A basis for product topology Consider $A=\{0,1\}$ with discrete topology, and let $X$ be a non empty set.\ We know that there is a natural bijection between $\{0,1\}^X$ and $P (X)$, power set of $X$. Now if we consider $\{0,1\}^X$ with product topology and consider $P(X)$ as a topological space with the natural bijection, I want to know with kind of subsets of $X$ form the set of open sets for this topology?
 A: The set $\{0,1\}^X$ can be viewed as the set of all functions from $X$ to $\{0,1\}$. The product topology says that the topology of $\{0,1\}^X$ are sets of functions which are are the same in a finite number of coordinates, i.e., $A$ is open if there exists a finite set $\{x_1,...,x_n\}\subset X$ such that if $f,g\in A$, then $f(x_i)=g(x_i)$ for all $i=1,...,n$, and $A$ consists of all such functions. Therefore, under the natural bijection, $A$ gets sent to that subset of $\mathcal P(X)$ consisting of all subsets of $X$ which contain $$\{x_i:f(x_i)=1\text{ for }f\in A,i=1,...,n\},$$ and do not contain $$\{x_i:f(x_i)=0\text{ for }f\in A,i=1,...,n\}.$$ An important point brought up by Najib Idrissi is that if $X$ is countably infinite, $\mathcal P(X)$ is homeomorphic to the Cantor space.
A: Thinking of $\wp(X)$ as a product for each $x\in X$ there is a projection $p_x:\wp(X)\to\{0,1\}$ prescribed $A\mapsto1$ if $x\in A$ and $A\mapsto0$ otherwise.
The topology on $\wp(X)$ is the coarsest topology such that every projection $p_x:\wp(X)\to\{0,1\}$ is continuous. 
We have $p_x^{-1}(\{1\})=\{A\in\wp(X)\mid x\in A\}$ and $p_x^{-1}(\{0\})=\{A\in\wp(X)\mid x\notin A\}$. These sort of sets form a subbase of the topology.
A base is then (as usual) the collection of finite intersections of elements of the subbase.
Note that: 
$$p_{x_{1}}^{-1}\left(\left\{ 1\right\} \right)\cap\cdots\cap p_{x_{n}}^{-1}\left(\left\{ 1\right\} \right)\cap p_{y_{1}}^{-1}\left(\left\{ 0\right\} \right)\cap\cdots\cap p_{y_{m}}^{-1}\left(\left\{ 0\right\} \right)=\left\{ A\in\wp\left(X\right)\mid x_{1},\dots,x_{n}\in A\wedge y_{1},\dots,y_{m}\notin A\right\}$$
The topology is the collection of sets that can be written as a union of elements of the base.
A: Let $\mathscr{P}$ be the family of ordered pairs of disjoint finite subsets of $X$. For each $\langle F,G\rangle\in\mathscr{P}$ let
$$\mathscr{B}(F,G)=\{S\in\wp(X):F\subseteq S\subseteq X\setminus G\}\;;$$
in words, $\mathscr{B}(F,G)$ is the family of subsets of $X$ that contain $F$ and are disjoint from $G$. Then
$$\mathfrak{B}=\{\mathscr{B}(F,G):\langle F,G\rangle\in\mathscr{P}\}$$
is a base for a topology $\mathfrak{T}$ on $\wp(X)$ such that $\langle\wp(X),\mathfrak{T}\rangle$ is homeomorphic to the product space $\{0,1\}^{|X|}$. I’ll leave this as an easy exercise.
Added: There are quite a few interesting ways to impose topologies on families of subsets of some set $X$, especially when $X$ is already equipped with a topology, and many of them are quite similar to what I’ve done here. For example, let $\langle X,\tau\rangle$ be a $T_1$ space, and let $\mathscr{F}$ be the family of non-empty finite subsets of $X$. For each $F\in\mathscr{F}$ and $U\in\tau$ let 
$$\mathscr{B}(F,U)=\{G\in\mathscr{F}:F\subseteq G\subseteq U\}\;,$$
and let $\mathfrak{B}$ be the family of all such sets $\mathscr{B}(F,U)$; $\mathfrak{B}$ is a base for the Pixley-Roy topology on $\mathscr{F}$.
If $\mathscr{I}$ is the family of infinite subsets of $\Bbb N$, for each finite $F\subseteq\Bbb N$ and $I\in\mathscr{I}$ let
$$\mathscr{B}(F,I)=\{J\in\mathscr{I}:F\subseteq J\subseteq I\}\;;$$
the family of all such sets $\mathscr{B}(F,I)$ is a base for the Ellentuck topology on $\mathscr{I}$. This topology, which is finer than that induced by the product topology $\mathfrak{T}$ on $\wp(\Bbb N)$, has proved useful in studying infinite Ramsey theory.
