Ordinals that satisfy $\alpha = \aleph_\alpha$ with cofinality $\kappa$

Question

I am trying to solve the following question:

Prove that for every regular cardinal, $\kappa \gt \aleph_0$, there is a exists an $\alpha$ with cofinality $\kappa$ such that $\alpha = \aleph_\alpha$

I tried to build $\alpha$ as the limit of $\kappa$ regular cardinals with various properties (this seems to sort out the requirement that $cof(\alpha) = \kappa$). I want to choose all of them to be weakly inaccessible but I'm not entirely sure it's "allowed" and if there isn't a simpler approach.

My attempt:

• Let $A = \langle\alpha_i\mid i < \kappa\rangle$, increasing series of weakly inaccessible cardinals, and let $\alpha = \bigcup_{i < \kappa}\alpha_i$
• $cof(\alpha) = \kappa$. Otherwise, we can define a bijection between $A$ and $\kappa$ and contradict $\kappa$'s regularity.
• We know that $\alpha \leq \aleph_\alpha$.
• Assume $\alpha < \aleph_\alpha$, so there is $i$ such that $\alpha < \aleph_{\alpha_i}$
• $\alpha_i$ is weakly inaccessible, hence it's a fixed point of the aleph function, which means $\alpha_i = \aleph_{\alpha_i}$
• But $\alpha < \aleph_{\alpha_i} = \alpha_i < \alpha$, and this is a contradiction.

Answer 1

You can't prove the existence of weakly inaccessible cardinals. You need to assume that. So using inaccessible cardinals is generally the wrong approach for this problem.

Instead, we define the following function:

• $F(0)=\aleph_0$,
• $F(\alpha+1)=\aleph_{F(\alpha)}$ (where $\aleph_{F(\alpha)}$ is $\aleph_{\beta}$ where $\beta$ is the least ordinal of cardinality $F(\alpha)$).
• $F(\delta)=\sup\{F(\alpha)\mid\alpha<\delta\}$ when $\delta$ is a limit ordinal.

It is not hard to prove that $F$ is normal, and that for every limit ordinal $\delta$ we have $\delta=\aleph_\delta$. By normality we have that every possible cofinality is realized at some limit ordinal.

Answer 2

You don't need inaccessible cardinals. The index $\alpha$ is the (ordinal) number of alephs below $\aleph_\alpha$; an aleph fixed point is just an uncountable cardinal $\kappa$ with $\kappa$ cardinals below it.

Take any infinite cardinal $\lambda_0$ to start with. Let $\lambda_1$ be an infinite cardinal with more than $\lambda_0$ cardinals below it; i.e., if $\lambda_0=\aleph_\alpha$, you can take $\lambda_1=\aleph_{\omega_{\alpha+1}}$ or anything bigger. Continue in this way for $\omega$ steps, so that $\lambda_{n+1}$ is a cardinal with more than $\lambda_n$ cardinals below it. It's easy to see that $\lambda=\sup\{\lambda_n:n\lt\omega\}$ is an aleph fixed point of cofinality $\omega$. Since your starter cardinal $\lambda_0$ was arbitrary, we see that there are arbitrarily large aleph fixed points (of cofinality $\omega$).

Now let $\kappa$ be any regular infinite cardinal. If $(\mu_\alpha:\alpha\lt\kappa)$ is a strictly increasing sequence of aleph fixed points, then $\mu=\sup\{\mu_\alpha:\alpha\lt\kappa\}$ is an aleph fixed point of cofinality $\kappa$.