Take the 2-minute tour ×
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.

Let $\alpha>\omega$ be an ordinal such that $2^\alpha$ = $\alpha$.

Then $\alpha$ is an epsilon number?

I have tried many different ways, but i can only work with the left side of $\alpha$(e.g, I have proved such ordinals satisfy $\omega$$\alpha$ = $\alpha$ and etc), but i think it's critical to work with right side of $\alpha$ in proof and i cant handle this.. Help

Plus i want to know even when the base is not 2 but finite, whether $\alpha$ is an epsilon number

share|improve this question
    
So far, i have proved that $\alpha$ is a limit ordinal. –  Katlus Jun 4 '12 at 8:08
    
A slightly more complete answer: Let $E = \{\omega\} \cup \{\epsilon_\alpha | \alpha $ an ordinal}. Then, for any $\beta$, the ordinals $\alpha$ such that $\beta^\alpha = \alpha$ is precisely $\{\alpha | \alpha \in E, \alpha > \beta\}$. –  Deedlit Sep 17 '12 at 9:29

1 Answer 1

up vote 2 down vote accepted

Lemma: $2^{\omega\alpha}=\omega^\alpha.$

Proof: Given $2^{\omega\alpha}=\omega^\alpha,$ we have $2^{\omega(\alpha+1)}=2^{\omega\alpha+\omega}=\omega^{\alpha+1},$ where the first equality is by definition of the function $\omega x$ and the second is by the hypothesis and calculation of the limit of $2^{\omega\alpha+n}.$

At limit ordinals, it's true because the composition of continuous functions is continuous.$\square$

Writing $\alpha$ in Cantor normal form to base $\omega,$ if $\alpha$ has any $\omega^n$ terms for finite $n,$ then $2^\alpha>\alpha$ by normality of $2^x$ and inspection:

Since $2^x$ is a normal function, we know $2^\alpha\geq\alpha.$ Addition of 1 in the exponent is multiplication by 2; addition of $\omega$ in the exponent is multiplication by $\omega;$ addition of $\omega^2$ yields multiplication by $\omega^\omega$ (all of these are meant as on the right). All of these produce larger ordinals.

If $\alpha$ does not have any $\omega^n$ terms, then $\alpha=\omega\alpha$ and so if $2^{\alpha}=\alpha$ then $\alpha=\omega^\alpha$ by the lemma; this is the defining property of an $\epsilon$-number.

Note that the same applies to any finite base.

share|improve this answer
    
Nice. Thank you :) –  Katlus Jun 4 '12 at 9:55

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.