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.


2 Answers 2


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.


If you already know some basic properties of the cofinality function, then this is a fairly quick deduction: $\DeclareMathOperator{\cf}{cf}$

Cofinality is unique. In other words, if $\sup A=\delta$, then $\cf(\delta)=\cf(\operatorname{otp}(A))$, where $\operatorname{otp}(A)$ is the unique ordinal which is order isomorphic to $A$.

Now set $A$ as the set $\{\alpha_i\mid i<\kappa\}$, since the function is strictly increasing, $\operatorname{otp}(A)=\kappa$. So by the above property we have $\cf(\kappa)=\cf(A)=\cf(\beta)$. But by the assumption that $\kappa$ is regular, we get $\cf(\kappa)=\kappa$.

  • $\begingroup$ Can you explain the blockquote / provide a reference? $\endgroup$ May 19, 2020 at 0:44
  • 1
    $\begingroup$ This is a very instructive exercise. Think about it for a bit. $\endgroup$
    – Asaf Karagila
    May 19, 2020 at 5:40

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