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I need to show that

if $\cal F$ is a normal filter on a regular uncountable cardinal $\kappa$, and if $\cal F$ contains all tail sets, i.e. all $$ C_\alpha=\{\beta \ : \ \alpha<\beta<\kappa\}$$ Then $\cal F$ is $\kappa$-complete

My thought

Let $X$ be a collection of less than $\kappa$ elements of $\cal F$, i.e. for some $\lambda<\kappa$: $$ X = \{X_\alpha,\ \alpha<\lambda\}$$ With $X_\alpha\in \cal F$ for all $\alpha$.

We need to show that $\bigcap_{\alpha<\lambda} X_\alpha \in \cal F$. Let $Y$ be the collection $X$ adding the tail set with $\alpha\geq\lambda$: $$Y = \{X_\alpha:\ \alpha<\lambda\} \cup \{C_\alpha : \ \alpha \geq \lambda\}$$

Lets define $Y_\alpha$ as $$Y_\alpha = \left\{ \begin{array}{ll} X_\alpha&\text{ if } \alpha<\lambda\\ C_\alpha&\text{ if } \alpha\geq\lambda \end{array}\right.$$

For all $\alpha$, $Y_\alpha\in\cal F$, then because $\cal F$ is normal, $$\triangle_{\alpha<\kappa} Y_\alpha \in \cal F$$ But by definition of the diagonal intersection $\triangle_{\alpha<\kappa} Y_\alpha = \{\beta<\kappa : \ \beta \in \bigcap_{\alpha<\beta}Y_\alpha\}$ we have $\triangle_{\alpha<\kappa}Y_\alpha \subseteq \bigcap_{\alpha<\lambda} X_\alpha$. Therefore, because $\cal F$ is a filter $$ \bigcap_{\alpha<\lambda} X_\alpha\in \cal F$$

My issue is with the statement $$\triangle_{\alpha<\kappa}Y_\alpha \subseteq \bigcap_{\alpha<\lambda} X_\alpha$$

Let $\beta \in \triangle_{\alpha<\kappa}Y_\alpha = \{\beta<\kappa : \ \beta \in \bigcap_{\alpha<\beta}Y_\alpha\}$

Clearly if $\lambda < \beta < \kappa$, then $\beta\in\bigcap_{\alpha<\beta}Y_\alpha$ and $\beta\in\bigcap_{\alpha<\lambda}Y_\alpha=\bigcap_{\alpha<\lambda}X_\alpha$

But if $\beta < \lambda$, how can I prove that $\beta\in\bigcap_{\alpha<\lambda}X_\alpha$ ?

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    $\begingroup$ Intersect $X_\alpha$ with $C_\lambda$ in your definition of $Y_\alpha.$ That is still in $\mathcal F$ of course, and it takes care of the last part for you. $\endgroup$ – spaceisdarkgreen Mar 19 at 11:19

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