Cofinality of two ordinals I would calculate cofinality of two ordinals:


*

*$Cof(\omega_2^{\omega_1})$

*$Cof(\aleph_{\omega^\omega})$


I know that a $cof(X) < X$, but i cannot solve it..
 A: The other part of the answer: the first cofinality is $\omega_1$. 
To see this, note that $f\colon \alpha\to \omega_2^{\alpha}$ is a cofinal ordinal function $\omega_1\to\omega_2^{\omega_1}$, meaning, it's strictly increasing and its range is cofinal in the codomain shown. So $cof(\omega_2^{\omega_1}) \le \omega_1$. If equality didn't hold, then we'd have to have $cof(\omega_2^{\omega_1}) = \omega$, because $cof(\alpha)$ is always a regular cardinal ($cof(cof(\alpha))=cof(\alpha)$) and the next smallest one is $\omega$. If that were so, there would be a cofinal function $g\colon\omega\to\omega_2^{\omega_1}$, and you could use $g$ and $f$ to define a cofinal function $\omega\to\omega_1$ — impossible, as $\omega_1$ is regular.
A: As a partial answer, the second cofinality is $\omega$. In general, 
$$
\mathop{\mathrm{cf}}(\aleph_\alpha)=\mathop{\mathrm{cf}}(\alpha)
$$
whenever $\alpha$ is a limit ordinal. Hence $\mathop{\mathrm{cf}}(\aleph_{\omega^\omega}) = \omega$. This can be seen directly since
$$
\aleph_{\omega^\omega} = \textstyle\bigcup\{\aleph_{\omega^n}: n\in \omega\}.
$$
Therefore, the countable sequence $\{\aleph_{\omega^n}: n\in \omega\}$ is cofinal in $\aleph_{\omega^\omega}$
