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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?

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

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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.

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