Can N be constructed?
in [Logic]
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From: Virgil on 11 Jun 2010 16:57 In article <0651cfdc-6f61-4cec-95c0-b5244cc7592e(a)s9g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Jun., 08:15, Virgil <Vir...(a)home.esc> wrote: > > In article > > <9d4e6d3a-f3c8-47a1-b209-db6519187...(a)w12g2000yqj.googlegroups.com>, > > > > �WM <mueck...(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. > > What would be the result of the construction as long as it is > unfinished? That is not Cantor's problem to solve. > > > > > > > > > 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. > > Look here for instance, if you want to have the opinion of an expert: > http://groups.google.de/group/sci.math.research/browse_frm/thread/4a899e543147 > 04ad# There are 'experts' on both sides of the issue, with physicists on one side and mathematicians on the other. But, for example, the mathematics of coding, on which a huge amount of modern commerce is deeply dependant, is entirely and exclusively non-physics. So there are important, even essential, areas of mathematics in which physics has absolutely no part.
From: Virgil on 11 Jun 2010 17:00 In article <fad1c8ec-7d56-408d-adba-66082d35c5fd(a)u7g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Jun., 09:16, Virgil <Vir...(a)home.esc> wrote: > > In article > > <55744d1f-57e0-48ee-9913-ebfdfc6d8...(a)u7g2000yqm.googlegroups.com>, > > > > > > > > > > > > �WM <mueck...(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 > > Try to return to logical thinking. If that means thinking like WM does, heaven forfend! There are enough matheologicians in > the world who argue on their beloved belief. > > If more than one are claimed to be required, then at least two > required must be shown. At least two are requires, but no particular ones, just any infinite set of them. When one says "at least two" that only refers to the numbr required and does not imply any particular ones are required. WM conflates "at least two" with some particular two, which is a mugs game.
From: Virgil on 11 Jun 2010 17:05 In article <67f71c97-e8bc-4d01-af37-5a5ecf8e4683(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Jun., 16:22, 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. > > > > 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. > > > > >> This is because every line is removed at some finite step > > >> in the sequence of infinite steps. There is no point in the > > >> sequence where a line (finite or otherwise) is not removed > > >> from the list. After all the steps, an infinite number of lines > > >> have been removed from the list. There are none left > > > > WM wrote: > > > That would be correct, unless a removal is not executed before the > > > next line has been established. > > > > You are contradicting yourself. You said above: > > | ... but remove always line number n after having constructed the > > | next line number n + 1. > > > > Now you are saying "unless a removal is not executed" at some point. > > > > > So are you saying that your construction rule does not actually > > apply to every step? Do your rules change, or perhaps thay are applied > > only randomly?- > > I say: If my rule applies to every step, then there is always one line > left. > Then it cannot be true that: "There are none left". WM conflates what is necessary during a process with what results after the process is ended. Each step is during the process, but that does not apply to after the whole thing is over. WM is, in effect, returning to something like the Zeno paradoxes, by arguing that one cannot move from one point to another because there must always be half the distance left cover.
From: Virgil on 11 Jun 2010 17:10 In article <82bdde12-e829-4e18-9b7c-1b67907d05d8(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Jun., 16:29, David R Tribble <da...(a)tribble.com> wrote: > > 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: > > >> 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. > > > > WM wrote: > > > That is wrong. Only every line *before the last one constructed* is > > > erased. > > > > Since your construction rule states that after every line is > > constructed (and a previous line is removed) another line is > > always constructed, there can be no "last one" constructed. > > According to your rule, there is always a next line constructed. > > > > Or are you saying that your construction rules do not apply to > > every line? > > > > > Only if "all lines" can be constructed, then all lines are > > > erased and are not erased. > > > > How can all lines be both erased and not erased? At different "times"! At any time during the process there are lines not yet erased, but having completed the process all lines are erased' In moving from point A to point B, at any time during the process there are points still to be covered but at the end of the process there are no more points to be covered. But according to WM, it is impossible to move from point A to point B. > > That cannot be. Therefore the precondition must be wrong. There are > not "all" lines. > > > Does your construction rule erase a line at each step or not? > > Yes. > > > Or perhaps your construction rule apply to some steps but not > > to others? > > It applies to every step, if there is every step. My rule applies to > every step that can be done. > > > > > This sheds some doubt on the assertion that all lines can be > > > constructed. > > > > That would mean that your construction rule must fail at some point. > > > > Would this be because your rule is somehow flawed, or is it > > because your rule simply stops working after some particular line? > > The reason is that infinity cannot be finished. There are at every > step infinitely many lines left. In moving from point A to point B, at any time during the process there are points still to be covered but at the end of the process there are no more points to be covered. But according to WM'a analysis, it must be impossible to move from point A to point B, as one can never cover all those points.
From: Virgil on 11 Jun 2010 17:11
In article <a4454a4e-8967-44ac-9e93-74aadc05d432(a)a30g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Jun., 16:33, David R Tribble <da...(a)tribble.com> wrote: > > WM wrote: > > > How would you get to an infinite number of steps when each step has > > > another finite number? > > > > If each step is followed by another finite number step, how would > > you stop at any finite step? > > It is not necessary to stop somewhere in order to remain in the finite > domain. > > Regards, WM In moving from point A to point B, at any time during the process there are points still to be covered but at the end of the process there are no more points to be covered. But according to WM, it is impossible to move from point A to point B. |