Supra Topology on Shift spaces

Question

For an alphabet $A=\lbrace 1,2,3,...,k-1\rbrace$ the set $A^\mathbb{N}=A\times A\times A\times\cdots$ consisting of all (one-sided) sequences of elements of $A$. The shift map $\sigma:A^\mathbb{N}\rightarrow A^\mathbb{N}$ is defined by $\sigma(x_0x_1x_2\cdots)=x_1x_2\cdots$, which is continuous on this space. We called $(A^\mathbb{N},\sigma)$ the full $A$- shift space. We defined a metric $d$ on $A^\mathbb{N}$ by $$ \begin{eqnarray} d(s,t)=\left\{ \begin{array}{ll} 0, & s = t,\nonumber \\ 2^{-j}, & s\neq t. \end{array}\right. \end{eqnarray}$$ where $j \in N$ is the minimal number such that $s_j \neq t_j$. The space $(A^\mathbb{N},d)$ is a topological space. So the basic open ball $W$ is any subset of the full $A-$ shift of the form $$ W=X_W \lbrace x\in \Sigma_k:x_0x_1x_2\cdots x_{k-1}=w\rbrace$$, where $w$ is an allowed block of length $k$. My question is how can we define a supra topology on $A^\mathbb{N}$ ?. Where the supra topology $\tau^*$ is a collection of subsets of the $P(X)$ such that $X$ and $\emptyset$ are in $\tau^*$ and $\tau^*$ is closed under arbitrary union.

Answer 1

The topology is defined by the metric.

As the shift operator is not needed to define the metric, calling this the shift space is a misnomer. This space is an infinite string space.

A base for this space is all sets of strings with a common prefix. Notice that the space you defined does not allow for finite strings.

You ask for a supra topology with such scant requirements, that any topology whatever is a supra topology.