14
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ Just a nit-pick: $\kappa$ must be an uncountable regular cardinal. $\endgroup$
    – Trevor Wilson
    Sep 14, 2012 at 15:55

1 Answer 1

Reset to default
16
$\begingroup$

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 $\mathbb{B}_I$).

Improve this answer
$\endgroup$
5
  • $\begingroup$ Hm, that is quite a wonderful insight. I have two followup questions now: (1) Is there yet a measure theoretic notion corresponding to that? (2) If we take an arbitrary ideal $I$, is there a reasonable way to describe the operation $\inf[X_\alpha]$ in $P(\kappa)/I$, or is that just a very nice property of normal ideals? $\endgroup$
    – Asaf Karagila ♦
    Sep 14, 2012 at 14:57
  • $\begingroup$ I do not think there is a nice notion of $inf[X_{a}]$ in arbitrary ideals $I$ since the $inf[X_{\alpha}]$ generally does not exist. For instance, in $P(\omega)/fin$ where $fin$ is the ideal consisting of finite sets, then the least upper bound of a countable sequence of elements almost never exists since there is no countable strictly increasing sequence of elements $(x_{n})$ in $P(\omega)/fin$ with a least upper bound. $\endgroup$
    – Joseph Van Name
    Sep 14, 2012 at 16:06
  • $\begingroup$ Regarding Asaf's questions: 1) I don't know, other than viewing elements of a Boolean algebra as being 0, 1, or "positive". 2) It's just a very nice property of normal ideals. For non-normal ideals on $\kappa$, sups and infs of $\kappa$-sized subsets of $\mathbb{B}_I$ do not necessarily exist, as shown by Joseph's example (his example was a non-normal ideal on $\kappa = \omega$, but more generally, for any regular $\kappa$ the ideal $J$ of bounded subsets of $\kappa$ is $\kappa$-complete, non-normal, and some $\kappa$-sized subsets of $\mathbb{B}_{J}$ fail to have sups). $\endgroup$
    – Sean Cox
    Sep 14, 2012 at 17:02
  • $\begingroup$ Ah, I see. Thanks Joseph and Sean. Is normality somehow equivalent to completeness (or some $\kappa^+$-closedness) of the quotient algebra? $\endgroup$
    – Asaf Karagila ♦
    Sep 14, 2012 at 17:33
  • $\begingroup$ Asaf, it is not equivalent. Suppose $I$ is a $\kappa^+$-saturated ideal on a regular $\kappa$. Let $f$ be a bijection between the limit ordinals and successor ordinals of $\kappa$. Then $f$ projects $I$ to a $\kappa$-complete, $\kappa^+$-saturated ideal $J$ for which the set of successor ordinals is measure one. $J$ is not normal but $P(\kappa)/J$ is a complete boolean algebra. $\endgroup$
    – Monroe Eskew
    Sep 14, 2012 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.