Why is $\mathrm{cf}(\alpha)$ a limit ordinal and $\mathrm{cf}(\alpha)\leq \alpha$?

Question

Defining the cofinality $\mathrm{cf}(\alpha)$ of a limit ordinal $\alpha$ as the least ordinal number $\theta$ such that $\alpha$ is the limit of an increasing sequence of ordinals of length $\theta$, why must $\mathrm{cf}(\alpha)$ be a limit ordinal and furthermore $\mathrm{cf}(\alpha)\le\alpha$?

Answer 1

Your definition of $\mathrm{cf}(\alpha) = \theta$ should say that $\alpha$ is a limit of an increasing sequence of length $\theta$ of ordinals less than $\alpha$. Otherwise every ordinal would have cofinality $1$, since $\alpha$ is the limit of the singleton sequence $\alpha$.

• $\mathrm{cf}(\alpha)$ is a limit ordinal: Suppose $(\beta_\gamma)_{\gamma\in \theta}$ is an increasing sequence of ordinals less than $\alpha$, with $\alpha = \sup_{\gamma\in \theta} \beta_\gamma$. If $\theta = 0$, then $\alpha = \sup \emptyset = 0$, contradicting the fact that $\alpha$ is a limit ordinal. If $\theta = \theta'+1$ is a successor ordinal, then $\alpha = \sup_{\gamma\in \theta} \beta_\gamma = \beta_{\theta'}$, since the sequence has a greatest element. But this contradicts $\beta_{\theta'} < \alpha$. Since $\theta$ is not zero or a successor ordinal, it must be a limit ordinal.

In fact, you can do better than this and prove that $\mathrm{cf}(\alpha)$ is an infinite cardinal (in the sense that no ordinal $\beta<\mathrm{cf}(\alpha)$ can be put in bijection with $\mathrm{cf}(\alpha)$). I'll leave this to you as an exercise.

• $\mathrm{cf}(\alpha) \leq \alpha$: We have $\alpha = \sup_{\beta\in \alpha} \beta$, so $\alpha$ is the limit of an increasing sequence of length $\alpha$ of ordinals less than $\alpha$. Since $\mathrm{cf}(\alpha)$ is the least such length, $\mathrm{cf}(\alpha)\leq \alpha$.