Timeline for Basic Algebraic Applications of Stationary Sets?
Current License: CC BY-SA 3.0
11 events
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| Aug 30, 2017 at 19:54 | comment | added | Joel David Hamkins | @PaceNielsen Yes, I had understood you to be asking that, but unfortunately, I don't know of any such elementary application of stationarity to algebra. Perhaps someone else has an example. | |
| Aug 30, 2017 at 19:43 | comment | added | Pace Nielsen | What I'm looking for is an application you could show a first-year graduate student to impress upon their minds the usefulness of stationary sets in the context of algebra (just as Zorn's lemma is used to impress them with the usefulness of the axiom of choice). | |
| Aug 30, 2017 at 19:41 | comment | added | Pace Nielsen | @JoelDavidHamkins Rephrasing your answer: Stationary sets (1) are connected to a natural quotient of the Boolean algebra $P(\kappa)$ when $\kappa$ is regular uncountable [more generally $P_{\kappa}(X)$ for appropriate $X$], (2) they provide for another way to measure size, and (3) their definition can be algebraically characterized. These are three great reasons, the first of which I should have thought of myself (having already read chapter 8 of Jech)! Still, is there an application of stationary sets in ZFC to solve an algebraic problem (perhaps similar in spirit to Whitehead's problem)? | |
| Aug 30, 2017 at 18:59 | comment | added | Joel David Hamkins | Yes, that's right. | |
| Aug 30, 2017 at 18:24 | comment | added | Gerhard Paseman | Qualifying your comment to "uncountable algebras of countable type", I agree. Gerhard "Thanks You For Your Perspective" Paseman, 2017.08.30. | |
| Aug 30, 2017 at 18:03 | comment | added | Joel David Hamkins | Perhaps it would help to point out that stationarity is inherently about uncountable objects, and uncountable algebras always have numerous countable subalgebras. We can think of the subalgebras as "points". So a stationary set is a positive-measure collection of such points: it has such a point (in fact many) for any given algebra. | |
| Aug 30, 2017 at 17:48 | comment | added | Gerhard Paseman | OK. For the benefit of the question (and because I am trying to jump-start part of my brain), I was hoping to elicit a response that was algebraic but not subalgebraic: sometimes you are in a context where the functions are not basic operations (internal to an algebra) but you get closed sets another way (say through a precursor to a Galois connection). Unfortunately, I have not worked obviously with stationary sets, so I may be asking the wrong question. Gerhard "Wants More Than Just Closure" Paseman, 2017.08.30. | |
| Aug 30, 2017 at 17:37 | comment | added | Joel David Hamkins | The way I think about it is that algebras give you a particular closed set: the collection of subalgebras, and stationary sets are those that work with any such closed set, so they contain a subalgebra for any given algebra. And there is an implicit closure operator for any algebra: closing a set under the algebraic operations. | |
| Aug 30, 2017 at 17:34 | comment | added | Gerhard Paseman | Interesting. Do you see a similar connection between stationary sets and closure operators on a set? Gerhard "Needs More Coffee To Imagine" Paseman, 2017.08.30. | |
| Aug 30, 2017 at 15:27 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 125 characters in body
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| Aug 30, 2017 at 15:15 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |