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My textbook Introduction to Set Theory by Hrbacek and Jech presents Hausdorff's Formula:

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and its corresponding proof:

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I am unable to deduce 3. from 1. and 2. as stated in the proof.

  1. Each function $f:\omega_\beta \to \omega_{\alpha+1}$ is bounded

  2. The definition of ordinal exponentiation: $\omega_{\alpha+1}^{\omega_\beta}=\sup \{\omega_{\alpha+1}^{\lambda} \mid \lambda< \omega_\beta\}=\bigcup_{\lambda< \omega_\beta}\omega_{\alpha+1}^{\lambda}$

  3. $\omega_{\alpha+1}^{\omega_\beta}=\bigcup_{\gamma < \omega_{\alpha+1}}\gamma^{\omega_\beta}$

Could you please elaborate on this point? Thank you for your help!

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1 Answer 1

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It is not ordinal exponentiation, it is cardinal exponentiation, i.e. $\omega_{\alpha+1}^{\omega_\beta}$ is the set of all functions $\omega_\beta\to \omega_{\alpha+1}.$ If every such function is bounded, then every such function is a function $\omega_\beta\to \gamma$ for some $\gamma < \omega_{\alpha}$ and hence $ \omega_{\alpha+1}^{\omega_\beta} = \bigcup_{\gamma < \omega_{\alpha+1}}\gamma^{\omega_\beta}.$

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  • $\begingroup$ Thank you so much for your answer! I am now clear. To avoid this confusion, it is better to denote the set of all functions $f:X \to Y$ by $^X Y$ rather than $Y^X$. Although this denotation is very uncommon, I have seen some users doing so. $\endgroup$
    – Akira
    Commented Dec 30, 2018 at 4:18

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