If $\mathfrak{c} = 2^\mathbb{N}$ why $|D^\mathfrak{c}|=\mathfrak{c}$ with $D = \{0, 1\}$. Let $\mathfrak{c} = 2^\mathbb{N}$ and $D^\mathfrak{c}$ be the topological product of $\mathfrak{c}$ copies
of the discrete Abelian group $D = \{0, 1\}$. If $D^\mathfrak{c}$ taken as a group is denoted by $G$,
and $D^\mathfrak{c}$ taken as a topological space is denoted by $X$.

Why $|G| = |X| = c$?

I think maybe it has to do with Boolean but I'm not sure. 

On the other hand I am correct or I am wrong that $|G| = |D^\mathfrak{c}|=|D|^\mathfrak{c} =2^\mathfrak{c}&gt; \mathfrak{c}$?

I am studying this example of Korovin Orbits in the book Topological groups and related structures of Mikhail Tkachenko.



Thank you for your Help!
It is the case that $|D^{\mathfrak{c}}|=2^{\mathfrak{c}}$. We can find a bijection from the product of $\mathfrak{c}$ copies of $D$, which you have written $D^{\mathfrak{c}} $, to the power set $2^{\mathfrak{c}}$ of $\mathfrak{c}$ as follows.

If $x\in D^{\mathfrak{c}} $, then $x$ essentially consists of a mapping from each element of $\mathfrak{c}$ to either $0$ or $1$, and we may write $x(c)$ to denote the value ($0$ or $1$) corresponding to $c\in\mathfrak{c}$. Let $S(x)$ be the set of all elements $c\in \mathfrak{c}$ for which $x(c)=1$. Then $S: D^{\mathfrak{c}} \mapsto 2^{\mathfrak{c}}$ is a bijection. In essence, $x$ acts as an “indicator function” for every subset of $\mathfrak{c}$ indicating (with a $1$) which elements of $\mathfrak{c}$ the subset contains.
It is the case that $|D^{\mathfrak{c}}|=2^{\mathfrak{c}}$. We can find a bijection from the product of $\mathfrak{c}$ copies of $D$, which you have written $D^{\mathfrak{c}} $, to the power set $2^{\mathfrak{c}}$ of $\mathfrak{c}$ as follows.

If $x\in D^{\mathfrak{c}} $, then $x$ essentially consists of a mapping from each element of $\mathfrak{c}$ to either $0$ or $1$, and we may write $x(c)$ to denote the value ($0$ or $1$) corresponding to $c\in\mathfrak{c}$. Let $S(x)$ be the set of all elements $c\in \mathfrak{c}$ for which $x(c)=1$. Then $S: D^{\mathfrak{c}} \mapsto 2^{\mathfrak{c}}$ is a bijection. In essence, $x$ acts as an “indicator function” for every subset of $\mathfrak{c}$ indicating (with a $1$) which elements of $\mathfrak{c}$ the subset contains.