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First of all a couple of definitions which might be different to the standard ones:

A function is called cofinal if $$f:\alpha\mapsto \beta$$is such that $$\sup\{ f(\gamma):\gamma<\alpha\}=\beta$$

Secondly, cf$(\kappa)$ by the $\min\{\alpha:\exists f: \alpha\mapsto \beta\}$ where $f$ is cofinal. An ordinal is called regular if $cf(\alpha)=\alpha$, and singular if it is not regular.

The problem now asks: Prove that there exist singular cardinals of every regular cofinality.

So say we are given $\kappa$ with the property that $\kappa=$cf$(\kappa)$, and then we can consider $\aleph_{\kappa+\kappa}$. Then we first show that there exists a cofinal function that maps $\kappa$ into $\aleph_{\kappa+\kappa}$, just let $f(\alpha)=\aleph_{\kappa+\alpha}$ by the way addition works we see that $\sup\{f(\alpha):\alpha<\kappa\}=\aleph_{\kappa+\kappa}$. However I do not know how to show that this is the least one that works. I am assuming that I must use the regularity of $\kappa$ somehow but I dont know how.

Thanks.

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Corrected: Suppose that there is some $\lambda<\kappa$ and a function $g:\lambda\to\aleph_{\kappa+\kappa}$ such that $\sup\{g(\xi):\xi\in\lambda\}=\aleph_{\kappa+\kappa}$. For each $\xi\in\lambda$ let $\alpha_\xi=\min\{\alpha<\kappa:g(\xi)<\aleph_{\kappa+\alpha}\}$. Then $\{\alpha_\xi:\xi\in\lambda\}$ is a subset of $\kappa$ of cardinality at most $\lambda$, $\lambda<\kappa$, and $\kappa$ is regular; can you see now how to use this to get a contradiction? Notice that $g(\eta)\le\sup\{\aleph_{\kappa+\alpha_\xi}:\xi\in\lambda\}$ for every $\eta\in\lambda$, since by definition $g(\eta)<\aleph_{\kappa+\alpha_\eta}$.

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But the indexing function is normal, so we might have that $\kappa=\aleph_\kappa$. –  Daniel Montealegre Mar 1 '12 at 23:13
    
The punch line is going to be that then you can construct a function from $\lambda$ to $\kappa$ which is cofinal and hence cf$(\kappa)\leq \lambda <\kappa$, and the way you will define such a function $h$ would be by $h(\xi)=\alpha_\xi$, but I dont see how $\sup\{\alpha_{xi}:\xi\in \lambda\}=\kappa$. –  Daniel Montealegre Mar 1 '12 at 23:25
    
this might be a dumb question, but why is $g(\xi)\leq \sup\{\alpha_{\xi}:\xi\in\alpha\}$ –  Daniel Montealegre Mar 2 '12 at 1:24
    
@Daniel: Damn! I fixed the missing symbol in one place but not in the other. That should be $\sup\{\aleph_{\alpha_\xi}:\xi\in\lambda\}$. I’ll do it right away. –  Brian M. Scott Mar 2 '12 at 1:33
    
Oh ok so $\sup\{g(\eta):\eta\in \lambda\}\leq\sup\{\aleph_{\alpha_{\xi}}:\xi\in \lambda\}<\aleph_\kappa$ contradicting the fact that $g$ is a cofinal function from $\lambda$ to $\aleph_\kappa$, but I still think that we should use $\aleph_{\kappa+\kappa}$ –  Daniel Montealegre Mar 2 '12 at 1:48
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Indeed as you thought $\aleph_{\kappa+\kappa}$ is a good candidate. The missed up point on the regularity of $\kappa$ comes in the fact that $\mathrm{cf}(\mu)$ is a regular cardinal for every $\mu$.

This means that if $\kappa$ is a singular cardinal then $\aleph_{\kappa+\kappa}$ will have the same cofinality as $\kappa$, which by the assumption of singularity is not $\kappa$ itself.


We will prove the following theorem:

Suppose $\kappa$ is an infinite cardinal, then $\mathrm{cf}(\kappa)=\mathrm{cf}(\mathrm{cf}(\kappa))$. In particular it shows that $\mathrm{cf}(\kappa)$ is a regular cardinal.

Denote $\mathrm{cf}(\kappa)=\lambda$ and $\mathrm{cf}(\lambda)=\mu$. Since $f(\alpha)=\alpha$ is cofinal in $\lambda$ we have that $\mu\le\lambda$. Suppose by contradiction that this is a strong inequality. This means that there exists a function $f\colon\mu\to\lambda$ which is cofinal in $\lambda$.

Since $\lambda=\mathrm{cf}(\kappa)$ we know there is some $h\colon\lambda\to\kappa$ which is cofinal in $\kappa$. We show that $(h\circ f)\colon\mu\to\kappa$ is also cofinal in $\kappa$:

Suppose $\beta<\kappa$ then there is some $\alpha<\lambda$ such that $\beta<h(\alpha)$, and since $f$ is cofinal in $\lambda$ we have some $\gamma<\mu$ such that $\alpha<f(\gamma)$. This means that $h(\alpha)<h(f(\gamma)$.

However we also know that $\beta<h(\alpha)<h(f(\gamma))$. Therefore $h\circ f$ is a cofinal function whose domain is strictly smaller than $\lambda$, in contradiction to our assumption that $\lambda=\mathrm{cf}(\kappa)$ was the minimal ordinal which can be mapped unboundedly into $\kappa$.

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why is cf$(\mu)$ regular? I am sorry but I barely learned the above concepts so know very little about them. –  Daniel Montealegre Mar 1 '12 at 23:36
    
@Daniel: I'd expect the first theorem after defining cofinality would be that $\mathrm{cf}(\mathrm{cf}(\kappa))=\mathrm{cf}(\kappa)$. The proof is not difficult, we have $\le$ trivially, suppose by contradiction that there is no equality, then $\mathrm{cf}(\mathrm{cf}(\kappa))<\mathrm{cf}(\kappa)$. Use this to generate a new cofinal function which shows that $\kappa$ has smaller cofinality then we originally assumed it has. From this theorem we have automatically that cofinalities are always regular cardinals. –  Asaf Karagila Mar 1 '12 at 23:40
    
The professor only said the definitions and said he might have it on my midterm tomorrow so I am trying to get comfortable with the concept and the book doesnt cover it. The reason why you say that $\mbox{cf}(\mbox{cf}(\beta))\leq\mbox{cf}(\beta)$ is because $\mbox{cf}(\alpha)\leq \alpha$ always holds through the identity? –  Daniel Montealegre Mar 1 '12 at 23:59
    
@Daniel: Indeed. The function $f(\alpha)=\alpha$ is unbounded and shows that you can only consider ordinals which are at most $\alpha$. –  Asaf Karagila Mar 2 '12 at 0:02
    
so to finish my proof that is in my post I ought to show that $\mbox{cf}(\aleph_{\kappa+\kappa})<\kappa$ would yield a contradiction, and the argument should be somewhat similar, assume not then create some cofinal function such that $\mbox{cf}(\kappa)<\kappa$, a contradiction. Is there a relation between the alephs and their index? (i.e., is there a relation between $\mbox{cf}(\aleph_\alpha)$ and $\alpha$) –  Daniel Montealegre Mar 2 '12 at 1:15
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