2

Let $A$ be an infinite set and let $S$ be the set of all bijections $A \rightarrow A$. Then if $\mid A \mid = \kappa$, then $\mid S \mid = 2^\kappa$.

I'm able to prove it for $A = \mathbb{N}$ by showing an injection $P(\mathbb{N}) \rightarrow S$, but how can I prove it for any set $A$?

Involutive Automorphism
  • 341

1 Answers1

8

It suffices to exhibit an injection $P(A) \to S$.

Let $X$ be a subset of $A$, and fix a derangement $d_X:A \setminus X \to A \setminus X$ (see this question for proof of existence of $d_X$). Now define the bijection $f_X:A \to A$ as follows:

$f(x)= \begin{cases} x, & x \in X \\[2ex] d_X(x), &x \notin X \end{cases} $

The map $X \mapsto f_X$ is our required injection.

Community
  • 1
TonyK
  • 64,559
  • 1
    The existence of $d_X$ assumes that $A \setminus X$ has at least two elements. So if $|A \setminus X| = 0$ or $1$, then there are no derangements. Fortunately, one can avoid this issue by setting up an injection from $\mathcal{P}(A \setminus {a,b})$ to $S$ for some $a,b \in A$. See the answer in https://mathoverflow.net/a/86578/133798 for this modification. Please be aware that the roles of $X$ and $A$ are interchanged in that answer. – Nuvishramun Jul 22 '22 at 16:30
  • Why does it suffice to show an injection? – jsmith Jul 31 '22 at 21:57
  • The argument here shows $|P(A)| \leq |S|$. To show an injection from $S$ to $P(A)$, observe that $|S| \leq |A^A|$, where $A^A$ is the set of all functions from $A$ to $A$. Now, $|A^A| \leq |P(A \times A)|$ because you can associate every function uniquely to its graph. Finally, $|A| = |A \times A|$ for an infinite set. Thus, $|S| \leq |A^A| \leq |P(A \times A)| = |P(A)|$. – Nuvishramun Aug 20 '22 at 16:00