Prove that product topology and metric topology are the same on $\{1,2,\ldots,m\}^{\mathbb{N}}$

Question

For $m\in\mathbb{N}$, define $\mathcal{A}_{m}:=\{1,2,\ldots,m\}$. Then consider the set $\mathcal{A}_{m}^{\mathbb{N}}$ consisting of sequences in $\mathcal{A}_{m}$. We equip $\mathcal{A}_{m}$ with the discrete topology. Hence, any subset $S\subset\mathcal{A}_{m}$ is open. Let $\tau$ be the (infinite) product topology on $\mathcal{A}_{m}^{\mathbb{N}}$, which is the smallest topology such that the projections $$\pi_{i}\colon\mathcal{A}_{m}^{\mathbb{N}}\to\mathcal{A}_{m},\qquad \pi_{i}(x):=x_{i}$$ are continuous for all $i\in\mathbb{N}$. In other words, $\tau$ is the topology that is generated by the basis that consists of all finite intersections of $\pi_{i}$-preimages of (open) subsets in $\mathcal{A}_{m}$. Now consider the metric $$d(x,y):=\begin{cases}0&x=y\\ 2^{-\min\{i\in\mathbb{N} \ | \ x_{i}\neq y_{i}\}}&x\neq y\end{cases}$$ on $\mathcal{A}_{m}^{\mathbb{N}}$. Let $\tau_{d}$ be the topology generated by the metric $d$, that is the topology generated by the basis consisting of sets of the form $B_{d}(x,\varepsilon):=\{y\in\mathcal{A}_{m}^{\mathbb{N}} \ | \ d(x,y)<\varepsilon\}$ for all $x\in\mathcal{A}_{m}^{\mathbb{N}}$ and $\varepsilon>0$. How do I prove that $\tau=\tau_{d}$?

Answer 1

Let $O$ be product open in $X:=\mathcal{A}_m^\mathbb{N}$, and we'll show it's open in the topology induced by $d$:

Let $x=(x_n) \in O$. So there must be a basic element, of the form $\bigcap_{j=1}^k \pi_{i_j}^{-1}[U_j]$ for some finite set of indices $\{i_1, \ldots, i_k\}, k \in \mathbb{N}$ and (open) subsets $U_j \subseteq \mathcal{A}_m$ for $j=1,\ldots, k$, such that

$$(x_n) \in \bigcap_{j=1}^k \pi_{i_j}^{-1}[U_j] \subseteq O\tag{1}$$

Let $M=\max(i_1,i_2,\ldots,i_k)$ and define $r=2^{-M} >0$.

Now, if $y=(y_n) \in X$ obeys $d(x,y) < r$, the only way that can happen is when either $x=y$ or $\min\{i \in \Bbb N: x_i \neq y_i\} > M$, which implies that $x_i=y_i$ for $i \le M$ (in either case).

So in particular, as all $i_j \le M$, for all $j=1,\ldots, k$, $y \in \pi_{i_j}^{-1}[U_j]$ as $x_{i_j}=y_{i_j}$.

So $(1)$ implies that $$B_d(x,r) \subseteq O$$

and as this holds for any $x\in O$, $O$ is $d$-open. So $\mathcal{T}_{\text{prod}} \subseteq \mathcal{T}_d$.

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On the other hand, if we consider any fixed $r>0$ and $x \in X$, find $M$ such that $2^{-M} < r$ and let $M$ be minimally so. Then $y \in B_d(x,r)$ iff $d(x,y) < r$ iff $x_i = y_i$ for all $i \le M$ and so we can write any ball as

$$B_d(x,r) = B_d(x,2^{-M}) = \bigcap_{i=1}^M \pi_i^{-1}[\{x_i\}]\tag{2}$$

and $(2)$ shows that all $d$-open balls are product open and so $\mathcal{T}_d \subseteq \mathcal{T}_{\text{prod}}$ (balls generate the metric topology), showing equality of topologies.