Showing that $cf(\aleph_\alpha) \leq cf(\alpha)$

Question

How can I prove this direction?

Let $\alpha$ be a limit ordinal. Then $cf(\aleph_\alpha) \leq cf(\alpha)$.

Answer 1

If $\alpha_i \uparrow \alpha$ then show $\aleph_{\alpha_i}\uparrow \aleph_{\alpha}$ too.

Answer 2

Assuming that by "$\aleph \alpha$" you mean $\aleph_\alpha$.

By definition $\aleph_\alpha = \bigcup_{\beta < \alpha} \aleph_\beta$. Now take $\langle \beta_\gamma : \gamma < \operatorname{cf}(\alpha) \rangle$ to be a cofinal sequence in $\alpha$, then $\langle \aleph_{\beta_\gamma} : \gamma < \operatorname{cf}(\alpha) \rangle$ is cofinal in $\aleph_\alpha$. So we see that $\operatorname{cf}(\aleph_\alpha) \leq \operatorname{cf}(\alpha)$.

Edit: as pointed out in the comments, we actually have equality. You can see this by using a similar trick: let $\langle \delta_\gamma : \gamma < \operatorname{cf}(\aleph_\alpha) \rangle$ be cofinal in $\aleph_\alpha$. Then for each $\gamma < \operatorname{cf}(\aleph_\alpha)$ we define $\beta_\gamma$ to be the least such that $\aleph_{\beta_\gamma} > \delta_\gamma$. Note that we then still have $\aleph_{\beta_\gamma} < \aleph_\alpha$. Now it is easy to check that $\langle \beta_\gamma : \gamma < \operatorname{cf}(\aleph_\alpha) \rangle$ is cofinal in $\alpha$, giving us $\operatorname{cf}(\alpha) \leq \operatorname{cf}(\aleph_\alpha)$ and thus combining with the previous result we have equality.