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Daedalian Member

 Posted: Wed Mar 06, 2002 9:36 pm    Post subject: 1 Anyone know something about order types? The problem I am annoyed at right now is proving that the rationals in the interval (0,1) have the same order type (under the normal < ordering) as Q (the whole set of rationals). It seems pretty obvious to me that they are of the same type, but I can't find a pretty (or ugly, for that matter) mapping that preserves order.
Lilifreid
DANGER!

 Posted: Wed Mar 06, 2002 10:16 pm    Post subject: 2 I can map (I think) the rationals in the interval (0,1) to the rationals in (1, infinity), but (0,1) to Q is a bit more troubling..
Daedalian Member

 Posted: Wed Mar 06, 2002 10:24 pm    Post subject: 3 But that doesn't preserve order, you have to flip the ordering...
Daedalian Member

 Posted: Wed Mar 06, 2002 10:24 pm    Post subject: 4 (Assuming you mean f(q)=1/q)
Daedalian Member

 Posted: Wed Mar 06, 2002 10:25 pm    Post subject: 5 That means we have one from (0,1) to (-infinity, -1), anyway.
Daedalian Member

 Posted: Wed Mar 06, 2002 10:27 pm    Post subject: 6 I think we can also tack on as many intervals as we want... I just don't know if we can tack on an infinity number of them.
Daedalian Member

 Posted: Wed Mar 06, 2002 10:34 pm    Post subject: 7 wait, don't have to. f1: (0,1)---> (-1,0) where f1(q)=q-1 f2: (0,1)---> (-infinity,-1) where f2(q)=-1/q f3: (0,1)---> (1,infinity) where f3(q)=-1/f1(q)=-1/(q-1) Then we could map (0,1/4) to (-infinity,-1), (1/4,1/2) to (-1,0), (1/2,3/4) to (0,1), and (3/4,1) to (1,infinity). That leaves out three points though.
Daedalian Member

 Posted: Wed Mar 06, 2002 10:36 pm    Post subject: 8 wait, no it doesn't. I can map 1/4 to -1, 1/2 to 0, and 3/4 to 1. Does that work?
Chuck
Daedalian Member

 Posted: Wed Mar 06, 2002 11:02 pm    Post subject: 9 I thought you could map the rationals to integers easily. You can then map any subset of the rationals to the integers by leaving the rationals that aren't in the subset out of the list. Since the rationals and any subset of the rationals map to the integers they must map to each other. Wouldn't this double mapping preserve order? You'd be removing nonqualifying rationals from the set of all rationals to make the subset. That doesn't change the ordering of those that are left.
CzarJ
Hot babe

 Posted: Thu Mar 07, 2002 5:15 am    Post subject: 10 *walks in* ... Um... I can prove that 2sin a - 1 = cos^4 a - sin^4 a.... ... *walks out* ------------------ Unslumping yourself is not easily done.
CrystyB
Misunderstood Guy

 Posted: Mon Mar 11, 2002 10:17 pm    Post subject: 11 mith, i really don't know a THING about order types. Intuitively though, yes it should be possible to "tack on" a countable infinity of intervals. Map (1/2,1) to the first (last if you can order the destination backwards), (1/3,1/2) to the second (second last), ... , (1/(n+1),1/n) and so on. And the points in between to the points in between the destination intervals. It works great in my imagination - i just hope i'm not being to intuitive about this whole thing. Chuck, the emphasis mith did not underline enough was the "preserves order" part. a
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