Two questions about continuous mappings of compact metric spaces into themselves.

Question

Let $M$ be a compact metric space with distance function $d$ and let $T$ be the set of all continuous mappings of $M$ into itself. If $f_1$ and $f_2$ are any pair of mappings in $T$, let $t$ be $$ t(f_1,f_2):=\sup_{x\in M}\{d(f_1(x),f_2(x))\}. $$ Is $t$ a metric for $T$? If so, is $T$-with distance function $t$-a compact metric space?

Answer 1

Yes, you can prove by directly verifying the axioms for a metric:

• $t(f,g)=0$ iff $f=g$ for all $f,g\in T$. Indeed, $$ t(f,f)=\sup_{x\in M}\{\underbrace{d(f(x),f(x))}_{=0\text{ since $d$ is a metric }}\}=\sup_{x\in M}\{0\}=0. $$ and $$ t(f,g)=0\Rightarrow\sup_{x\in M}\{d(f(x),g(x))\}=0\Rightarrow d(f(x),g(x))=0~\forall x\in M\Rightarrow f=g. $$
• $t(f,g)=t(g,f)$ for all $f,g\in T$. Indeed, $$ t(f,g)=\sup_{x\in M}\{d(f(x),g(x))\}=\sup_{x\in M}\{d(g(x),f(x))\}=t(g,f), $$ where I have used the fact that $d$ is a metric in the second equality.
• $t(f,g)\leq t(f,h)+t(h,g)$ for all $f,g,h\in M$. Indeed, $$ t(f,g)=\sup_{x\in M}\{d(f(x),g(x))\}\leq\sup_{x\in M}\{d(f(x),h(x))+d(h(x),g(x))\}=\sup_{x\in M}\{d(f(x),h(x))\}+\sup_{x\in M}\{d(h(x),g(x))\}=t(f,h)+t(h,g) $$ As far as the second part of the question is concerned, it is sufficient to consider $M:=[0,1]$, which gives rise to a counterexample.