order topology vs product topology for an infinite product Let $A$ be a finite set with the discrete topology (i.e. every subset is open), and $\mathbb{N}$ denote the set of natural numbers.
Is the product topology on $A^\mathbb{N}$ equivalent to the order topology on $A^\mathbb{N}$ induced by the lexicographical ordering?
 A: Here’s an outline of an argument that will work.
Let $A=\{0,\dots,n-1\}$. Points of $A^\mathbb{N}$ are functions from $\mathbb{N}$ to $A$. For $x\in A^\mathbb{N}$ and $k\in\mathbb{N}$ let $B(x,k)=\big\{y\in A^\mathbb{N}:y\upharpoonright\{0,\dots,k\}=x\upharpoonright\{0,\dots,k\}\big\}$.


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*Show that $\mathscr{B}=\{B(x,k):x\in A^\mathbb{N}\text{ and }k\in\mathbb{N}\}$ is a base for the product topology of $A^\mathbb{N}$.

*Express each $B(x,k)\in\mathscr{B}$ as an open interval in the lexicographic order on $A^\mathbb{N}$. (It’s actually enough to show that each $B(x,k)\in\mathscr{B}$ is a union of open intervals in the lexicographic order, but that’s not significantly easier.) Conclude that every set open in the product topology is also open in the order topology.

*Show that every open interval in the lexicographic order is a union of members of $\mathscr{B}$. (Given an open interval $(x,y)$ and a point $z\in(x,y)$, show how to find $B(z,k)\subseteq(x,y)$.) Conclude that every open set in the order topology is open in the product topology.

