Equivalent condition for and infinite $ \kappa $ to be regular

Question

I'm trying to figure out why the following is true:

Suppose $ \kappa $ is an infinite cardinal. Then the following conditions are equivalent:

1) $\kappa $ is regular

2) for every cardinal $ \lambda < \kappa $ and a non-decreasing sequence of ordinals $ (x_\alpha : \alpha < \kappa)$ such that $ x_\alpha < \lambda $ for all $ \alpha < \kappa $ it holds that $ \sup\limits_{\alpha < \kappa}x_\alpha < \lambda $

I know that $ \kappa $ is regular iff for every $ \lambda < \kappa $ and a strictly increasing $ x : \lambda \rightarrow \kappa $ it holds that $ \sup\limits_{\alpha < \lambda} x_\alpha < \kappa$, but the other condition is somehow an inverse to that. How do I approach the solution? It simply seems untrue for my inexperienced eyes

Edit: non-decreasing

Answer 1

Suppose that $\lambda<\kappa$ and $\langle x_\alpha:\alpha<\kappa\rangle$ is a non-decreasing $\kappa$-sequence in $\lambda$. Suppose further that for each $\xi<\lambda$ there is an $\alpha(\xi)<\kappa$ such that $x_{\alpha(\xi)}>\eta$. For each $\eta<\lambda$ let $\beta(\eta)=\sup_{\xi\le\eta}\alpha(\xi)$; then $\langle\beta(\eta):\eta<\lambda\rangle$ is a non-decreasing $\lambda$-sequence in $\kappa$ such that $\langle x_{\beta(\eta)}:\eta<\lambda\rangle$ is cofinal in $\lambda$. Let $\mu=\sup_{\eta<\lambda}\beta(\eta)$; then either

• $\mu=\kappa$ and $\langle\beta(\eta):\eta<\lambda\rangle$ is cofinal in $\kappa$, or
• $x_\alpha=x_\mu$ for $\mu\le\alpha<\kappa$, and $\sup_{\alpha<\kappa}x_\alpha=x_\mu<\lambda$.