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 [quote="dead mith chap"]But that doesn't preserve order, you have to flip the ordering...[/quote]
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CrystyB
Posted: Mon Mar 11, 2002 10:17 pm    Post subject: 1

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<b means they map to some numbers in the same relation, f(a)<f(b). That might give way to some interesting problems, imo.

((ow and in case you want to temper my math narcissism sometime, just remind me that i thought exp(x) was a map from rational (0,1) to rational (1,inf)... ))
CzarJ
Posted: Thu Mar 07, 2002 5:15 am    Post subject: 0

*walks in*
...

Um... I can prove that 2sin a - 1 = cos^4 a - sin^4 a....

...

*walks out*

------------------
Unslumping yourself is not easily done.
Chuck
Posted: Wed Mar 06, 2002 11:02 pm    Post subject: -1

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.
Posted: Wed Mar 06, 2002 10:36 pm    Post subject: -2

wait, no it doesn't. I can map 1/4 to -1, 1/2 to 0, and 3/4 to 1. Does that work?
Posted: Wed Mar 06, 2002 10:34 pm    Post subject: -3

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.

Posted: Wed Mar 06, 2002 10:27 pm    Post subject: -4

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.
Posted: Wed Mar 06, 2002 10:25 pm    Post subject: -5

That means we have one from (0,1) to (-infinity, -1), anyway.
Posted: Wed Mar 06, 2002 10:24 pm    Post subject: -6

(Assuming you mean f(q)=1/q)
Posted: Wed Mar 06, 2002 10:24 pm    Post subject: -7

But that doesn't preserve order, you have to flip the ordering...
Lilifreid
Posted: Wed Mar 06, 2002 10:16 pm    Post subject: -8

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..