Topological group, a big exercise!

Question

Let $(G,d)$ be a group that is also a compact metric space, and G is topological (The maps $G \times G \to G, (x,y) \to xy$ and $G \to G, x \to x^{-1}$ are continuous). And then we consider $(G,\rho)$, where $\rho(x,y)=\inf\limits_{g \in G} d(gx,gy)$ is assumed to be a metric.

The first question consists of showing that $\rho$ is $G$ left invariant. (if $h \in G$, $$\rho(hx,hy)=\inf\limits_{g \in G} d(ghx,ghy)=\inf\limits_{k \in G} d(kx,ky)=\rho(x,y),$$ since the map $g \to gh$ is a bijection on $G$).

I am clueless for the remaining question that asks us to show that the the identity mapping from $(G,d)$ to $(G,\rho)$ is Lipschitz and from $(G,\rho )$ to $(G,d)$ is uniformly continuous. (For the second, it is enough to prove continuity, and the rest follows by compactness). Edited: The first part is done, thanks to a small but clever comment by John Hughes. As for the second part, I tried to think a little, but came nowhere, as I thought that, fixing $x,y$ in $G$, and we consider $f(g)=d(gx,gy)$. I want to prove that this map is continuous, and if true, this might make it easier, as $\rho(x,y)$ will be attained by compactness.

Answer 1

To see that $\rho$ is a metric, if $x=y$ then $$\rho(x,x)=d(ex,ex)=d(x,x)=0$$ and if $\rho(x,y)=0$, then $x=y$ follows from @user254665's comment to my answer. $\rho(x,y)=\rho(y,x)$ is clear from symmetry. For the triangle inequality, we have for $x,y,z\in G$ \begin{align} \rho(x,y)+\rho(y,z) &= \inf_{g\in G} d(gx, gy) + \inf_{h\in G} d(hy, hz)\\ &\geqslant \inf_{g\in G} \left(d(gx, gy) + d(gy, gz)\right)\\ \end{align} since $$\{d(gx,gy)+d(hy,hz):g,h\in G\}\subset \{d(gx,gy)+d(gy,gz):g\in G\} $$ and from the triangle inequality of $d$ we obtain $$d(gx,gy) + d(gy,gz)\geqslant d(gx,gz) $$ for all $g\in G$, which implies that $$\rho(x,y)+\rho(y,z)\geqslant \rho(x,z). $$

As for showing that $\rho$ is $G$-left invariant, your proof looks correct to me.

To show that the identity map from $(G,\rho)$ to $(G,d)$ is $1$-Lipschitz, @John Hughes's hint is sufficient.

As for the last question, I am not sure. But due to compactness of $(G,d)$, it is sufficient to show that a sequence $\{x_n\}$ which converges with respect to $\rho$ has a Cauchy subsequence with respect to $d$.

Answer 2

ANSWER TO THE LAST PART.....We need these basic facts about the compact metric space $(G,d)$ :

(1) Any metric for a compact metric space is a complete metric .

(2) If $k \in N$ and $(x_{j,n})_{n \in N}$ is a sequence of members of $G$ for $j=1,...,k$ then there is a strictly increasing $f:N \to N$ such that for each $j=1,...,k$ the sequence $(x_{j,f(n)})_{n \in N}$ is convergent to a limit.

(3) If $(x_n)_{n \in N}$ and $(y_n)_{n \in N}$ are sequences in $G$ converging respectively to $x$ and $y$ then $(x_n y_n)_{n \in N}$ converges to $x y$. To prove this last assertion, let $V$ be any nbhd of $x y$ . Then $ U=\{(a,b) \in G^2 :a b \in V \}$ is a nbhd of $(x,y)$ in $ G^2$ , so there are open sets $A,B$ of $ G$ with $x \in A$ and $y \in B$ and $A\times B \subset U$. Now $A^*= \{n :x_n \not \in A\}$ and $B^*= \{n :y_n \not \in B\}$ are finite, so $\{n :x_n y_n \not \in U\} \subset A^*\cup B^*$, which is finite....... Now to show that $id_G$ is uniformly continuous from $(G,\rho)$ to $ (G,d)$ by contradiction, suppose it is not. Then for some $r>0$ there are sequences $(x_n)_n$ and $ (y_n)_n $ in G such that $$\lim_{n \to \infty} b_n=0 \text{ where } b_n=\rho(x_n,y_n),$$ $$\text{ and } d(x_n,y_n)\ge r \text{ for all } n.$$ Observe that $b_n=\rho(x_n,y_n)\ne 0$, because $\rho$ is a metric and $d(x_n,y_n)\ge r >0 \implies x_n \ne y_n$. So by the def'n of $\rho $ take $g_n \in G$ with $d(g_n x_n,g_n y_n)<2 b_n$ .Now by (2) take a strictly increasing $f:N\to N$ where with respect to the metric $d$ we have $$x=\lim_{n\to \infty}x_{f(n)},$$ $$ y=\lim_{n\to \infty}y_{f(n)},$$ $$g=\lim_{n \to \infty}g_{f(n)}.$$ Therefore by (3) we have $$gx=\lim _{n \to \infty} g_{f(n)} x_{f(n)},$$ $$ gy=\lim_{n \to \infty} g_{f(n)} y_{f(n)} .$$ But $$gx=gy$$ because $$\lim_{n \to \infty} d(g_{f(n)} x_{f(n)},g_{f(n)} y_{f(n)})\le \lim_{n \to \infty} 2 b_{f(n)}=0.$$ So $$x=y.$$ Which is absurd because $$d(x_{f(n)}, y_{f(n)})\ge r$$ for all $n$.