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Let [J]: Jech "Set theory" (Millenium edition)

Let $\kappa$ a limit ordinal.

From [J], T.3.11, p. 33 we have that $\kappa<\kappa^{cf(\kappa)}$.

I improved that proof, and obtain :

$\kappa < cf(\kappa)^{cf(\kappa)}$

PROOF: Let $\gamma:=cf(\kappa)$ and $c: \gamma\to \kappa$ cofinal, ordered, injective. Give a family $F=${$f_k|k<\kappa$} of functions $f_k: cf(\kappa)\to cf(\kappa)$ define $f: cf(\kappa)\to cf(\kappa)$ as follow:

$f(\delta)$ is the minimum $\beta$ such that $\beta\neq f_k(\delta)\ \forall k < c(\delta)$.

We have that $f(\delta\dot{+}1)\neq f_{c(\delta)}(\delta\dot{+}1)$ (where $\dot{+}$ is the ordinal sum, or successor).

Then $f\not\in F$. (end of Proof)

From [J], Cor.5.12 we have that $cf(2^{\aleph_\alpha})>\aleph_\alpha$ for any ordinal $\alpha$.

Form the cardinal identity $\lambda^\lambda=2^\lambda$ ([J], L.5.6) and from above follow that $2^{cf(\aleph_\alpha)}>\aleph_\alpha$

My question is:

How are related $2^{cf(\aleph_\alpha)}$ and $cf(2^{\aleph_\alpha})$ ?

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1 Answer 1

Your improved version of Konig's theorem is not correct. For example, the cofinality of $\aleph_\omega$ is $\omega$, but under the CH, $\omega^\omega$ is much less than $\aleph_\omega$. So it isn't generally true that $\kappa\lt\text{cof}(\kappa)^{\text{cof}(\kappa)}$.

One doesn't need CH to provide a counterexample, in light of the fact that there are arbitrarily large cardinals with cofinality $\omega$, and they can't all be less than $\omega^\omega$.

The flaw in your proof is that you define $f(\delta)$ to be different from $c(\delta)$ many values, but if $c(\delta)$ is bigger than $\text{cof}(\kappa)$, then there may be no values left.

Regarding your last question, the cofinality of $\aleph_\alpha$ is $\aleph_\alpha$, when this is a successor cardinal, and is $\text{cof}(\alpha)$, when it is a limit cardinal. So $2^{\text{cof}(\aleph_\alpha)}$ is similarly either $2^{\aleph_\alpha}$ or $2^{\text{cof}(\alpha)}$ depending on whether you are in the successor or limit case. Meanwhile, the cofinality of $2^{\aleph_\alpha}$ is always at least $\aleph_{\alpha+1}$ by Konig's theorem. But it can be less than $2^{\aleph_\alpha}$ if the GCH fails badly. So it is possible for $2^{\text{cof}(\aleph_\alpha)}$ to be larger than $\text{cof}(2^{\aleph_\alpha})$. Meanwhile, it is also possible for them to be equal, for example for successor cardinals under the GCH, and it is also possible for it to be less, such as the case of a singular limit cardinal $\aleph_\alpha$ under the GCH.

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thank you very much! –  Buschi Sergio Jun 17 '12 at 9:29

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