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

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.
• Are you revising material for an exam? – Asaf Karagila Jan 10 '14 at 12:21
• @AsafKaragila yes, why? – Hila Jan 10 '14 at 15:23
• What's the fun in that? – Asaf Karagila Jan 10 '14 at 15:31
• Whatever floats your boat... – Hila Jan 10 '14 at 15:32
• My boat is a submarine. – Asaf Karagila Jan 10 '14 at 15:36

You don't need inaccessible cardinals or normal functions. 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 $\kappa$.

• So... you don't need normal functions, but what you did here was essentially proving that a certain function is normal, then used the fact that its range is a club. As long as you don't need it, though! – Asaf Karagila Jan 11 '14 at 7:19
• "an aleph fixed point is just an infinite cardinal $\kappa$ with $\kappa$ cardinals below it." This can't be quite right, since $\aleph_0$ has precisely $\aleph_0$ cardinals strictly below it, however it is not a fixed point of $\aleph$. Perhaps adding uncountability to the claim makes it true? i.e. Is it true that an aleph fixed point is just an uncountable cardinal $\kappa$ with $\kappa$ cardinals below it? – goblin Apr 29 '14 at 7:35
• @user18921, good point. An aleph fixed point is an infinite cardinal $\kappa$ with $\kappa$ infinite cardinals (i.e. alephs) below it. Seeing as there are only countably many finite cardinals, we can afford to ignore them if $\kappa$ is uncountable. In other words, you are right on all counts. – bof Apr 29 '14 at 8:47

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.

• Or, just show that there are arbitrarily large aleph fixed points (say, of cofinality $\omega$). If $\kappa$ is regular, then the supremum of the first $\kappa$ aleph fixed points is an aleph fixed point of cofinality $\kappa$. – bof Jan 10 '14 at 11:27
• @Asaf Nice, we haven't learned about normal functions. It's a good thing to know. Thanks! – Hila Jan 10 '14 at 11:31
• First, do you know how to find an aleph fixed point of cofinality $\omega$? – bof Jan 10 '14 at 11:37
• @Hila: Even if you haven't learned about normal functions yet, this is essentially the very method you need to use. Just limit yourself to the first $\kappa$ steps, or the first $\kappa$ steps above $\kappa$ itself. – Asaf Karagila Jan 10 '14 at 11:43
• @bof: Really this is just the same proof for the same amount of effort. – Asaf Karagila Jan 10 '14 at 11:44