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


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


So how does $\aleph_2$ enter the equation?


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.

  • $\begingroup$ Thanks for the quick answer! How is it clear that it is at least $\aleph_2$? Just because there's $\aleph_2$ elements in the sum? $\endgroup$ – Dan Saattrup Nielsen May 17 '13 at 21:35
  • $\begingroup$ @Leidem: Yes. You’re looking at (the cardinality of) a disjoint union of $\aleph_2$ sets, only one of which is empty. $\endgroup$ – Brian M. Scott May 17 '13 at 21:37
  • $\begingroup$ Ah yes of course, that makes total sense now. Thanks a bunch! $\endgroup$ – Dan Saattrup Nielsen May 17 '13 at 21:39
  • $\begingroup$ @Leidem: You’re welcome! $\endgroup$ – Brian M. Scott May 17 '13 at 21:41

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 $\beta =0$).

  • $\begingroup$ Thanks for the answer. Why is it important that $\kappa$ is regular? Does it have something to do with the boundedness of the range of functions $f\in{^\lambda\kappa}$? $\endgroup$ – Dan Saattrup Nielsen May 20 '13 at 11:30
  • $\begingroup$ @Leidem Yes, it does: if $\kappa$ is regular, its cofinality, $cf(k)$, is $\kappa$ itself and therefore you can't have a cofinal map from a cardinal $\lambda< \kappa$ to $\kappa$ (otherwise $cf(k)\leq \lambda$), i.e. given any cardinal $\lambda< \kappa$ there are no functions $\lambda\rightarrow \kappa$ with unbounded range. Boundness of functions is the key point to be able to write down $\kappa$ as that union (it is exactly what that means). $\endgroup$ – Marco Vergura May 20 '13 at 12:06
  • $\begingroup$ Ah yeah, that makes sense, thank you! On a small note, I think you meant $\alpha<\kappa$ in the union. $\endgroup$ – Dan Saattrup Nielsen May 20 '13 at 13:35
  • $\begingroup$ Yes, for sure, thanks! I'll edit! $\endgroup$ – Marco Vergura May 20 '13 at 13:40

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