Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 5 down vote favorite

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?

Thanks!

share|improve this question
1  
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 at 3:00

1 Answer

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 $(2^\beta)^\beta=2^\beta<\alpha\leq\alpha^\beta$.

share|improve this answer
1  
I'm not sure what your counterexample is. Can you give your $\alpha$ and $\beta$ explicitly? – Chris Eagle Feb 16 at 13:50
What does this have to do with the question? – Chris Eagle Feb 16 at 13:53
1  
@Adar: I added to the answer. – Asaf Karagila Feb 16 at 14:03
1  
@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 at 14:16
1  
Okay, thank you very much!! – Adar Hefer Feb 16 at 14:18
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.