0

Let $(K,d)$ a compact metric space and $Hom(K)$ denotes the set of all homeomorphisms on $K$. Define a map $\rho:Hom(K) \times Hom(K) \rightarrow \mathbb{R}$ by

$$\rho(\phi_1,\phi_2)=sup\{d(\phi_1(x),\phi_2(x)),d(\phi_1^{-1}(x),\phi_2^{-1}(x))\mid x\in K\}.$$

I want to show that $\rho$ is a metric on $Hom(K)$. I have proved all the conditions except the triangle inequality. Any help is appreciated.

LoveMath
  • 109

1 Answers1

1

For any $\varepsilon>0$, according to the definition of the supremum, there exists $x_0\in K$, such that $$\rho(\phi_1,\phi_3)\leq d(\phi_1(x_0),\phi_3(x_0))+\varepsilon\quad {\rm or} \quad\rho(\phi_1,\phi_3)\leq d(\phi_1^{-1}(x_0),\phi_3^{-1}(x_0))+\varepsilon.$$ Therefore, \begin{equation*} \begin{aligned} \rho(\phi_1,\phi_3)&\leq d(\phi_1(x_0),\phi_3(x_0))+\varepsilon\\ &\leq d(\phi_1(x_0),\phi_2(x_0))+d(\phi_2(x_0),\phi_3(x_0))+\varepsilon\\ &\leq \rho(\phi_1,\phi_2)+\rho(\phi_2,\phi_3)+\varepsilon. \end{aligned} \end{equation*} or \begin{equation*} \begin{aligned} \rho(\phi_1,\phi_3)&\leq d(\phi_1^{-1}(x_0),\phi_3^{-1}(x_0))+\varepsilon\\ &\leq d(\phi_1^{-1}(x_0),\phi_2^{-1}(x_0))+d(\phi_2^{-1}(x_0),\phi_3^{-1}(x_0))+\varepsilon\\ &\leq \rho(\phi_1,\phi_2)+\rho(\phi_2,\phi_3)+\varepsilon. \end{aligned} \end{equation*} Because of the arbitrariness of $\varepsilon$, we obtain the triangle inequality.

HeroZhang001
  • 2,201
  • Indeed, as seen here, the triangle inequality also holds for non-compact $K$; strictly speaking, compactness is neede “only” to ensure finiteness of $\rho$. Nevertheless, one can make use of compactness for the triangle inequality as well if desired - and get rid of the need to use $\epsilon$. – Hagen von Eitzen Jan 29 '23 at 07:14