I am trying to prove the following Lemma:
Lemma: Let $\alpha$ be a limit ordinal. Then $cf(\aleph_{\alpha}) = cf(\alpha)$.
I am using the following definitions:
Definition: Let $(\mathbb{P}, \preceq)$ be a partially ordered set. A subset $A \subseteq \mathbb{P}$ is said to be cofinal if for all $p \in \mathbb{P}$ there exists $q \in A$ such that $p \preceq q$.
Definition: Let $\alpha$ be an ordinal. The cofinality of $\alpha$ is the least ordinal $\lambda$ such that there exists a function $f: \lambda \to \alpha$ whose range is cofinal in $(\alpha, \leq)$. We denote the cofinality of $\alpha$ by $cf(\alpha)$.
There is also a useful lemma:
Lemma: Let $\alpha$ be a non-zero ordinal number. Then there exists a strictly increasing function $f: cf(\alpha) \to \alpha$ whose range is cofinal in $(\alpha,\leq)$.
My idea is to prove the lemma in question is to prove that $cf(\aleph_{\alpha}) \leq cf(\alpha)$ and $cf(\alpha) \leq cf(\aleph_{\alpha})$. My idea for the first inequality is as follows:
Proof idea: Let $f: cf(\alpha) \to \alpha$ be a strictly increasing function such that $ran(f)$ is cofinal in $(\alpha,\leq)$. Consider the function $g: \alpha \to \aleph_{\alpha}$ such that $g(\theta)=\aleph_{\theta}$ for all $\theta \in \alpha$.
If I can show that the range of $g \circ f: cf(\alpha) \to \aleph_{\alpha}$ is cofinal in $(\aleph_{\alpha},\leq)$, then $cf(\aleph_{\alpha}) \leq cf(\alpha)$.
Let $p \in \aleph_{\alpha}$. How can I show that there is $q \in ran(g \circ f)$ such that $p \leq q$?
Thank you!