# The cofinality of $\aleph_{\omega\cdot9+3}$

I am studying for a test and I was able to find the cofinality 3 of the 4 ones given, but am having a lot of trouble with the 4th. the 3 first ones are:

1. $\newcommand{\cf}{\operatorname{cf}}\cf(\aleph_\omega)=\omega$
2. $\cf(\aleph_{\omega^2})=\omega$
3. $\cf(\omega_1)=\omega_1$

I do not know how to calcuate $\cf(\aleph_{\omega\cdot9 + 3})$. I am very tempted to say it is simply $\omega$.

Thank you.

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Hint: Note that $\aleph_{\alpha+1}$ is regular, for any ordinal $\alpha$. In particular $\alpha=\omega\cdot 9+2$.
Like you, I initially thought $\omega$ is enough. So I tried the sequence $\aleph_{n\cdot 9+3}$, it turns out all of them are strictly smaller than $\aleph_{\omega}$, which violates the conditions of a cofinality.
No,it's $\aleph_{\omega\cdot 9+3}$. You can search cofinality on this website, Asaf Karagila have excellent expositions on this topic.Besides, on Page 49 of Jech's Set Theory 3ED, there's a proof. – Metta World Peace Nov 26 '12 at 22:02