singular cardinals of every regular cofinality

Question

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.

Answer 1

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 $g(\eta)<\aleph_{\kappa+\alpha_\eta}$.

Answer 2

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.

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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 $\kappa$.