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I'm trying to prove that if ${\kappa}$ is an infinite cardinal, then there are $2^{\kappa}$ bijective functions from ${\kappa}$ to ${\kappa}$. I would greatly appreciate any tips. Thank you.

MathUser123
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    A function from $\kappa$ to $\kappa$ is a subset of $\kappa \times \kappa$. So that gives you an upper bound. For a lower bound you could split $\kappa$ into $\kappa$-many two-element sets and consider functions that either fix or permute each pair. – user49640 May 29 '17 at 22:45

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I would do it differently.

The number of bijections is clearly not more than $\kappa^\kappa=2^\kappa$.

Let $S$ be the set of subsets of $\kappa$ that leave more than $1$ element in the complement. Clearly $S$ has the same cardinality as $2^\kappa$.

To show that there are at least $2^\kappa$ bijections we give an injection from $S$ to the set of bijections of $\kappa$. To do this we send each $s\in S$ to a bijection $\sigma$ of $\kappa$ such that $s$ is the set of fixed points of $\sigma$.

DanielWainfleet
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Asinomás
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We know that there are at most $\kappa^\kappa\le(2^\kappa)^\kappa=2^{\kappa^2}=2^\kappa$ bijections from $\kappa$ to $\kappa;$ we have to show that there are at least $2^\kappa$ bijections. Since $|\{0,1\}\times\kappa|=2\kappa=\kappa,$ it will suffice to exhibit $2^\kappa$ bijections from $\{0,1\}\times\kappa$ to $\{0,1\}\times\kappa,$ which we do as follows.

For each set $S\subseteq\kappa$ define $$f(\langle i,\ \alpha\rangle)=\begin{cases} \langle1-i,\ \alpha\rangle\ \text{ if }\ \alpha\in S,\\ \ \ \ \ \ \ \ \langle i,\ \alpha\rangle\ \text{ if }\ \alpha\notin S.\\ \end{cases}$$

bof
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