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I am working through a proof on cardinals I found and can't reason some of the steps.

The proposition is that there is an $\aleph$-fixed point, i.e. there is an ordinal $\alpha$ (which is necessarily a cardinal), so that $\aleph_{\alpha} = \alpha$. The proof goes as follows:

Let $\alpha_{0} = \aleph_{0}$ (or any other cardinal), $\alpha_{n + 1} = \aleph_{\alpha_{n}}$, and $\alpha = \sup \{ \alpha_{n} \mid n \in \omega \}$. Now if $\alpha = \alpha_{n}$ for some $n$, then $\alpha = \alpha_{n+1} = \aleph_{\alpha_{n}} = \aleph_{\alpha}$. Otherwise $\alpha$ is a limit ordinal and we have that $\aleph_{\alpha} = \sup \{ \aleph_{\xi} \mid \xi < \alpha\} = \sup\{ \aleph_{\alpha_{n}} \mid n \in \omega \} = \sup \{\alpha_{n + 1} \mid n \in \omega \} = \alpha$.

Now the limit case makes sense to me, but why on earth can we state that if $\alpha = \alpha_{n+1}$, then $$ \alpha = \alpha_{n+1} = \aleph_{\alpha_{n}} = \aleph_{\alpha}. $$

I suppose the main issue I am having is why does $\alpha = \alpha_{n}$ imply that $\alpha = \alpha_{n+1}$.

After that, it is really just a matter of applying definitions.

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I think the key piece is that the $\alpha_n$s are increasing: $\alpha_n\le\alpha_{n+1}$.

More broadly, the following is true:

$\aleph_\beta\ge\beta$, for all $\beta$.

So we know $\alpha_n\le\alpha_{n+1}\le\alpha$ (the latter inequality since $\alpha$ is the sup of the $\alpha_n$s), so if $\alpha_n=\alpha$ then $\alpha_n=\alpha_{n+1}$ - and indeed $\alpha_n=\alpha_k$ for all $k\ge n$.

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  • $\begingroup$ Thanks, this is more clear now. $\endgroup$ – Oiler Apr 18 '16 at 15:30
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If $\alpha=\alpha_n$, then $\alpha_{n+1}=\aleph_{\alpha_n}$ is no greater than $\alpha_n=\aleph_{\alpha_{n-1}}$ because of the $\sup$ in the definition of $\alpha$

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$\alpha\mapsto \aleph_{\alpha}$ is a normal ordinal function: it's (strictly) increasing, and it's "continuous at limit ordinals", in the sense that $$ \aleph_{\lambda} = \sup_{\alpha < \lambda} \aleph_{\alpha}. \tag{limit $\lambda$} $$ Every normal ordinal function $f$ has fixed points, in fact unboundedly many if its domain is unbounded. Suppose $\gamma \in dom(f)$. Let $\alpha_0 = \gamma$ and $\alpha_{n+1} = f(\alpha_n)$; let $\alpha = \sup_n f(\alpha_n)$. Then $\alpha$ is a limit, as $(\alpha_n)_{n<\omega}$ is a strictly increasing sequence with limit $\alpha$. By definition of a normal function at limits, $f(\alpha) = \sup_{\xi < \alpha} f(\xi) = \sup_n f(\alpha_n) = \sup_n \alpha_{n+1} = \alpha$.

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  • $\begingroup$ @Holo: What is the edit about? $\endgroup$ – Asaf Karagila Aug 3 at 12:21
  • $\begingroup$ @AsafKaragila I miss-clicked downvote by mistake, and notice few minutes later, so I edit to remove my downvote $\endgroup$ – ℋolo Aug 3 at 12:34

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