \times \mathbb{B}(X;N)$. Choose a suitable metric for each Cartesian product.
I'm really lost in this one, my plan was to use the metric $$d'(f,g) = \operatorname{Sup}_{x \in X}d(f(x),g(x))$$ on $\mathbb{B}(X; M \times N)$ but I can't visualize the map $\phi: \mathbb{B}(X;M \times N) \rightarrow \mathbb{B}(X;M) \times \mathbb{B}(X;N)$ so then I can choose a metric $d$ in $M \times N$ to $\phi$ be an isometry.
Any help or hint is appreciated.
I think you can do the following. Suppose $f\in\mathbb{B}(X;M\times N)$. Then $f(x)=(f_1(x),f_2(x))$ with $f_1(x)\in M$ and $f_2(x)\in N$. Then you can consider the map $$ f\to (f_1,f_2)\in\mathbb{B}(X;M)\times\mathbb{B}(X;N). $$ On $\mathbb{B}(X;M)$ and $\mathbb{B}(X;M)$ you can take the sup metrics $$ d_1(g,h)=\sup_X d_M(g(x),h(x));\quad d_2(g,h)=\sup_X d_N(g(x),h(x)) $$ On $\mathbb{B}(X;M)\times\mathbb{B}(X;N)$ you can choose the distance $$ \tilde d((g,h),(j,k))=d_1(g,j)+d_2(h,k) $$ Now if you define the distance on $\mathbb{B}(X;M\times N)$ as $$ \hat d(f,g)=d_1(f_1,g_1)+d_2(f_2,g_2) $$ where $f_1,f_2$ and $g_1,g_2$ are the "components" of $f$ and $g$, you clearly have an isometry.