A variant of diagonal intersection I am doing an exercise about a variant of diagonal intersection：$\bigtriangleup_{s\in {\left[\kappa\right]}^{<\omega}}X_s=\{\alpha\in\kappa:\alpha\in\bigcap_{s\in {\left[\alpha\right]}^{<\omega}}X_s\}$.
Here $\kappa$ is a cardinal, $X_s\in U$, where $U$ is a normal filter (We can assume $U$ is $\kappa$-complete or ultra if needed).
The problem is showing $\bigtriangleup_{s\in {\left[\kappa\right]}^{<\omega}}X_s\in U$. I tried to construct a useful bijection from ${\left[\kappa\right]}^{<\omega}$ to $\kappa$ but failed, can someone give me a hint? Thanks in advance.
 A: Observe that your diagonal intersection is equal to
$$\bigcap_{n<\omega}\left\{ \alpha<\kappa\> \middle| \> \alpha\in \bigcap_{s\in [\alpha]^n}X_s\right\}.$$
Now I claim that $U$ is closed under the diagonal intersection over $[\kappa]^n$, that is, if $\{Y_s\mid s\in[\kappa]^n\}\subseteq U$ is a family of sets, then $\triangle_{s\in[\kappa]^n}Y_s = \{\alpha<\kappa\mid \alpha\in\bigcap_{s\in[\alpha]^n}Y_s\}\in U$.
The case $n=1$ is obvious (it is just a usual diagonal intersection.) Assume that our claim holds for $n$. Let $\{Y_s\mid s\in[\kappa]^{n+1}\}\subseteq U$ be a given family of sets.
We can see that
$$\triangle_{s\in [\kappa]^{n+1}}Y_s= \{\alpha<\kappa\mid \alpha\in \bigcap_{t\in[\alpha]^n}\bigcap_{\max t<\beta<\alpha} Y_{t\cup\{\beta\}}\},$$
which is equal to $\triangle_{t\in[\kappa]^n}\{\alpha<\kappa\mid \alpha\in \bigcap_{\max t<\beta<\alpha} Y_{t\cup\{\beta\}}\}$. You can see that each
$$\left\{\alpha<\kappa\>\middle|\> \alpha\in \bigcap_{\max t<\beta<\alpha} Y_{t\cup\{\beta\}}\right\}$$
is a member of $U$, since it is a diagonal intersection of sets of $U$.
