# Strictly increasing sequence of ordinals indexed by a regular cardinal: $\text{cf}(\bigcup\{\alpha_i:i\in\kappa\})=\kappa.$

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.

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

$$g(\eta):=\sup\left[\{i<\kappa:\alpha_i Note that $$g(\eta)<\kappa$$ by regularity of $$\kappa$$ and $$\eta<\lambda\le \kappa$$.

Obviously, $$\xi<\eta$$ implies $$g(\xi).

Given $$i<\kappa$$, $$\alpha_i<\beta$$, so $$\alpha_i for some $$\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 $$\cf(\kappa)=\kappa$$.

• Can you explain the blockquote / provide a reference? May 19 '20 at 0:44
• This is a very instructive exercise. Think about it for a bit. May 19 '20 at 5:40