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Here is a problem:

Show that if $\Omega$ is an infinite set, then $|\operatorname{Sym}(\Omega)|=2^{|\Omega|}$.

I have worked on a problem related to a group that is $S=\bigcup_{n=1}^{\infty } S_n$. Does it make sense we speak about the relation between $S$ and $\operatorname{Sym}(\Omega)$ when $\Omega$ is an infinite set. Moreover, I know that $|\operatorname{Sym}(\Omega)|=|\Omega|^{|\Omega|}$. How we can reach from $|\Omega|^{|\Omega|}$ to $2^{|\Omega|}$. Thanks.

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If $\Omega$ is infinite, the two have the same cardinality. –  André Nicolas Jul 11 '12 at 17:01
    
@AndréNicolas: Isn't Sym$(\Omega)\subset S$? –  B. S. Jul 11 '12 at 17:27
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We can have two infinite sets $A$ and $B$, with $A$ a proper subset of $B$, but with $A$ and $B$ of the same cardinality. for example (Galileo) let $B$ be the integers, and $A$ the even integers. –  André Nicolas Jul 11 '12 at 17:48

2 Answers 2

up vote 4 down vote accepted

You have $|\Omega|^{|\Omega|} \geq 2^{|\Omega|}$, since $\Omega$ is infinite. Then, $|\Omega| < 2^{|\Omega|}$ (Cantor's theorem) so $|\Omega|^{|\Omega|} \leq (2^{|\Omega|})^{|\Omega|}=2^{|\Omega|.|\Omega|}=2^{|\Omega|}$, since $\Omega$ is infinite.

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$\textbf{Lemma}$ : For cardinals, $2 \leq \kappa \leq \lambda$ where $\lambda$ is infinite, then $\kappa^\lambda = 2^\lambda$.

It is clear that $2^\lambda \leq \kappa^\lambda$. However $\kappa^\lambda \leq (2^\kappa)^\lambda = 2^{\kappa \cdot \lambda} = 2^{\lambda}$, where $\lambda \cdot \kappa = \text{max}\{\lambda, \kappa\}$, which can be proved.

Now taking $\kappa = \lambda = |\Omega|$ in the lemma, you have $2^{|\Omega|} = |\Omega|^{|\Omega|}$.

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