Generalizing an identity of Hausdorff: $\aleph_1^{\aleph_0}=2^{\aleph_0}$

Question

In a set theory class, I learned that $\aleph_1^{\aleph_0}=2^{\aleph_0}$, which was proven by a clever counting argument.

I was wondering if it was possible to generalize this. I.e., does $\aleph_2^{\aleph_1}=2^{\aleph_1}$, or more generally $$\aleph_{\beta+1}^{\aleph_\beta}=2^{\aleph_{\beta}}$$ for other ordinals?

Answer 1

$2\leq \aleph_{\beta+1}$ implies $2^{\aleph_\beta} \leq \aleph_{\beta+1}^{\aleph_\beta}$.

$\aleph_{\beta+1} \leq 2^{\aleph_\beta}$ implies $\aleph_{\beta+1}^{\aleph_\beta} \leq (2^{\aleph_\beta})^{\aleph_\beta} = 2^{\aleph_\beta\cdot \aleph_\beta} = 2^{\aleph_\beta}$.

So $\aleph_{\beta+1}^{\aleph_\beta} = 2^{\aleph_\beta}$.

Answer 2

In fact, a stonger equality is known as Hausdorff formula $$\aleph_{\alpha+1}^{\aleph_\beta} = \aleph_\alpha^{\aleph_\beta}\cdot\aleph_{\alpha+1}.$$

If $\aleph_\beta\ge \aleph_{\alpha+1}$, then the equality is trivial. We will show this equality by using cofinality argument when $\aleph_\beta<\aleph_{\alpha+1}$.

Let $f:\omega_\beta\to\omega_{\alpha+1}$ be a function. (A set of all such a function has cardinality $\aleph_{\alpha+1}^{\aleph_\beta}$.) Since $\omega_{\alpha+1}$ is regular and $\omega_\beta<\omega_{\alpha+1}$, $f$ is bounded. Therefore, the range of $f$ is a subset of for some $\xi<\omega_{\alpha+1}$. In sum, we have $${^{\omega_{\beta}}}\omega_{\alpha+1}= \bigcup_{\xi<\omega_{\alpha+1}} {^{\omega_\beta}}\xi$$ and $\aleph_{\alpha+1}^{\aleph_\beta} = \sum_{\xi<\omega_{\alpha+1}} |\xi|^{\aleph_\beta} = \aleph_{\alpha+1}\cdot \aleph_\alpha^{\aleph_\beta}$.