# An explicit bijection between the power set $\mathcal P \left({\mathbb{N}}\right)$ and $2^\mathbb{N}$. [duplicate]

I know how to show that these two have the same cardinality and from that there must be a bijection between them.

Can anyone help with an explicit bijection between these sets?

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## marked as duplicate by Asaf Karagila, 5PM, Chris Eagle, Martin Sleziak, rschwiebFeb 6 '13 at 14:44

$\mathcal P \left({\mathbb{N}}\right)$ and $2^\mathbb{N}$ both denote the power set of $\mathbb{N}$ so I'm not sure what you're asking. –  user61031 Feb 6 '13 at 7:29
Usually $2^X$ denotes the set of functions $X \to 2$, which is naturally isomorphic to $\mathcal{P}(X)$ (in classical logic anyway...) but is not the same thing. –  Trevor Wilson Feb 6 '13 at 7:50
Hint: Think about, given $E\subseteq\mathbb{N}$, the indicator function $\chi_E:\mathbb{N}\to 2$ given by $\chi_E(n)=1$ if $n\in E$ and $0$ otherwise.
HINT: Consider indicator (or characteristic) functions of subsets of $\Bbb N$.