The topology generated by the metric is the product topology of discrete space {0,1} Let $X=\{0,1\}^{\mathbb{N}}$. For $x=x_1x_2x_3\cdots$ and $y=y_1y_2y_3\cdots$ in $X$, define 
\begin{align*}
d(x,y)=2^{-\textrm{min}\{n\in \mathbb{N}:x_n\neq y_n \}}
\end{align*}
I showed that $(X,d)$ is a metric space.
I want to show that the topology generated by the metric $d$ is the product topology of discrete space $\{0,1\}$.
I thought that this problem is associated with bases.. 
Any help is appreicated!!
Thank you! 
 A: A basis for the topology induced by the metric $d$ is the family of all open balls $B\left((x_n)_n, 2^{-m}\right)$ for $m \in \mathbb{N}$.
A basis for the product topology of discrete spaces is $$\{x_1\} \times \{x_2\} \times\cdots \times \{x_m\} \times \{0,1\} \times \{0,1\} \times \cdots$$
for some $x_1, \ldots, x_m \in \{0,1\}$ and $m \in \mathbb{N}$.
Those two basis are in fact the same:
\begin{align}
(y_n)_n \in B\left((x_n)_n, 2^{-m}\right) &\iff d\left((x_n)_n, (y_n)_n\right) < 2^{-m}\\
&\iff  2^{-\min\{n\in\mathbb{N} : x_n \ne y_n\}} < 2^{-m}\\
&\iff \min\{n\in\mathbb{N} : x_n \ne y_n\} > m\\
&\iff x_n = y_n \,\text{for all } n = 1, \ldots, m\\ 
&\iff (y_n)_n \in \{x_1\} \times \{x_2\} \times\cdots \times \{x_{m}\} \times \{0,1\} \times \{0,1\} \times \cdots
\end{align}
Therefore, the two topologies are equal.
A: Let $T_d$ be the topology given by the metric $d$ and $T_p$ the product topology you defined. One way to prove that they coincide is to show the following things:


*

*A sub-basis (it's actually a basis) of $T_d$ is given by all the open balls centered on $x$ of radius $2^{-n}$, where $n$ ranges in $\mathbb{N}$ and $x$ in $X$.

*A sub-basis of $T_p$ is given by all the $\pi_n^{-1}(k)$, where $n$ ranges in $\mathbb{N}$ and $k$ in $\{0,1\}$ and $\pi_n$ is the projection on the $n-$th coordinate.

*a) All the balls of 1. are in $T_p$. b) All the sets of 2. are in $T_d$.

*Convince yourself that this is enough :)


Of course, 3. is the most difficult part. In a) you should try to write $B_{2^{-n}}(x)$ as a finite intersection of elements of the form $\pi_m^{-1}(k)$, $m \le n$. In b) you should try to write $\pi_n^{-1}(k)$ as a (finite) union of elements of the form $B_{2^{-n}}(x)$.
A: You can also use the universal property of the product topology.
Let $Y$ be a topological space and $f_n\colon Y\longrightarrow \{0,1\}$ a continuous map for all $n\in\mathbb{N}$. Clearly there is exactly one map $f\colon Y\longrightarrow \{0,1\}^{\mathbb{N}}$ such that $f(y)_n=f_n(y)$. We only have to show that $f$ is continuous with respect to $d$.
For that purpose let $y\in Y$ and $\epsilon>0$. Choose $N\in\mathbb{N}$ such that $2^{-(N+1)}<\epsilon$. Since each $f_n$ is continuous and the topology on $\{0,1\}$ is discrete, there exist neighborhoods $U_1,\dots,U_N$ of $Y$ such that $f_n(y')=f_n(y)$ for $y'\in U_n$, $n\leq N$.
Let $U=\bigcap_{n=1}^N U_n$. For all $y'\in U$ we have $f_n(y')=f_n(y)$ for $n\leq N$, hence 
$$
d(f(y),f(y'))\leq 2^{-(N+1)}<\epsilon.
$$
Thus $f$ is continuous.
