Consider $X = \{1, 2, \ldots, n\}^{\Bbb Z}$. We can endow $X$ with a topology in at least two ways:
- Let $\tau_1$ be the product topology on $X$, due to the discrete topology on $\{1, 2, \ldots, n\}$.
- Let $\tau_2$ be the metric topology on $X$, where the metric $d: X\times X \to \Bbb R$ is given by $d(x,y) = 2^{-k}$ if $k = \max\{j: x_i = y_i \text{ for } |i| \le j\}$, and $d(x,x) = 0$.
I'd like to show that $\tau_1= \tau_2$.
$\tau_1$ is generated by the basis $$\mathcal B_1 = \left\{\prod_{i\in \Bbb Z} U_i: U_i \subset \{1, \ldots, n\}, U_i = \{1, \ldots, n\} \text{ for all but finitely many } i\in \Bbb Z \right\}$$
and $\tau_2$ is generated by the basis consisting of open balls in $X$ (with respect to $d$). The following proposition from point-set topology should be helpful:
Proposition. $\tau \supset \tau'$, i.e., $\tau$ is finer than $\tau'$ if and only if for all $x\in X$, and each basic open set $B'$ of $\tau'$ containing $x$, there exists a basic open set $B$ of $\tau$ such that $x\in B\subset B'$.
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Now, pick any $x\in X$. I shall try to show $\tau_1 \supset \tau_2$ first. Suppose $x\in B(y,r)$, where $y\in X$ and $r > 0$. We want $B\in \mathcal B_1$ such that $x\in B \subset B(y,r)$.
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To finish, we need $\tau_2 \supset \tau_1$. Pick any $x\in X$, and a basic open set $\prod_{i\in \Bbb Z} U_i$ containing $x$ (note that $U_i = \{1, \ldots, n\}$ for all but finitely many $i$). It shall suffice to find $r > 0$ such that $x\in B(x,r) \subset \prod_{i\in \Bbb Z} U_i$. There exists $m \ge 1$ such that for every $|i| \ge m$, $U_i$ is $\{1, \ldots, n\}$.
As it stands, both directions are incomplete. Please note that I ask for pointers and hints to complete my work and not necessarily a detailed solution (which may be found in possible duplicates).
Best Answer
Specific hints:
Fix a $y$ and an $r>0$. There is a $k$, $2^{-k}<r$. We can find a $\mathcal{B}_1$-neighbourhood of $y$ by setting finitely many coordinates $i$ equal to $y_i$, and letting all higher coordinates vary. Think about the specific meaning of $d(x,y)<2^{-k}$. You might also want to reverse engineer the solution to the other direction.
For the other direction, say $J$ is the largest index for which $U_J\subsetneq\{1,\cdots,n\}$. How does $B(x,2^{-J-1})$ relate to the $\mathcal{B}_1$-neighbourhood of $x$?
Overkill abstraction:
It is a general theorem that the countable topological product of metric spaces $(X_i;\rho_i)_{i\in\Bbb N}$ is metrizable via the metric: $$\varrho(a,b)=\sum_{i=1}^\infty2^{-i}\color{blue}{\frac{\rho_i(a_i,b_i)}{1+\rho_i(a_i,b_i)}}$$
Really, we only need a bounded $\color{blue}{blue}$ metric that yields the same topology on $X_i$ as $\rho_i$. The ratio I give there is the 'canonical' way to do this, but any set of uniformly bounded metrics with equivalent topology will do. To that end, let's replace the $\color{blue}{blue}$ with some uniformly bounded metrics $\rho'_i$ on $X_i$ such that $(X_i;\rho_i)$ has the same topology as $(X_i;\rho'_i)$: $$\varrho(a,b)=\sum_{i=1}^\infty2^{-i}\color{blue}{\rho'_i(a_i,b_i)}$$
The theorem can be found in Royden's real analysis, exercise $12$, chapter $16$ (available on the web).
You can try this if you want. You need to check $\varrho$ is indeed a metric, and you need to check the topology is the same as the product topology. You should also check that the 'canonical' choice of $\rho'_i$ I introduced at the start does actually induce the same topology.
This applies to your problem by allowing the discrete metric on each $X_i=\{1,\cdots,n\}$. Well, not directly, but $d$ is equivalent to the induced $\varrho$.