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I was given the following task in one exam. The task was as follows:

$$\text{Simplify } 2^{\omega_1} \text{ using the following lemma: }1 < \alpha, \beta < \gamma \Rightarrow \alpha^\beta < \alpha^\gamma.$$

I would be grateful for any of your hints.

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this is ordinal arithmetic right? – Apostolos May 31 '11 at 15:10
Yes, added to tags. – ShyUser May 31 '11 at 15:14
up vote 4 down vote accepted

You first need to observe that if $\beta$ is a countable ordinal then $2^\beta$ is countable as well.

Work with induction: For the successor step observe that $2^{\beta+1}=2^\beta\cdot 2$ which is countable if $2^\beta$ is countable (which is the induction hypothesis). For the limit stage the limit of a sequence of ordinals is their union and since $\beta$ is countable then the countable union of countable sets is countable.

So $2^{\omega_1}$ is the limit of a strictly increasing (by the lemma given) $\omega_1$-sequence of countable ordinals and thus it is $\omega_1$.

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Thanks, that helped me a lot. :) I have drawn the same conclusion, but sadly not using the given lemma. – ShyUser May 31 '11 at 15:57

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