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 $cf(\alpha) \leq cf(\aleph_{\alpha})$. I have already proven that $cf(\aleph_{\alpha}) \leq cf(\alpha)$. My idea for the other direction is as follows:
Proof idea: By definition there exists a function $f:cf(\aleph_{\alpha}) \to \aleph_{\alpha}$ such that $ran(f)$ is cofinal in $(\aleph_{\alpha},\leq)$. Let $x \in \alpha$, then there is $\theta \in cf(\aleph_{\alpha})$ such that $f(\theta) \geq x$. So, {$f(\beta): \beta \in cf(\aleph_{\alpha}), f(\theta) \geq x$} is non-empty, and thus has a least element $\gamma$.
Now if I can prove that $\gamma \in \alpha$, then I could define the function
$h=${$(x,y) \in f: f(x) \in \alpha$} $\cup$ {$(x,y) \in dom(f) \times \alpha: f(x) \not \in \alpha, y=0$}.
Clearly, $dom(h) = cf(\aleph_{\alpha})$, $ran(h) \subseteq \alpha$ and $ran(h)$ is cofinal in $(\alpha,\leq)$. It follows that $cf(\alpha) \leq cf(\aleph_{\alpha})$.
How can I prove that the least element $\gamma \in \alpha$ for any $x \in \alpha$?
Thank you!
Let $\beta= \operatorname{cf}(\aleph_{\alpha})$ and let $f:\beta \to \aleph_{\alpha}$ be strictly increasing and cofinal.
For $\xi \in \beta$, define $g(\xi)=\eta$, where $\vert f(\xi) \vert = \aleph_{\eta}$. (If $f(\xi) \lt \omega$, define $g(\xi)=0$.) Then $g:\beta \to \alpha$ is cofinal because $\alpha$ is limit and $f$ is cofinal in $\aleph_{\alpha}$, so $\operatorname{cf}(\alpha) \leq \beta$.