If CH assumed, can we prove this?

Question

$$\aleph_2^{\aleph_0}=\aleph_2$$

Appreciate your help

Answer 1

Yes, you can. Any function $f:\omega\to\omega_2$ is bounded, so ${}^\omega\omega_2=\bigcup_{\alpha<\omega_2}{}^\omega\alpha$, and therefore

$$\aleph_2\le\aleph_2^{\aleph_0}=\left|{}^\omega\omega_2\right|\le\sum_{\alpha<\omega_2}|\alpha|^\omega=\aleph_2\cdot\aleph_1^{\aleph_0}=\aleph_2\cdot2^{\aleph_0\cdot\aleph_0}=\aleph_2\cdot2^{\aleph_0}=\aleph_2\cdot\aleph_1=\aleph_2\;.$$

Answer 2

One can use Hausdorff's formula (if it is for their disposal),

$$\aleph_{\alpha+1}^{\aleph_\beta}=\aleph_\alpha^{\aleph_\beta}\cdot\aleph_{\alpha+1}$$

From there we have: $$\aleph_2^{\aleph_0}=\aleph_1^{\aleph_0}\cdot\aleph_2=\left(2^{\aleph_0}\right)^{\aleph_0}\cdot\aleph_2=2^{\aleph_0}\cdot\aleph_2=\aleph_1\cdot\aleph_2=\aleph_2.$$

One may also derive Bernstein's formula: $\aleph_n^{\aleph_\beta}=2^{\aleph_\beta}\cdot\aleph_n$, from which the result is even more immediate.