# A metric and discrete topology

Let $\Sigma=\{1,2,...,n\}$ and $\Omega=\Sigma^\mathbb{N}$ be the set of infinite sequence of n digits. Define a metric $d$ on $\Omega$ by $d(\omega,\tau)=2^{-|\omega\wedge\tau|}$ where $|\omega\wedge\tau|$ is the length of the longest common prefix of $\omega$ and $\tau$.

Show that the metric topology on $\Sigma^\mathbb{N}:=\Sigma\times \Sigma\times...$ given by $d$ is the same as the product topology $\Sigma^\mathbb{N}$ where $\Sigma$ is taken with the discrete topology.

Attempt: Since the product topology is discrete, it is enough to show that for any $a\in\Sigma^\mathbb{N}$, there is a number $\epsilon>0$ such that $B_d(a,\epsilon)=\{a\}$ where $B_d(a,\epsilon):=\{x\in\Sigma^\mathbb{N}:d(x,a)<\epsilon\}$. But it seems to me impossible since we can always find an element $\neq a$ around $a$. I don't know where my mistake is. Thanks for your help!

• The product topology for an infinite product of spaces with more than one point is not discrete. – Matt Samuel Dec 7 '14 at 20:09
• I didn't realize it! Thanks! – Ergin Suer Dec 7 '14 at 20:10

Given a sequence $\omega=(\omega_1,\omega_2,\ldots)\in\Omega$, a basic open neighbourhood of $\omega$ in $\Omega$ with the product topology is of the form $(\omega_1,\ldots,\omega_n)\times\Sigma\times\Sigma\cdots$, and this set consists of the sequences $\tau=(\tau_1,\tau_2,\ldots)$ such that $d(\omega,\tau)\leq2^{-n}$.