(CH) implies $\omega_2 = \omega_2^\omega$: reference?

Question

Just by chance I found out that (CH) implies $\omega_2 = \omega_2^\omega$, which I found quite surprising, since this an almost trivial conclusion of $2^{\omega_1} = \omega_2$ rather than of $2^{\omega} = \omega_1$. So, I'm wondering, if this implication is known, and if yes, I would appreciate to get a reference for this. BTW: I'm always working in (ZFC). I think in particular the above mentioned proof uses a lot of choice.

Answer 1

Since $\omega_2$ is regular, any function from $\omega$ to $\omega_2$ is bounded. Thus $\omega_2^\omega=\bigcup_{\alpha<\omega_2}\alpha^\omega$. The cardinality of this is just $\omega_2\cdot\sup_{\alpha<\omega_2}|\alpha^{\omega}|$. Since $|\alpha^{\omega}|\le|\omega_1^{\omega}|=\omega_1$ under $\mathsf{CH}$ for $\alpha<\omega_2$, $\omega_2^\omega=\omega_2\cdot\omega_1=\omega_2$.

Answer 2

Since you ask for a reference, let me point out that this is essentially a special case of the Hausdorff formula $$\aleph_{\alpha+1}^{\aleph_\beta} = \aleph_\alpha^{\aleph_\beta}\cdot \aleph_{\alpha+1}$$

Indeed, assuming CH we have $\aleph_2^{\aleph_0} = \aleph_1^{\aleph_0}\cdot \aleph_2 = \aleph_1\cdot \aleph_2 = \aleph_2$.

Proceeding by induction, we find that that under CH, $\aleph_n^{\aleph_0} = \aleph_n$ for all finite $n$.

A reference for the Hausdorff formula is Jech's Set Theory, Equation (5.22) on p. 57.