Ytukyg

I'm trying to show that club filter of $kappa$ is $kappa$-complete for uncountable regular cardinal $kappa$:

Let $kappa$ be uncountable regular cardinal, let $C(kappa)$ be the club filter generated by $kappa$.

To show that $C(kappa)$ is $kappa$-complete it is enough to show that for every sequence $langle alpha_imid i<gammarangle$ with $gamma<kappa$ and $alpha_i$ club of $kappa$ the set $alpha=bigcap_{i<gamma}alpha_i$ is a club of $kappa$.

Showing that $alpha$ is close is easy:

If $beta_n$ is a sequence in $alpha$ then for every $i$ we that $beta_n$ is a sequence in $alpha_i$, because $alpha_i$ is a club it is close hence $lim beta_n$ is in $alpha_i$ for every $i$ so it is also in $alpha$

But I have a problem with unboundedness, my guess will be for each $lambda<kappa$ to find a sequence for each $alpha_i$ such that all of the sequences converge to some $mu>lambda$, that way $beta<muinalpha$. I think I'll have to use the fact $kappa$ is regular, but I don't know how to proceed and prove my idea.

So, is my idea correct, and if yes how can I proceed?

Here is a possible approach, to begin with you need to prove that the intersection of two club sets in $kappa$ is still a club set, which I'll leave for you since it's an easier case.

Once this is done we can prove that if $kappa$ is regular, $gamma<kappa$ and $(C_alphamid alpha<gamma)$ is a sequence of clubs in $kappa$ then $bigcap C_i$ is also a club in $kappa$, by induction on $gamma$.

The successor case follows easily from the fact I stated above about the intersection of two clubs, so we concentrate on the case in which $gamma$ is a limit ordinal.
Note that by replacing $C_alpha$ with $bigcap_{lambdaleqalpha} C_lambda$ we can assume that $C_0supseteq C_1supseteq C_2supseteqcdots$, since we didn't change the intersection, that is $$bigcap_{alpha<gamma}C_alpha=bigcap_{alpha<gamma}bigcap_{lambdaleqalpha}C_{lambda}.$$

We can now construct the needed sequence of length $gamma$. We fix $beta<kappa$, pick $beta_0>beta$ and for every $lambda<gamma$ we pick $beta_lambdain C_lambda$ with $beta_lambda>sup{beta_ximid xi<lambda}$ (which exist since every $C_alpha$ is unbounded). Since $kappa$ is regular we have that the sup of this sequence, call it $hat{beta}$ is still smaller than $kappa$, furthermore $hat{beta}$ is a limit point of every $C_alpha$, namely it is the limit of the sequence $(beta_{nu}mid alphaleqnu<gamma)subseteq C_alpha$, so $hat{beta}in C_alpha$ for every $alpha$ and $hat{beta}inbigcap C_alpha$.

Note that we don't really need regularity here, if $kappa$ is singular, as long as $operatorname{cof}(kappa)>omega$, the club filter on $kappa$ is still $operatorname{cof}(kappa)$ complete, by exactly the same argument, replacing $gamma<kappa$ with $gamma<operatorname{cof}(kappa)$.