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Background

The beth function is defined recursively by: $\beth_0 = \aleph_0$, $\beth_{\alpha + 1} = 2^{\beth_\alpha}$, and $\beth_\lambda = \bigcup_{\alpha < \lambda} \beth_\alpha$. Since the beth function is strictly increasing and continuous, it is guaranteed to have arbitrarily large fixed points by the fixed-point theorem on normal functions.

The cofinality of an ordinal $\alpha$ is the smallest ordinal $\beta$ such that there are unbounded increasing functions $f : \beta \to \alpha$.

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Yes. The definable class $C$ of fixed points $\beth_\kappa = \kappa$ is closed and unbounded and therefore contains ordinals of all possible cofinalities. Specifically, let $\kappa_\alpha$ denote the $\alpha$-th element of $C$. Note that $\kappa_\delta = \sup_{\alpha<\delta} \kappa_\alpha$ for every limit ordinal $\delta$. If $\lambda$ is a regular cardinal then $\kappa_\lambda$ has cofinality $\lambda$.

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Yes, the κth fixed point will have the same cofinality as κ, since the κ many earlier fixed points are cofinal in κ.

More generally, for any function f from the ordinals to the ordinals, it is not difficult to see that the collection of ordinals α for which f " α subset α, that is, the closure points of f, will form a closed unbounded class C. And for any limit ordinal β, the β-th element of C has the same cofinality as β, since it is the limit of the β many smaller elements of C.

In particular, if we consider the beth function, where α maps to Bethα, then β is a closure point of this function if and only if Bethα < β for all α < β, and this implies Bethβ = β. So the club C in this case consists of the Beth fixed points.

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I think this is our record for closest back-to-back answers. :) –  François G. Dorais Mar 6 '10 at 18:08
    
I think you're right! –  Joel David Hamkins Mar 6 '10 at 18:19

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