An uncountable countable set
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From: Virgil on 16 Jul 2006 13:01 In article <1153039232.374101.300670(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > In an infinite tree one can show that the endless paths are uncountable" > > > We know that. Therefore we have the following circulus falsus: > > |N| << |P(N)| > || || > |{edges}| >= |{paths}| > > Regards, WM "Mueckenh" can have as many false circles as he likes in his false world, but they none of them exist in mine.
From: Virgil on 16 Jul 2006 13:07 In article <1153039371.156834.8960(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > > Each ordinal has a successor, even limit ordinals like N, and any > > > > ordinal can be indexed by the members of its successor. > > > > > > Each ordinal can be indexed by itself. 1 is indexed by 1. 1,2 is > > > indexed by 1 and 2. 1,2,3 is indexed by 1,2 and 3 and so on. But all > > > naturals are not indexed by all naturals? Extremely ridiculous! > > > Extremely! > > > Every ordinal has successors, whether in ZF or NBG or most other set > > theories, and to index any of them it is sufficient to use at the > > members of any successor. > > But the successor of 4 does not consist of the ingredients of 4 only. In NBG 4 = {0,1,2,3} and succ(4) = {0,1,2,3,4}. > So the successor of all natural numbers cannot consist of only all > natural numbers only. Wrong. "Mueckenh" missed one. the successor of any set contains that set as a member, at least in NBG. > This is obvious and therefore Tony Orlow is one > of the few persons who realize he truth and you and many others are not > among them. Then let "mueckenh' join TO in Trollheim and leave those of us in the real world alone.
From: Virgil on 16 Jul 2006 13:15 In article <1153050854.529809.266820(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In order to justify the "0." part of "0.1" 0r "0.11" , etc., > > one must have a natural number base greater than 1, otherwise the "0." > > is redundant. > > "0." := "Look, here comes a unary representation of a whole number." " " := "Look, here comes a unary representation of a whole number." Then 0.1, 0.2, 0.3, ... are decimal representations of a whole number where "0." := "Look, here comes a decimal representation of a whole number." As no other base seems to need any such flags to say "look", unary does not either. > > > > > > > > > > The result is the same. Hence, if 0.111... exists, > > > > > it is in he list and is not in the list. > > > > It exists but not as any of its predecessors. > > > > > How then can it be the *+ sum over its predecessors? In unary , summation is by concatenation of counters. When one concatenates all the 1's in all {0.1, 0.11, 0.111,...}, one gets 0.111... .
From: Virgil on 16 Jul 2006 13:20 In article <1153051034.733810.190150(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Hartley schrieb: > > > > >> Can you provide a proof of that statement? Because you state just above: > > >> "Cantor's conclusion is correct, with no doubt, if infinite sets like > > >> the set of my transpositions do exist." Which contradicts what you state > > >> here. > > > > > >No. Cantor obviously was correct, if infinite sets would exist. But > > >they do not. So he was wrong. And you are wrong, as I have shown above > > >by Cantor's text which you misinterpreted grossly. > > > > Cantor's proposition (as you interpret it) leads directly to a > > contradiction in standard set theory. So it is disproved. To show that > > set theory is inconsistent you must also show that it can be proved. You > > cannot do this by arguing over quotations from Cantor. He does not > > appear to have offered a proof and he was not infallible. You claim it > > is obvious, so prove it yourself, with no more quotations, (except > > references to standard theorems}. > > Obvious is that no transposition of two elements of a well-ordered set > can destroy the well-order. > Obvious is that, if infinite sets do exist and can be exhausted, the > set of my transpositions does exist and can be exhausted. The issue is what would result. Since at any finite stage one still has infinitely many infinitely descending sequences of rationals by magnitude, what evidence is there that "exhaustion" would produce anything else?
From: Virgil on 16 Jul 2006 13:25
In article <1153051163.295503.224950(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1152978585.858028.263870(a)p79g2000cwp.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > > In usual mathematics 0.999... is assumed to exist. > > > > > > > > It is not assumed. There is a definition floating around giving a > > > > meaning > > > > to that notation. > > > > Without the definition that notation makes no sense. > > > > > > And with that definition it makes no sense either, as the definition is > > > of he same kind as the squared circle. > > > > You do not believe in limits? Otherwise, why does the definition not make > > sense? Quote the definition for precisely that notation: > > 0.999... = lim{n -> oo} sum{k = 1..n} 9.10^(-k) > > what part of that definition makes no sense? > > "n --> oo" because there is no natural number oo. If one takes x --> y as indicating a limit, y is specifically excluded from any part in the limit process, so that "mueckenh"'s argument fails. > > It is not that I fight against the identification of 0.999... with 1 or > against the identification of the limit of the series of > Gregory-Leibniz with pi/4 for computational purposes. But the assertion > that the digits of these limits could be used to construct a diagonal > number is simply nonsense. Except that it is only nonsense when one grants all of "muecknh"'s peculiar premises, which we do not. |