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If $A$ is denumerable, is the set of all injective functions $A\to A$ equipotent with $2^A$?

I have proved $\aleph_0^{\aleph_0} = 2^{\aleph_0}$.

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This question asks (among other things) the same thing for $A=\mathbb N$; replacing arbitrary infinite countable set by $\mathbb N$ of course does not change the cardinality. –  Martin Sleziak May 12 '12 at 13:25

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up vote 5 down vote accepted

Let $I$ be the set of such injective functions. Partition $\mathbb{N}$ into countably many disjoint countably infinite sets $N_0,N_1,\ldots$. Every element of $\prod_n N_n$ is an injective function. Clearly, $$\prod_n\{0,1\}=2^\mathbb{N}\leq\prod_n N_n\leq I\leq\mathbb{N}^\mathbb{N}\leq 2^\mathbb{N}.$$ Since the ordering is antisymmetric and transitive, $|I|=|2^\mathbb{N}|$.

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