# Cardinality of the countably infinite product of a two-point set $\{0,1\}$?

I'm so confused about cardinalities of some sets. What is the countable infinite product of a two points set $\{0,1\}$? Does it have the same cardinality as the real number $\mathbb R$? Or is the infinite product just countable?

Could anyone give me the answer?

• think of the binary representation of the numbers in $[0,1]$ – David Holden Jan 2 '15 at 5:51
• I want to know why you think it might be countable. – Asaf Karagila Jan 2 '15 at 11:32

Let $X$ be a set and consider the set $\{0,1\}^X$ of all maps $X\to\{0,1\}$. Then there is a bijection $$f\colon \mathcal{P}(X)\to\{0,1\}^X$$ ($\mathcal{P}(X)$ is the power set of $X$) defined by sending each subset $A$ of $X$ to its characteristic function $$\chi_A(x)=\begin{cases} 1 & \text{if x\in A}\\ 0 & \text{if x\notin A} \end{cases}$$ Thus $\{0,1\}^X$ has the same cardinality of $\mathcal{P}(X)$.
The countably infinite product of the set $\{0,1\}$ is simply the set of all infinite binary strings which Cantor showed to be uncountable in the classic diagonalization argument.