A dumb formula
Feb. 5th, 2013 12:17 amSo. As you probably know, I'm currently writing a paper in which I prove that a certain subset of the real numbers is well-ordered with order type ωω. And yes it should have been done like a month ago (or probably earlier). Real life has introduced some delays.
Anyway. The point is, in order to do this, I need to cite various set theory/order theory/point set topology stuff, to handle ordinals embedded in larger totally ordered sets (which here is always the reals). I have a section of the paper where I introduce all the stuff I need from there. I'd held off on actually writing it until pretty recently though. I didn't actually know where I'd cite the stuff from; I figured it wouldn't be too hard to find.
So Jeff suggested I ask Andreas Blass where I might find citations for these things. His response on hearing just what I wanted was, yes, that's certainly easy, but I'm actually not certain you'll find it in the literature. Try the following really old books; they cared a lot about this sort of thing around the turn of the century.
So I got the books Andreas suggested out from the library, and, unfortunately, they mostly didn't help. So, I'm having to write most of the proofs myself.
Anyway. The point of this entry was a particular thing. Say we have an ordinal α, and we let β be the set of all limit ordinals less than α. How can we express β in terms of α? Sounds easy, right? I mean it's basically just "dividing by ω", right?
Well basically yes, but if you actually want the order type on the nose, well, it's surprisingly easy to screw up. Now actually all I really need is that if α<ωn+1 (n finite), then β<ωn, so I don't need to prove a whole formula. (Actually, I guess if you just replaced n+1 with 1+n, this statement would still be true for n infinite.) But today I sat down and figured out just what the correct formula was (because you know I coudln't find this listed anywhere) so I thought I'd record it here for reference.
First off, if α is finite, there are no limit points. Otherwise...
Say α has Cantor normal form ωγ_kaγ_k + ... + ωγ_0aγ_0, and assume that γ_0=0 (we're going to allow the a_i to be 0). Define α/ω (this may have a standard meaning but I forget, oh well) to mean ω(γ_k)-1aγ_k + ... + ω(γ_1)-1aγ_1. (When I write γ-1 for an ordinal γ, I mean subtracting the 1 off the beginning. So for a finite ordinal this is subtracting 1, and for an infinite ordinal it does nothing.)
Then the order type of β is α/ω-1 if a_0=0, and is (α/ω-1)+1 if a_0>0. (Yeah, that notation is kind of crappy -- it looks like (γ-1)+1 should just be γ (and often it is). But we're subtracting 1 off the beginning, not the end, so rather we have 1+(γ-1)=γ, while (γ-1)+1 is only γ if γ is finite. I guess what we need here is some additive equivalent of the \ that gets used for division-on-the-left in certain contexts. Oh well.
-Harry
Anyway. The point is, in order to do this, I need to cite various set theory/order theory/point set topology stuff, to handle ordinals embedded in larger totally ordered sets (which here is always the reals). I have a section of the paper where I introduce all the stuff I need from there. I'd held off on actually writing it until pretty recently though. I didn't actually know where I'd cite the stuff from; I figured it wouldn't be too hard to find.
So Jeff suggested I ask Andreas Blass where I might find citations for these things. His response on hearing just what I wanted was, yes, that's certainly easy, but I'm actually not certain you'll find it in the literature. Try the following really old books; they cared a lot about this sort of thing around the turn of the century.
So I got the books Andreas suggested out from the library, and, unfortunately, they mostly didn't help. So, I'm having to write most of the proofs myself.
Anyway. The point of this entry was a particular thing. Say we have an ordinal α, and we let β be the set of all limit ordinals less than α. How can we express β in terms of α? Sounds easy, right? I mean it's basically just "dividing by ω", right?
Well basically yes, but if you actually want the order type on the nose, well, it's surprisingly easy to screw up. Now actually all I really need is that if α<ωn+1 (n finite), then β<ωn, so I don't need to prove a whole formula. (Actually, I guess if you just replaced n+1 with 1+n, this statement would still be true for n infinite.) But today I sat down and figured out just what the correct formula was (because you know I coudln't find this listed anywhere) so I thought I'd record it here for reference.
First off, if α is finite, there are no limit points. Otherwise...
Say α has Cantor normal form ωγ_kaγ_k + ... + ωγ_0aγ_0, and assume that γ_0=0 (we're going to allow the a_i to be 0). Define α/ω (this may have a standard meaning but I forget, oh well) to mean ω(γ_k)-1aγ_k + ... + ω(γ_1)-1aγ_1. (When I write γ-1 for an ordinal γ, I mean subtracting the 1 off the beginning. So for a finite ordinal this is subtracting 1, and for an infinite ordinal it does nothing.)
Then the order type of β is α/ω-1 if a_0=0, and is (α/ω-1)+1 if a_0>0. (Yeah, that notation is kind of crappy -- it looks like (γ-1)+1 should just be γ (and often it is). But we're subtracting 1 off the beginning, not the end, so rather we have 1+(γ-1)=γ, while (γ-1)+1 is only γ if γ is finite. I guess what we need here is some additive equivalent of the \ that gets used for division-on-the-left in certain contexts. Oh well.
-Harry
