Intersection of two clubs.

Question

I am currently rereading the proof of Lemma 136. given above and I understand almost every part of the proof, except for the next one.

Namely, I am aware that for such constructed $\beta$ author is trying first to make sure that this $\beta$ satisfies exactly the same properties as $\lambda$ in the "closedness" definition of Definition 135.

This is why he wants $\beta$ to be a limit ordinal, "smaller" then $\kappa$ and such that $\beta$ $\cap$ $C$ is unbounded in $\beta$ as well as $\beta$ $\cap$ $D$ unbounded in $\beta$.

What I don't understand is why does $cof(\kappa)\geq \omega_1$ imply that $\beta < \kappa$.

Answer 1

By construction, $\beta_0,\beta_1,\beta_2,\ldots$ is an increasing sequence of length $\omega$, consisting of ordinals $<\kappa$. The definition of cofinality states that if $\mathrm{cf}(\kappa) = \lambda$, then no sequence of ordinals $<\kappa$ of length $<\lambda$ can limit to $\kappa$. Since $\kappa$ has uncountable cofinality, no countable sequence of ordinals $<\kappa$ can limit to $\kappa$. Since each $\beta_n$ is $<\kappa$, the limit of the $\beta_n$ is at most $\kappa$; since the sequence is countable, the limit can't be $\kappa$, so the limit must be $<\kappa$. That limit is $\beta$, so $\beta < \kappa$.

Answer 2

$\beta = \sup_{n < \omega} \beta_n$. Since all the $\beta_n$ are below $\kappa$ and $\kappa$ is of uncountable cofinality, it follows (by the definition of cofinality) that the set $\{ \beta_n \mid n < \omega \}$ is not $<$-cofinal in $\kappa$ and hence that its least upper bound - $\beta$ - is strictlty below $\kappa$.