Cantor Confusion
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From: Dik T. Winter on 12 Jan 2007 08:14 In article <32631958.1168598767888.JavaMail.jakarta(a)nitrogen.mathforum.org> Andy Smith <andy(a)phoenixsystems.co.uk> writes: > > > So you can have an infinite set with a first and > > last member? > > > > Sure. Consider the following sets of real numbers: > > > > {x| 0 <= x <= 1} > > {x| 0 < x <= 1} > > {x| 0 < x < 1} > > Perhaps I should have asked, can you have a "countably > infinite" ordered set with a first and last member? Take the above sets with x rational. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Andy Smith on 11 Jan 2007 23:48 > On Fri, 12 Jan 2007 05:45:37 EST, Andy Smith wrote: > > > Perhaps I should have asked, can you have a > "countably > > infinite" ordered set with a first and last member? > > Sure. Any countable successor ordinal will do. For > example, w+1, where w > represents omega, the first transfinite ordinal. The > members of w+1 are > > 0, 1, 2, ..., w. > > Its first (smallest) member is 0, and its last > (largest) is w. > > Thank you Dave (and others). This is becoming reminiscent of scenarios where the Devil offers you 3 wishes which all turn to worms because of a lack of exact specificity in the wishes. Have any of you considered theology as a possible career path? I'll try again. Can you have a well ordered countable infinite set a0,a1,... such that : it has a defined finite first and last member and a(i+1) > a(i) for all i<last i ? No, you can't, because there can be no last member of a countably infinite set. So what about the set of zero crossings of sin(pi/x), -1>=x,x=<1 ? These are ordered by virtue of construction. We can try to label them a(0),a(1) .... but cannot then get past the zero crossing at 0, as far as the labelling is concerned - there is no last crossing before that at 0. Even if we label the crossing at 0 as a(w), then we cannot label any subsequent terms - there is no next zero crossing after that at 0. So the crossings a(0),a(1).... and maybe a(w) can be labelled; the rest (for zero crossings with x>0) can't. That doesn't bother me other than aesthetically. But this is an infinite ordered sequence with a defined beginning and a defined end. So, suppose I associate the zero crossings of this series with an infinite number of bit positions of an integer - then we have all the finite numbers - and then numbers represented by bbb...bbb , where the ellipsis "..." representes the zero crossings between the nth zero crossing from x>-1 and some mth zero crossing from x<+1? So we can label the (infinite) last term a(L), the previous term a(L-1), down to the crossing at x = 0, which is a(L-w) = a(w)? Kill me now. --- Andy
From: Dave Seaman on 12 Jan 2007 10:13 On Fri, 12 Jan 2007 09:48:54 EST, Andy Smith wrote: >> On Fri, 12 Jan 2007 05:45:37 EST, Andy Smith wrote: >> > Perhaps I should have asked, can you have a >> "countably >> > infinite" ordered set with a first and last member? >> Sure. Any countable successor ordinal will do. For >> example, w+1, where w >> represents omega, the first transfinite ordinal. The >> members of w+1 are >> 0, 1, 2, ..., w. >> Its first (smallest) member is 0, and its last >> (largest) is w. > Thank you Dave (and others). > This is becoming reminiscent of scenarios where the Devil > offers you 3 wishes which all turn to worms because of a > lack of exact specificity in the wishes. > Have any of you considered theology as a possible career path? > I'll try again. > Can you have a well ordered countable infinite set a0,a1,... such that : > it has a defined finite first and last member > and > a(i+1) > a(i) for all i<last i That holds for the w+1 example above. > No, you can't, because there can be no last member of > a countably infinite set. Wrong; w+1 is a countably infinite set and its last member is w. > So what about the set of zero crossings of sin(pi/x), > -1>=x,x=<1 ? What about them? What point are you trying to make? > These are ordered by virtue of construction. > We can try to label them a(0),a(1) .... > but cannot then get past the zero crossing at 0, as > far as the labelling is concerned - there is no last > crossing before that at 0. Even if we label the crossing > at 0 as a(w), then we cannot label any subsequent terms > - there is no next zero crossing after that at 0. The crossings are totally ordered but they are not well ordered, nor is the reverse ordering well ordered. > So the crossings a(0),a(1).... and maybe a(w) can be > labelled; the rest (for zero crossings with x>0) can't. > That doesn't bother me other than aesthetically. > But this is an infinite ordered sequence with a defined > beginning and a defined end. > So, suppose I associate the zero crossings of this series > with an infinite number of bit positions of an integer - then we have all the > finite numbers - and then numbers represented by bbb...bbb > , where the ellipsis "..." representes the zero crossings between the nth zero crossing from x>-1 and some mth zero crossing from x<+1? The fact that two ordered sets have the same cardinality does not imply that they have the same order type. You need to describe more carefully how you are associating those two sets, keeping in mind that each bit position of an integer is finite. > So we can label the (infinite) last term a(L), the > previous term a(L-1), down to the crossing at x = 0, which is > a(L-w) = a(w)? There is no order-preserving mapping from any ordinal to the set of crossings. It's not a well ordered set. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Dave Seaman on 12 Jan 2007 10:34 On Fri, 12 Jan 2007 15:13:52 +0000 (UTC), Dave Seaman wrote: > On Fri, 12 Jan 2007 09:48:54 EST, Andy Smith wrote: >> Can you have a well ordered countable infinite set a0,a1,... such that : >> it has a defined finite first and last member >> and >> a(i+1) > a(i) for all i<last i > That holds for the w+1 example above. My mistake; w is not finite. How about if we replace w by -1? 0, 1, 2, ..., -1. >> No, you can't, because there can be no last member of >> a countably infinite set. > Wrong; w+1 is a countably infinite set and its last member is w. And here we have -1 as the finite last member. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Andy Smith on 12 Jan 2007 00:52
> > > I'll try again. > > > Can you have a well ordered countable infinite set > a0,a1,... such that : > > > it has a defined finite first and last member > > > and > > > a(i+1) > a(i) for all i<last i > > That holds for the w+1 example above. > w is not finite though. > > > No, you can't, because there can be no last member > of > > a countably infinite set. > > Wrong; w+1 is a countably infinite set and its last > member is w. > So can I label the first w+1 zero crossings a(0),a(1) ... a(w)? > > The crossings are totally ordered but they are not > well ordered, nor is > the reverse ordering well ordered. > How does one define "well ordered" as distinct from "totally ordered", please? > > The fact that two ordered sets have the same > cardinality does not imply > that they have the same order type. You need to > describe more carefully > how you are associating those two sets, keeping in > mind that each bit > position of an integer is finite. > Well the point I was suggesting was that the infinite sequence of crossings is an ordered set, with a start and an end and even if one can't label each term with an ascending integer, their locations exist on the real line as much as any other fairyland construction. (and, hence, maybe, other infinite sequences e.g. 11...11 with a beginning and an end might exist and have some meaning?) What can you say then about an ordered countably infinite set that has a well defined beginning and an end - or does the fact that it have a beginning and an end automatically disqualify it from being a legitimate object? --- Andy |