Suppose that for all $\alpha<\kappa$ we have that $A\_\alpha\subseteq\kappa$. We define the diagonal intersection to be $$\bigtriangleup\_{\alpha<\kappa}A\_\alpha = \left\lbrace\xi<\kappa\ \middle|\ \xi\in\bigcap\_{i<\xi}A\_i\right\rbrace$$


One of the most surprising theorems in basic set theory, I think, is that if $A\_\alpha$ is closed and unbounded (and $\kappa$ is regular and uncountable) then this diagonal intersection is also a closed and unbounded set.


Looking at it from a measure theoretic point of view now, clubs correspond to sets of measure one. Is there any measure theoretic operation which corresponds to diagonal intersections?


Are there possibly other analogies in mathematics which can be used to describe this construction in a rather simple way that non-set theorists could relate to?


Furthermore, it is quite clear that changing the order of the $A\_\alpha$ or taking a subsequence can completely change the resulting set. Is there some invariance? For example, up to order the result is unique modulo a non-stationary set?



Intuition behind the diagonal intersection
The diagonal intersection corresponds to the infimum in the boolean algebra $\mathbb{B}\_I:=P(\kappa)/I$ (where $I$ is the nonstationary ideal, or more generally where $I$ is any normal ideal on $\kappa$). More precisely: if $Z \subset P(\kappa)/I$ and $|Z| = \kappa$, then $Z$ has an infimum in $\mathbb{B}\_I$, and this infimum is exactly ``the'' diagonal intersection of representatives from the members of $Z$. (This diagonal intersection does not depend on the particular $\kappa$-enumeration of $Z$ or the choice of representatives from the equivalence classes in $Z$; they'll all yield the same element of $\mathbb{B}\_I$).


