Cantor Confusion
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From: David Marcus on 12 Jan 2007 19:28 Virgil wrote: > In article <MPG.2011c51e73b463e1989aef(a)news.rcn.com>, > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > > Dave Seaman wrote: > > > On Fri, 12 Jan 2007 11:48:38 EST, Andy Smith wrote: > > > > > > > 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. > > > > Normally, a "sequence" doesn't have a last element. So, the rationals > > are an example of a countably infinite set (not sequence) that has a > > finite start and end point (where "start" and "end" mean smallest and > > largest). > > If you mean "those" rationals, the ones in [0,1], yes. Yes, thanks. > But "the" rationals could be misinterpreted to mean all rationals, > particularly if the lead in sentence gets snipped. -- David Marcus
From: Virgil on 12 Jan 2007 19:30 In article <11087446.1168647563782.JavaMail.jakarta(a)nitrogen.mathforum.org>, Andy Smith <andy(a)phoenixsystems.co.uk> wrote: > > > > Here is a question for you: Suppose you have a > > countably infinite > > ordered set X. So there is a bijection f:N -> X. Does > > it have to be true > > that m < n implies that f(m) < f(n)? > > > At the risk of appearing an idiot, I would say yes. > > Because X is ordered such that x(n) > x(m) for all n > m. > > Am I missing something? Apparently! Since there is nothing requiring the order of X to be compatible with the order of N, and orderings of countable sets exist which are incompatible with the natural ordering of N, there is no requirement that f be order preserving. For example, consider X as the set of rationals with the standard rational order, for which there exist bijections from N to X which are clearly can n to be order preserving
From: David Marcus on 12 Jan 2007 19:37 Dik T. Winter wrote: > In article <1168511880.370120.180940(a)p59g2000hsd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Conclusion: Every finite binary tree contains a finite set of path. > > Right. > > > The > > countable union of finite sets is countable. > > RIght again. > > > The set of paths is > > countable. > > Wrong. The union of the set of paths of the finite trees is *not* the > set of paths in the union of the finite trees. Dyslexia. -- David Marcus
From: Michael Press on 12 Jan 2007 19:58 In article <1168516308.352303.113260(a)k58g2000hse.googlegroups.com> , cbrown(a)cbrownsystems.com wrote: > Now, "your" naturals may be different than "my" naturals because "my" > naturals obey certain additional idiosyncratic (from "your", sadly, > limited viewpoint) properties. E.g., that primes of the form 6*n + 1 > are red, and primes of the form 6*n - 1 are green; and that therefore I > have a theorem that no yellow primes exist. Hey! When did they repaint 2? -- Michael Press
From: David Marcus on 12 Jan 2007 19:59
Virgil wrote: > In article > <11087446.1168647563782.JavaMail.jakarta(a)nitrogen.mathforum.org>, > Andy Smith <andy(a)phoenixsystems.co.uk> wrote: > > > > > > > Here is a question for you: Suppose you have a > > > countably infinite > > > ordered set X. So there is a bijection f:N -> X. Does > > > it have to be true > > > that m < n implies that f(m) < f(n)? > > > > > At the risk of appearing an idiot, I would say yes. > > > > Because X is ordered such that x(n) > x(m) for all n > m. > > > > Am I missing something? > > Apparently! > > Since there is nothing requiring the order of X to be compatible with > the order of N, and orderings of countable sets exist which are > incompatible with the natural ordering of N, there is no requirement > that f be order preserving. > > For example, consider X as the set of rationals with the standard > rational order, for which there exist bijections from N to X which are > clearly can n to be order preserving I think you meant to write "... which clearly cannot be order preserving". -- David Marcus |