Can N be constructed?
in [Logic]
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From: WM on 11 Jun 2010 01:55 On 11 Jun., 02:32, David R Tribble <da...(a)tribble.com> wrote: > WM wrote: > >> Proof: Construct the above list, but remove always line number n after > >> having constructed the next line number n + 1. > > William Hughes wrote: > >> After any finite number of steps you get a line from the list. > > David R Tribble wrote: > >> Yes, that much is certain. > > William Hughes wrote: > >> After an infinite number of steps you get a line from the list. > >> The line you get after an infinite number of steps is not > >> a line from the list. > > David R Tribble wrote: > >> No, I don't think so. After an infinite number of steps, where > >> at each step a (finite) line is removed from the list, you end > >> up with no lines at all. > > William Hughes wrote: > > Well this depends on defining what you "end up with" > > > If your definition is (the very reasonable) "you end up > > with any lines that have been written down but not erased", > > then you end up with no lines as every line you write down > > gets erased. > > > However, I think in this context saying that "you end up > > with the limit line 111..." is better. However, this > > definition has its problems. The main one is that you > > "end up with" a line that you never write down. > > > Note, however, that in neither case do you end up with > > a line from the list. > > Yes, exactly. You can't end up with a line that was erased > at some point, nor with a line that was never written at any > point. > > It's obvious from the beginning that the line 111... does not > exist in WM's list, so it cannot possibly be the line that you > end up with. Saying so makes as much sense as saying that > you end up with the line 101010..., or or saying that infinity can be finished. Regards, WM
From: Virgil on 11 Jun 2010 01:57 In article <971fddc7-e936-4c49-a4e2-f86be6b2cc59(a)u26g2000yqu.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 14:39, jbriggs444 <jbriggs...(a)gmail.com> wrote: > > > >�Re-cast > > the problem in terms of limits > > Here it is: Limits and Dedekind cuts exist by finite definitions only. > Their number remains always countable (even finite). Therefore > uncountability is nonsense also from this point of view. > > Regards, WM The definitions of infiniteness are all finite, but that does not prohibit either the definitions or the defienda from existing.
From: Virgil on 11 Jun 2010 02:03 In article <fcb349af-80d7-4094-a780-eba75ff1b50f(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 21:44, Virgil <Vir...(a)home.esc> wrote: > > > > That would be correct, unless a removal is not executed before the > > > next line has been established. > > > > > Therefore the set of remaining lines cannot be empty. > > > > Then which lines are left?- > > My proof does not show which line is left. It also does not show that any line is left. > But it shows that finished > infinity is a non-mathematical notion. On the contrary, what can be defined cannot be barred from mathematical consideration, and "finished" infinity can be and has been defined. > Of course it is always the last > line that is left, and it is impossible to get rid of a last line Except that the process requires passing by every line so that there cannot be a last one. So if tWh finds a last line, WM is talking of a different process.
From: Virgil on 11 Jun 2010 02:15 In article <9d4e6d3a-f3c8-47a1-b209-db6519187ecb(a)w12g2000yqj.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 22:03, Virgil <Vir...(a)home.esc> wrote: > > > > That is just the case when constructing N or completeing Cantor's > > > list. Without completing the list, the diagonal argument is invalid. > > > > Cantor merely says that no completed list of binary sequences can > > contain all binary sequences, which is obviously true. > > He says first of all that an infinite list cna be completed. That is > matheology. Actually that is NOT what Cantor says in his "diagonal" argument. What Cantor does say is that for every list of infinite binary sequences presented he can construct a binary sequence not in that list. > > > > Thus if the SET of all completed binary sequences exists at all, > > It exists in the brains of some cranks and crackpots. That's all. It may not exist in the cranky and crackpot brain of WM, but that is WM's problem, since mathematics with such setsgets along quite nicely without WMs interference. > This assertion would be correct, but it is not Cantor's and not that > of his followers. > > > > This so upsets WM that he reacts by denying existence of the set of all > > infinite binary sequences. > > > > But in some axiom systems such a set must exist. > > > > > Of course that is nonsense, but it is assumed to be possible by set > > > theorists. > > > > When WM equates the reality of an axiom system with physical reality, he � > > becomes foolish. > > Mathematics is physics (V. A. Arnold). Physics is Mathematics (Virgil) There is much more to mathematics than mere physics, at least in the eyes of all but physicists and their slaves.
From: Virgil on 11 Jun 2010 03:16
In article <55744d1f-57e0-48ee-9913-ebfdfc6d85e1(a)u7g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 22:06, Virgil <Vir...(a)home.esc> wrote: > > In article > > <bc98ca3a-1b0a-439e-b4c6-8a4a0ec1b...(a)w12g2000yqj.googlegroups.com>, > > > > > > > > > > > > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 10 Jun., 16:13, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > After any finite number of steps the set of remaining lines > > > > cannot be empty. > > > > > No. After any possible step the set of remaining lines cannot be > > > empty. > > > > > > Look! Over There! A Pink Elephant! > > > > > > After an infinite number of steps the set of remaining lines > > > > cannot be empty. > > > > > How would you get to an infinite number of steps when each step has > > > another finite number? > > > > There is nothing in the relevant axiom system which requires accessing > > infinite cases only by such step-by-step operations on finite cases. > > > > In fact one such infinite case is built into those axioms > > The infinite list > 1 > 11 > 111 > ... > > need not be built step by step. Nevertheless the proof stands that > there are not two (or more) lines which are necessary to contain all > 1's of the list. More than two lines will work, if it is enough more!!! EVERY infinite set of lines does the trick, but no finite set of lines does. So there is no "smallest" set of lines which works, but there are still infinitely many sets of lines which will work. |