$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_2\cdot\aleph_1^{\aleph_0}$

Question

I've seen this statement in multiple posts (e.g. here and here), but I can't seem to understand it. I can see why

$$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_1^{\aleph_0},$$

by noting that every $\alpha<\omega_2$, so $|\alpha|\leq\aleph_1$, meaning

$$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\max_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_1^{\aleph_0}.$$

So how does $\aleph_2$ enter the equation?

Answer 1

The sum has $\aleph_2$ terms, each of which is at most $\aleph_1^{\aleph_0}$, so it’s bounded above by $\aleph_2\cdot\aleph_1^{\aleph_0}$. On the other hand, it’s clearly at least $\aleph_2$ and at least $\aleph_1^{\aleph_0}$, so its bounded below by their maximum, which is simply their product.

Answer 2

We can prove something quite more general. Given a regular cardinal $\kappa$, noticing that each function $\lambda\rightarrow \kappa$ for $\lambda<\kappa$ ($\lambda$ cardinal) must have bounded range in $\kappa$, we can write $$\kappa^{\lambda}=\bigcup_{\alpha <\kappa} \alpha^{\lambda},$$ where $\alpha$ ranges over ordinals. Therefore we get easily $$\kappa^{\lambda}=\sum_{\alpha< \kappa}\vert \alpha\vert^{\lambda}.$$ Using this fact, since each successor cardinal is regular, we can prove (for all $\alpha,\beta$ ordinals) the Hausdorff's Formula: $$\aleph_{\alpha +1}^{\aleph_{\beta}}=\aleph_{\alpha +1}\cdot\aleph_{\alpha}^{\aleph_{\beta}}.$$ (The question is a particular case for $\alpha =1$ and $\beta =0$).