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I am throwing yet another one of my solutions out here for the internets to debug and for future set-theory students.

Let $\aleph_\delta$ a weakly inaccessible cardinal. Prove that $A =\{\alpha < \aleph_\delta \mid \alpha = \aleph_\alpha\}$ is a club set.

Any comments on my solution are welcome.

Thanks!

Asaf Karagila
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HNHN
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1 Answers1

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First we will show that $A$ is closed.

Let $\langle\alpha_i \mid i < \lambda\rangle \subset A$, and let $\alpha_\lambda = \bigcup_{i < \lambda}\alpha_i$. We want to show $\alpha_\lambda = \aleph_{\alpha_\lambda}$. If $\alpha_\lambda < \aleph_{\alpha_\lambda}$, there is $\beta < \alpha_\lambda$ such that $|\alpha_\lambda| = \aleph_\beta$. Hence there is $\alpha_i$ such that $\alpha_\lambda \leq \aleph_{\alpha_i}$. But $\alpha_\lambda \leq \aleph_{\alpha_i} = \alpha_i < \alpha_\lambda$.

Now we will show that $A$ is not bounded. To show that, I will try to apply what I learned from this question.

Let $\gamma < \aleph_\delta$. $\aleph_\delta$ is weakly inaccessible, so it is a fixed point of the $\aleph$ function, hence: $\gamma \leq \aleph_\gamma < \aleph_\delta$. We define the following series:

  • $\gamma_0 = \aleph_\gamma$
  • $\gamma_{n+1} = \aleph_{\gamma_n}$
  • $\gamma_\omega = \bigcup_{n < \omega}\gamma_n$

From the same reasons as before, for every $n$, $\gamma_n < \aleph_\delta$. Also, $\aleph_0 < \aleph_\delta$ and $\aleph_\delta$ is regular, therefor $\gamma_\omega < \aleph_\delta$, so if we show that $\gamma_\omega = \aleph_{\gamma_\omega}$, we win.

But this is true because if $\gamma_\omega < \aleph_{\gamma_\omega}$, we get that for some $n$, $\gamma_\omega \leq \aleph_{\gamma_n} < \aleph_{\gamma_{n+1}} = \gamma_{n+2} < \gamma_\omega$.

Victory.

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HNHN
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    I think the same proof shows that the set of fixed points for the $\aleph$ function below any fixed point $\delta = \aleph_\delta$ such that $cf(\delta) > \aleph_0$ is club in $\delta$. – William Jan 20 '14 at 11:54
  • @William Looks like that, yes... You think there's a problem with my proof, or just that we could weaken the conditions in the question? – HNHN Jan 20 '14 at 12:06
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    No, the proof is correct. I just wanted to point out that this phenomenon is not special to the large cardinal. The question you linked to above shows that such $\delta$, I mentioned above, exist but large cardinals may not. Just to mention, this result is not true for fixed point of countable cofinality by looking at the first fixed point. – William Jan 20 '14 at 12:11
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    The proof looks fine. This is really the key point in my answer to the linked question. The fixed points of the $\aleph$ form a club [class] in the cardinals, therefore at any limit point (i.e. a fixed point which is a limit of fixed points) the intersection is a club. Of course that we want to talk about limit points which are of uncountable cofinality in order to talk about clubs to begin with. The second part of your proof, as noted by @William, shows that every fixed point of uncountably cofinality is a limit of fixed points. – Asaf Karagila Jan 20 '14 at 12:22