length of a normal not cofinal sequence

Question

Let $\kappa$ an infinite cardinal. Does every normal sequence $\langle\alpha_\xi\rangle \subseteq \kappa$ with $\sup \alpha_\xi<\kappa$ is of length $<cf\kappa$ ? I think yes but I have some difficulty to prove it rigorously...

Answer 1

If the cardinal is regular then the answer is yes. Every bounded normal sequence has less than $\kappa$ elements so it has to have order type of $<\kappa=\operatorname{cf}(\kappa)$.

If $\kappa$ is singular, then for every ordinal $\lambda$ such that $\operatorname{cf}(\kappa)<\lambda<\kappa$, $\lambda$ is a sequence which is longer then the cofinality, bounded, and certainly normal.

Answer 2

No. The normal sequence $\langle i \rangle_{i \leq \omega}$ in $\aleph_\omega$ is bounded in $\aleph_\omega$ but has length $\omega + 1 > \omega = \mathrm{cf} ( \aleph_\omega )$. (You should now see how to construct bounded normal sequences in $\aleph_\omega$ of length $\omega_n$ for all $n \in \omega$.)