Topology in the sequences space It is easy to see that $\mathbb{R}^w$, or the set of real-valued sequences, is a metric space with the distance $D(x,y) =  \sup \limits_{n} \frac{d(x_n, y_n)}{n}$ where $d(x_n, y_n) = min (|x_n-y_n|,1)$.
How do I prove that the induced topology by this distance is the same as the product topology of $\mathbb{R}^w$?
 A: We must prove that every open ball is such metric is an open set in the product topology, and that every element of the basis of the product topology is open in the topology induced by such metric.
(i)- Every open ball is open in the product topology.
Proof: Let $x:=(x_i)_{i\in\mathbb{N}}$ be an element of $\mathbb{R}^{\omega}$ and $r>0$. We wish to prove that $B(x,r)$(the open ball of center $x$ and radius $r$) is open in the product topology. If $r>1$, $B(x,r)=\mathbb{R}^{\omega}$ and there is nothing to prove. If $r\leq1$, define:
$$n:=\min(i\in\mathbb{N}:1/i<r)$$
Notice that if $m\geq n$, then $\frac{d(x_m,y_m)}{m}<\frac{1}{m}\leq\frac{1}{n}$ for every $y\in\mathbb{R}^{\omega}$. This implies that:
$$B(x,r)=(x-r,x+r)\times(x-2r,x+2r)\dots\times(x-(n-1)r,x+(n-1)r)\times\mathbb{R}\times\mathbb{R}\times\dots$$
Which is, by definition, open in the product topology.
(ii)- Every element of the product topology basis is open in the topology induced by such metric.
Proof: Let $A$ be an element of the product topology basis and $x=(x_i)_{i\in\mathbb{N}}\in A$, so there exists an $n\in\mathbb{N}$ such that:
$$A=A_1\times A_2\times\dots\times A_n\times\mathbb{R}\times\mathbb{R}\times\dots$$
Where each $A_i$ is open in $\mathbb{R}$. For $i\leq n$, use the fact that each $A_i$ is open to get an $1>r_i>0$ such that $(x-r_i,x+r_i)\in A_i$. Now define:
$$r:=\min_{1\leq i\leq n}\frac{r_i}{i}$$
By the deduction of $B(x,r)$ in the first proof, we see that $B(x,r)\subseteq A$ and, thus, $A$ is open in the topology induced by the metric $D$.
