Let $\mathcal{M}=\prod_{\mathbb{Z}}\{1,\ldots,k\}$, with finite $k$. On this space we define the following family of metrics:

Let $a>1$ and define $\rho_a:\mathcal{M}\times \mathcal{M}\rightarrow [0,\infty)$ as $$\rho_a(x,y)=\left\{ \begin{array}{ll} a^{-N} & \text{if $x\neq y$ and $N$ is maximal and } x_{[-N,N]}=y_{[-N,N]}\\ 0 & x=y \end{array}\right. $$ where $x_{[-N,N]}$ indicate the subword $x_{-N}...x_{N}$ of the word $x=(x_i)_{i\in\mathbb{Z}}\in\mathcal{M}$.

The work I am reading claims that the metrics $\rho_a$ are all equivalent. However, I am sceptical about it: let $a>b>1$, then clearly $\rho_a(x,y)\leq \rho_b(x,y)$ for all $x,y\in\mathcal{M}$, but how do we see that there exists a $c>0$ independent of $x,y$ such that $\rho_a(x,y)\geq c \rho_b(x,y)$? The quotient $\rho_b(x,y)/\rho_a(x,y)$ can be arbitrarily large, for example if $b=a^\gamma$ for $0<\gamma<1$.

What am I missing here?

Thanks a lot for the help!


For two metrics to be equivalent, it's not necessary that each is bounded by a constant times the other (this is a stronger condition) -- they simply have to generate the same topology. In this case you can verify that any ball $B_r(x) = \{y \in \mathcal{M} : \rho_a(x, y) < r\}$ under one $\rho_a$ is still a ball under any other $\rho_b$, so all $\rho_a$ are in fact equivalent.


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