# 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?

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What must you assume about $A$ in order for the question even to make sense? – Brian M. Scott Nov 17 '11 at 2:15
Brian, A is is ordered if that is what you are thinking about. – Cantor Nov 17 '11 at 15:12
GEdgar, here's what I'm thinking: on $A$, discrete and order topologies are equivalent. So the product topology on $A^\mathbb{N}$ is the same whether we tak $A$ to have order topology or discrete topology. It remains to show $A^\mathbb{N}$ with the lexicographical topology is equivalent to the product $A^\mathbb{N}$ assuming order topology for $A$. – Cantor Nov 17 '11 at 15:23

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 $B(x,k)=\big\{y\in A^\mathbb{N}:y\upharpoonright\{0,\dots,k\}=x\upharpoonright\{0,\dots,k\}\big\}$.
1. 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 $A^\mathbb{N}$.
2. 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.
3. 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.