Suppose I have the following space: X={0,1}$^\infty$ as my topological space.
I know that the following metrics is generates the discrete topology on it:
d(x,y)= 0 if x=y otherwise 1
Now I want to shat that my space is also metrizable under the product topology , But I couldn't find any metrics that isn't similar to the one above that generates the product topology on this space
One metric for the product topology is $ d((x_n),(y_n))= \sum\limits_{n=1}^{\infty} \frac {|x_n-y_n|} {2^{n}}$.
Let $U$ be an open set in the product topology containing $(x_n)$. Then there exists an integer $N$ and open sets $U_i, i \leq N$ such that $(x_n) \in U_1 \times U_2\times ...\times U_N$ and $U_1 \times U_2\times ... \subset U$. Note that we can find positive numbers $r_i, i \leq N$ such that $(x_i-r_r,x_i+r_i) \subset U_i$ for $i \leq N$. Now suppose $d((x_n),(y_n)) <r2^{-N}$ where $r$ is the minimum of $r_1,r_2,...,r_N$. I will let you verify that $(y_n) \in U_1 \times U_2\times ... $ so $(y_n) \in U$.
Now consider any open ball $B((x_n),r))$ for the metric $d$. There exists $N$ such that $\sum_{n> N} \frac 1 {2^{n}} <\frac r 2$.
We shall find $s>0$ such that $(x_1-s,x_1+s)\times ... \times (x_N-s,x_N+s)\times ...$ (which is open in the product topology and contains $(x_n)$ ) is contained in $B((x_n),r))$. For this it is enough to choose $s$ such that $s \sum\limits_{n=1}^{N} 2^{-n} <\frac r 2$ which is certainly possible. This completes the proof.
The discrete space $\{0,1\}$ is metrisable and I show here that a countable product of metrisable spaces (in the product topology) is metrisable again (and I give a working formula for its metric). It's essentially Kavi's proof in more detail.
