I am stuck on the following problem: Let $\kappa$ be a regular cardinal and $(\alpha_i)_{i\in\kappa}$ be a strictly increasing sequence of ordinals. Prove that $\text{cf}(\beta)=\kappa,$ where $\beta=\bigcup\{\alpha_i:i\in\kappa\}.$
I am thinking of separately showing the $\le$ and $\ge$ inequalities. First, to show the $\le$ inequality in the case $\kappa$ is infinite (leaving the finite case for later), we can use the following Criterion: Let $\alpha$ be a limit ordinal, and let $C\subseteq\alpha.$ Then $C$ is cofinal in $\alpha$ iff $\cup C=\alpha.$
Indeed, we can show (by contradiction) that $\beta$ is a limit ordinal. In addition, since $\kappa$ is infinite, for each $i\in\kappa$ we have $i+1<\kappa$ and thus $\alpha_i<\alpha_{i+1}\le\beta,$ so $\alpha_i\in\beta.$ Thus, since $\beta=\bigcup\{\alpha_i:i\in\kappa\},$ we have by the Criterion that $\{\alpha_i:i\in\kappa\}$ is cofinal in $\kappa,$ so $\text{cf}(\beta)\le\kappa.$
However, I am stuck on showing the other ($\ge$) inequality.
Best Answer
The reverse inequality is a bit trickier. We can argue as follows:
Put $\lambda=\text{cf}(\beta)\le \kappa$. Suppose $f:\lambda \to \beta$ is strictly increasing and cofinal (i.e. its range is cofinal).
Define $g:\lambda\to\kappa$ by recursion: Suppose $\langle g(\xi):\xi<\eta\rangle $ has already been defined, where $\eta<\lambda$. Let
$$ g(\eta):=\sup\left[\{i<\kappa:\alpha_i<f(\eta)\}\cup\{g(\xi)+1:\xi<\eta\}\right] $$ Note that $g(\eta)<\kappa$ by regularity of $\kappa$ and $\eta<\lambda\le \kappa$.
Obviously, $\xi<\eta$ implies $g(\xi)<g(\eta)$.
Given $i<\kappa$, $\alpha_i<\beta$, so $\alpha_i<f(\xi)$ for some $\xi<\lambda$. But then $i\le g(\xi)$ by definition of $g$. This shows that $g$ has cofinal range in $\kappa$. But then $\kappa\le \lambda$ by regularity, and we're done.