# Set of all injective functions $A\to A$

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

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

#### Solutions Collecting From Web of "Set of all injective functions $A\to A$"

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, $|I|=|2^\mathbb{N}|$.