Club filter of $\kappa$ is $\kappa$-complete 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 $\langle \alpha_i\mid i<\gamma\rangle$ 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 $\beta<\mu\in\alpha$. 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?
 A: 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 $(C_\alpha\mid \alpha<\gamma)$ 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 $\bigcap_{\lambda\leq\alpha} C_\lambda$ we can assume that $C_0\supseteq C_1\supseteq C_2\supseteq\cdots$, since we didn't change the intersection, that is $$\bigcap_{\alpha<\gamma}C_\alpha=\bigcap_{\alpha<\gamma}\bigcap_{\lambda\leq\alpha}C_{\lambda}.$$
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 $\beta_\lambda\in C_\lambda$ with $\beta_\lambda>\sup\{\beta_\xi\mid \xi<\lambda\}$ (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 $(\beta_{\nu}\mid \alpha\leq\nu<\gamma)\subseteq C_\alpha$, so $\hat{\beta}\in C_\alpha$ for every $\alpha$ and $\hat{\beta}\in\bigcap C_\alpha$.
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 $\gamma<\operatorname{cof}(\kappa)$.
