smallest cardinal greater than an infinite ordinal is a regular cardinal

Question

let $\alpha$ be an infinite ordinal, and $\alpha^+$ be the smallest cardinal greater than $\alpha$. Show that $\kappa^+$ is a regular cardinal.

This is for homework, but I'm not really sure where to begin.

Answer 1

Suppose that $\alpha^+$ is singular; then there are a cardinal $\lambda<\alpha^+$ and a sequence $\langle\alpha_\xi:\xi<\lambda\rangle$ such that $\alpha^+=\sup_{\xi<\lambda}\alpha_\xi$. Show that there must be a $\xi<\lambda$ such that $\alpha<|\alpha_\xi|<\alpha^+$, contradicting the definition of $\alpha^+$. If after some thought this isn’t enough, I’ve spoiler-protected a further hint below; mouse-over to see it.

HINT: You know that $\lambda\le|\alpha|$, so if $|\alpha_\xi|\le\alpha$ for all $\xi<\lambda$, then what can you say about $|\sup_{\xi<\lambda}\alpha_\xi|$?