I'm trying the following question to revise for a set theory exam:

Let $\omega_1$ be the first uncountable ordinal. Show the existence of an ordinal $\beta$ such that $\omega_1^\beta = \beta$

The hint I am given is that if $\alpha \le \beta$ then $\omega_1^\alpha \le \omega_1^\beta$ and for any $\alpha$, $\omega_1^\alpha < \omega_1^{\alpha+}$, where $\alpha+$ is the successor ordinal of $\alpha$.

I don't really know how to go about this; any pointers in the right direction would be appreciated.

  • 1
    $\begingroup$ Look up how to find an $\epsilon$ number greater than a given ordinal, which you can find in most any elementary (i.e. pre-forcing) set theory text. $\endgroup$ – Dave L. Renfro May 8 '17 at 15:38
  • $\begingroup$ Are you assuming the generalized continuum hypothesis? $\endgroup$ – Jef L May 8 '17 at 15:38
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    $\begingroup$ @JefLaga This is ordinal exponentiation, for which cardinality assumptions are irrelevant. $\endgroup$ – Andrés E. Caicedo May 8 '17 at 15:51

You can use the following fact: if $f: \text{ORD} \rightarrow \text{ORD} $ is a 'function' (i.e. assigns to each ordinal a unique ordinal) which is strictly increasing and continuous (i.e. for every limit ordinal $\lambda$ we have $f(\lambda) = \sup(\{f(\gamma) \mid \gamma < \lambda\})$ then $f$ has a fixed point: there exists an ordinal $\alpha$ such that $f(\alpha) = \alpha$. The 'function' $f(\alpha) = \omega_1^{\alpha}$ is seen to be strictly increasing by your hint (plus a small argument) and it is also continuous (by definition of exponentiation by transfinite induction).

The proof of the quoted fact is a standard exercise in ordinal arithmetic and I assume you can find it in most texts.


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