The topology generated by the metric is the product topology of discrete space {0,1}

Question

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!

Answer 1

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.

Answer 2

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:

1. 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 $X$.
2. 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.
3. a) All the balls of 1. are in $T_p$. b) All the sets of 2. are in $T_d$.
4. 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 $B_{2^{-n}}(x)$.

Answer 3

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.