Cantor Confusion
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From: Dave Seaman on 12 Jan 2007 11:26 On Fri, 12 Jan 2007 10:52:48 EST, Andy Smith wrote: >> > 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. So replace w by -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. > So can I label the first w+1 zero crossings a(0),a(1) ... a(w)? You don't have a "first w+1" crossings. The order type is not the same. >> 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? A set is totally ordered if it satisfies the trichotomy law. That is, if x and y are members of the set, then exactly one of the following statements holds: (1) x < y, (2) x > y, (3) x = y. A set is well ordered if every nonempty subset has a least member. Your set of crossings is not well ordered because the subset of crossings for x > 0 does not have a least element. Every ordinal is well ordered. Each natural number, being a finite ordinal, is likewise well ordered. The set of all natural numbers is identical to the ordinal w and is therefore well ordered. Every well ordered set is necessarily totally ordered. Some examples of totally ordered sets that are not well ordered are: (1) Your set of crossings. (2) The reals in [0,1]. (3) The rationals in [0,1]. (4) The integers. >> 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?) Certainly the crossings exist. That has never been in dispute. You are free to write digit sequences such as 11...11, but they have no particular meaning as numbers until you explain what you mean by them. > 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? Anything that can be rigorously defined is a legitimate mathematical object. The set w+1 is an ordered countably infinite set that has a well defined beginning and an end. -- 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 01:30 > > >> 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. > But -1 (= a(last i)) is not > its predecessor ? This is not an ordered set as I meant it. --- Andy
From: Andy Smith on 12 Jan 2007 01:48 > > > > My mistake; w is not finite. How about if we > replace > > w by -1? > > > > 0, 1, 2, ..., -1. > > > But -1 (= a(last i)) is not > its predecessor ? > This is not an ordered set as I meant it. > Just to elaborate a bit, what I am trying to show is that, in contrast to Peano, which says that if you have an ordered sequence such that each element has a successor => there is no maximum element, and the sequence is infinite, that you can have an ordered infinite sequence which has a finite start and end point. For infinite, read "countably infinite". This is different to e.g. 1/2,3/4, 5/8 ... because (as discussed re Zeno) such a sequence does not have a well defined last member, just a limit point that is never reached. Andy
From: Dave Seaman on 12 Jan 2007 12:23 On Fri, 12 Jan 2007 11:30:32 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. >> > But -1 (= a(last i)) is not > its predecessor ? > This is not an ordered set as I meant it. You asked that a(i+1) > a(i) for all i < last i, and indeed it is so. Your criterion doesn't mention anything about the last i having a predecessor. -- 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 12:27
On Fri, 12 Jan 2007 11:48:38 EST, Andy Smith wrote: >> > >> > My mistake; w is not finite. How about if we >> replace >> > w by -1? >> > >> > 0, 1, 2, ..., -1. >> > >> But -1 (= a(last i)) is not > its predecessor ? >> This is not an ordered set as I meant it. >> > Just to elaborate a bit, what I am trying to show > is that, in contrast to Peano, which says that > if you have an ordered sequence such that each element > has a successor => there is no maximum element, and the > sequence is infinite, Peano doesn't say that, and the ordinal w+1 is a counterexample. Another counterexample is any finite cyclic group. You didn't say there had to be a first element that is not a successor of anything. > that you can have an ordered > infinite sequence which has a finite start and end point. > For infinite, read "countably infinite". Of course. The rationals in [0,1], for example. > This is different to e.g. 1/2,3/4, 5/8 ... because > (as discussed re Zeno) such a sequence does not have a > well defined last member, just a limit point that is never > reached. As opposed to the rationals in [0,1], where the last element is 1. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/> |