constructible = unconstructible
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From: WM on 11 Jun 2010 11:01 On 11 Jun., 14:34, Gus Gassmann <horand.gassm...(a)googlemail.com> wrote: > > But I can estimate that all together form a set that is not > > larger than a countable set. > > By making a false continuity assumption. Big deal. What kind of discontinuity do you expect??? Infinity means going on and on and on. Do you believe that there is an abyss "after all"? Do you think along the lines of the medieval sophists who expected to fall from the earth after passing its edge? Well, your reasoning sounds like that. Don't worry. There is no abyss and there is no dragon and there is no discontinuity . Regards, WM
From: Virgil on 11 Jun 2010 17:17 In article <e43610dc-a225-47b4-b560-30f26d0b6c98(a)h13g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Jun., 05:02, Owen Jacobson <angrybald...(a)gmail.com> wrote: > > On 2010-06-10 09:30:27 -0400, WM said: > > > > > > > > > > > > > On 10 Jun., 14:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > >> On Jun 10, 7:34 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > >> <snip> > > > > >>> if we divide 1 by 9, then 0.111... does exist according > > >>> to the above definition. > > >>> If we write it in a single line, nobody will disagree. > > >>> If we write it such that every next 1 is placed in a new line, like > > > > >>> 0.1 > > >>> 0.11 > > >>> 0.111 > > >>> ... > > > > >>> then everybody will disagree that all 1 are in one line but all 1 > > >>> "must be there somehow". > > > > >> Yes and outside Wolkenmuekenheim, everyone is correct. > > >> However inside Wolenmuekenheim there is a 1 in 0.111... > > >> that is not in the list. > > > > > What is flaming good for? My proof uses complete induction. One can > > > believe in it or not. > > > > Your proof reaches nonsensical conclusions in the context in which you > > present it. Therefore there must be a flaw in it. Finding this flaw is > > decidedly *your* problem. > > The flaw has been found long time ago. It consists in the assumption > that infinity can be finished, entertained by set theorists. 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, because no such "infinity" can ever be completed.
From: Virgil on 11 Jun 2010 17:18 In article <0e920678-bb6c-4d8d-a4b1-11db0da2235a(a)q12g2000yqj.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 17:20, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Jun 10, 12:07�pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 10 Jun., 16:41, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > <snip> > > > > > > then neither an uncomputable number > > > > nor a list of all computable numbers exist. > > > > > I showed you a computed list that contains all computable numbers (and > > > some other names). So yout assertion is false. > > > > You did not show me a list that contains all computable > > numbers and only computable numbers. �Showing a computable > > list that contains "all computable numbers > > (and some other names)" is not enough to demonstrate > > the existence of a computable list that contains all > > computable numbers and only computable numbers. > > Why should I try to find such a list when I know that this list would > be countable? > > That's the same with the paths of the binary tree. I cannot enumerate > them. But I can estimate that all together form a set that is not > larger than a countable set. > > Regards, WM WM claims that no infinity can be "finished" 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. So according to WM, it is impossible to move from point A to point B.
From: Virgil on 11 Jun 2010 17:20 In article <04133839-b1dc-4555-a651-6e529471b526(a)u26g2000yqu.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Jun., 14:34, Gus Gassmann <horand.gassm...(a)googlemail.com> > wrote: > > > > But I can estimate that all together form a set that is not > > > larger than a countable set. > > > > By making a false continuity assumption. Big deal. > > What kind of discontinuity do you expect??? > Infinity means going on and on and on. > > Do you believe that there is an abyss "after all"? Do you think along > the lines of the medieval sophists who expected to fall from the earth > after passing its edge? > > Well, your reasoning sounds like that. Don't worry. There is no abyss > and there is no dragon and there is no discontinuity . > > 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, such infinities are impossible to complete, so it is impossible to move from point A to point B.
From: Owen Jacobson on 11 Jun 2010 19:35
On 2010-06-11 07:18:58 -0400, WM said: > On 11 Jun., 05:02, Owen Jacobson <angrybald...(a)gmail.com> wrote: >> On 2010-06-10 09:30:27 -0400, WM said: >>> On 10 Jun., 14:10, William Hughes <wpihug...(a)hotmail.com> wrote: >>>> On Jun 10, 7:34 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: >> >>>> <snip> >> >>>>> if we divide 1 by 9, then 0.111... does exist according >>>>> to the above definition. >>>>> If we write it in a single line, nobody will disagree. >>>>> If we write it such that every next 1 is placed in a new line, like >> >>>>> 0.1 >>>>> 0.11 >>>>> 0.111 >>>>> ... >> >>>>> then everybody will disagree that all 1 are in one line but all 1 >>>>> "must be there somehow". >> >>>> Yes and outside Wolkenmuekenheim, everyone is correct. >>>> However inside Wolenmuekenheim there is a 1 in 0.111... >>>> that is not in the list. >> >>> What is flaming good for? My proof uses complete induction. One can >>> believe in it or not. >> >> Let's frame your argument slightly differently. Presuppose N exists, >> and has the properties normally associated with it, for the sake of the >> following. >> >> For all elements n of N, let F_n be a function from N to {0, 1} as follows: >> >> � F_n(x) = 1 if x <= n >> � F_n(x) = 0 if x > n >> >> Let G be a function from N to {0, 1} as follows: >> >> � G(x) = 1 >> >> Let L be a function from N to {F_0, F_1, ...} as follows: >> >> � L(x) = F_x >> >> Which of the following statements accurately characterizes your argument? >> >> 1. If N exists, then there exists some natural number z in N such that >> L(z) = F_z = G. > > Yes. But z does not and cannot exist. Therefore N cannot exists as a > completed entity. Can you frame an existence argument for a number z in N with the properties above, either by deriving it directly, or by showing that it must exist, in terms of ZFC? Try to avoid analogies: you don't actually need binary trees or decimal expansions to frame your argument, and your argument would be much clearer without them. Framing your argument in terms of functions would help me, obviously, but framing it in terms of sets and subsets would work as well. It's not at all obvious from the arguments you've presented that z must exist. It's very obvious that you're convinced it must exist, but your argument is so buried in your "tree" analogy and in invented terms like "cover" that it's very hard to follow. >> 2. G cannot exist even given the presupposition that N exists as above. > > Given N then G must exist. G consists merely of indices n attached to > the ever repeating 1. > > The result of my proof is that G cannot exist as an infinite sequence. > G exists as a finite definition "1/9" or "0.111..." (this is a finite > definition using 8 symbols) or "G". Forget about "represents". I'm trying to eliminate analogies and comparisons from your argument so that I can understand it properly. G is a function and a function only, with specific, easily-determined properties. It's not a sequence; it's not a fraction; it's not a path through a tree. It's just a function. > But there are only countably many finite definitons. That was never in dispute. Look up "computable numbers," for example. -o |