Let $\alpha$ be a limit ordinal. Let $f\colon\beta\to\alpha$ be a cofinal function, that is, a function so that $f(\beta)$ is unbounded in $\alpha$. That is, we have $(\forall \gamma < \alpha)(\exists \eta < \beta)(\gamma \leq f(\eta))$.
It is true that $\bigcup_{\eta < \beta} f(\eta) = \alpha$? If yes, why?
On
Suppose $\gamma<\alpha$. Since $ran(f)$ is unbounded in $\alpha$, we can find some $\hat{\gamma}\in\beta$ with $f(\hat{\gamma})>\gamma$ (why could I write "$>$" instead of "$\ge$" here?). Now:
How are $\gamma$ and $f(\hat{\gamma})$ related in terms of "$\in$" (rather than "$>$")?
Why does this mean that $\bigcup_{\eta\in\beta}f(\eta)=\alpha$?
Apparently, yes. It is clear that $\bigcup_{\eta < \beta} f(\eta) \subseteq \alpha$.
Let $\gamma < \alpha$. Since $f$ is cofinal, let $\eta < \beta$ so that $\gamma \leq f(\eta)$. Then $\gamma < f(\eta) + 1$. Note that since $f(\eta) < \alpha$, $f(\eta) + 1 \leq \alpha$. Since $\alpha$ is a limit ordinal, $f(\eta) + 1 \neq \alpha$, hence $f(\eta) + 1 < \alpha$. Again, since $f$ is cofinal, let $\xi < \beta$ so that $f(\eta) + 1 \leq f(\xi)$. Then $$\gamma < f(\eta) + 1 \leq f(\xi).$$