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Is $\alpha^{\beta}=2^{\beta}$ for all infinite cardinalities $\alpha, \beta$?

I was wondering about this since I've encountered examples of times where this holds, but I can't seem to prove it myself and I'm pretty skeptical that it's true in general.

Can anyone shed light on this please? What are the limitations of this?


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My discrete math professor told us when we were freshmen "If you tried to prove something, and you couldn't, and then you found out it's wrong; then you did a good job not lying to yourself." (Although he said that on the continuing complaints that the exercise sheets are plagued with typos and mistakes...) – Asaf Karagila Feb 17 '13 at 3:00
up vote 11 down vote accepted

Not at all. For example let $\alpha=2^{2^\beta}$ then we have: $$\alpha^\beta=\left(2^{2^\beta}\right)^\beta=2^{2^\beta\cdot\beta}=2^{2^\beta}=\alpha.$$

If $2<\alpha\leq 2^\beta$ then $2^\beta\leq\alpha^\beta\leq2^\beta$ and we have equality. Otherwise $2^\beta<\alpha$ and $(2^\beta)^\beta=2^\beta<\alpha\leq\alpha^\beta$.

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I'm not sure what your counterexample is. Can you give your $\alpha$ and $\beta$ explicitly? – Chris Eagle Feb 16 '13 at 13:50
What does this have to do with the question? – Chris Eagle Feb 16 '13 at 13:53
@Adar: I added to the answer. – Asaf Karagila Feb 16 '13 at 14:03
@Adar: What do you mean? I assume that $\alpha$ satisfies these conditions. Cardinal exponentiation is weakly-increasing. That means that $\kappa<\lambda$ then $\kappa^\mu\leq\lambda^\mu$. – Asaf Karagila Feb 16 '13 at 14:16
Okay, thank you very much!! – Adar Hefer Feb 16 '13 at 14:18

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