1. Help With Cardinals

Hi, I've been asked to show the following assertion;

${|A|}^2 = |A| > 1 \rightarrow 2 \cdot |A| = |A| \wedge {|A|}^{|A|} = 2^{|A|}$

I've managed to show the $2 \cdot |A| = |A|$ part, but I'm having trouble with the next part, the ${|A|}^{|A|} = 2^{|A|}$. Am I right in thinking that ${|A|}^{|A|}$ is the number of functions $A \rightarrow A$, and is this on the right lines, or am I looking way off?

Any help would be appreciated!

2. Originally Posted by jackprestonuk
Hi, I've been asked to show the following assertion;

${|A|}^2 = |A| > 1 \rightarrow 2 \cdot |A| = |A| \wedge {|A|}^{|A|} = 2^{|A|}$

I've managed to show the $2 \cdot |A| = |A|$ part, but I'm having trouble with the next part, the ${|A|}^{|A|} = 2^{|A|}$. Am I right in thinking that ${|A|}^{|A|}$ is the number of functions $A \rightarrow A$, and is this on the right lines, or am I looking way off?

Any help would be appreciated!
I suppose we're talking of cardinals here, so let us put $\alpha:=|A|\,,\,\,so \,\,\,\alpha=\alpha^2\,\,\,and\,\,\,\alpha \neq 1\Longrightarrow \alpha\geq \aleph_0$ , and thus by Cantor's Theorem:

$\alpha^\alpha\leq \left(2^\alpha\right)^\alpha=2^{\alpha\cdot\alpha} =2^\alpha$ , and since $\alpha^\alpha\geq 2^\alpha$ is straightforward, applying Cantor-Schroeder-Bernstein we're done.

Tonio

3. Ah, thanks very much, that's really helpful!
In the meantime I'd come up with this argument; I wonder if it works just as well?

Set $|A| =: a \in \mathbb{N}$. Then $2^a \subseteq a^a \subseteq \mathcal{P}(a \times a)$, and so $2^{|A|} = 2^a \leq a^a \leq |\mathcal{P}(a \times a)| = 2^{a^2} = 2^a$. We have that $a^a = |A|^{|A|}$ and so we have our equality by Cantor Bernstein.

Apologies if sometimes I did or didn't put |...| in there, I still get a little confused when dealing with cardinals whether, say, $a^a$ represents the set of functions from the natural number $a$ to itself, or simply the cardinal number $a^a$ (that is, effectively, the number of such functions), or can stand for both...