An uncountable countable set
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From: mueckenh on 6 Jul 2006 14:01 Dik T. Winter schrieb: > In article <1152193390.830398.326300(a)q16g2000cwq.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > Either the diagonal number 0.111... is not distinguished from all > > > > finitely large numbers of the list > > > > 0. > > > > 0.1 > > > > 0.11 > > > > 0.111 > > > > ... > > > > then Cantor's proof fails. > > If we assume that that list contains natural numbers in some unary notation > (as I think you do) than: > > > > > Or 0.111... is distinguished from all finitely large numbers of the > > 0.111... is not a natural number in unary notation. So it is inherently > different from all elements of the list. OK. it is the unary representaton of omega. > > > "To be different" means for all unary representations of n > > An : 0.111... - n =/= 0 > > How do you propose to define that subtraction when 0.111... is not a > natural number? I chose just this number because it is also the decimal representation of 1/9. To be continued. Regards, WM >
From: Virgil on 6 Jul 2006 14:06 In article <1152183861.949387.268390(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > All sequences, even all single transpositions which I apply can be > > > enumerated by natural numbers, just like the lines of Cantor's list. > > > OK? And as long as I can enumerate, the number reached is finite. OK? > > > > But the numbers which are reachable include values larger than any given > > natural. > > How should that become possible??? It doesn't have to become, it is. > All transpositions can be enumerated > by *finite* natural numbers. (Perhaps larger than all you can write > down, but that is not he question.) No one is disputing that you are only applying countably many transpostions. What everyone else is disputing is that the effect is what is "mueckenh" claims it to be. > > > To assume a limit on what is achievable is counterfactual. > > To assume that the realm of finite numbers is left and that somewhere > "infinity" is entered is counterfactual. We only assume that even infinite sets can be exhausted (entirely used up). > > > > > Hence I do not need more than a finite set of transpositions, though > > > its cardinality cannot be given. > > > > You, in fact, need more than any finite number can supply. > > I may need more than any fixed number of transpositions. More than any finite number of transpositions. > But the notion > of countability includes the meaning that all elements can be > enumerated by natural numbers. Only by using ALL natural numbers, which is more than any finite number of naturals. > > > > > > > This case is in general denoted by > > > "countably infinite" - but the number oo does *not* appear. Please > > > consider these facts before you demand something to be proved for "oo". > > > > WE do not require that it be proved for "oo", which is not a natural, > > but that it be proved for "for all n in N". > > So we agree on this point. > > > > You method produces exceptions to "for all n in N", by implying that > > there is some remote member of N beyond which it is not necessary to > > proceed in order to establish "for all n in N". > > Completely wrong. All I say is that every natural number is finite. And WE say that there are more than any finite number of them. Or, more briefly but equivalently, that there are infinitely many of them.
From: Virgil on 6 Jul 2006 14:21 In article <1152191465.028598.312840(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152133235.172340.249300(a)v61g2000cwv.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > What I think is that there has to be a function exactly defined before > > > > K exits and when you have defined that you already have that function, > > > > so K is not in the range of the function. The P(N) is different because > > > > it contains all possible K:s. > > > > > > Insn't that a bit too much naive intuition? > > > > That is hard logic. > > | > o > / \ > o o > / \ / \ > > There are more paths than edges. That is hard logic. Only when there is some maximal natural number of edges in paths, as in your diagram. That is also hard logic. When every path contains more than any finite number of edges, then there are "more" paths than edges. That is even harder logic. I , and others, have previously demonstrated a bijection between the set of all infinite paths in a complete binary tree and the set of all subsets of N, P(N). I, and others, have also demonstrated a bijection between the set of edges (or the set of nodes) and N. So that "mueckenh"'s claim that there are "more" edges than paths, is a claim that there is a surjection from N to P(N) (or equivalently a an injection from P(N) to N) but no injection from N to P(N). The trivial injection from N to P(N), f(x) = {x}, proves "mueckenh" wrong.
From: Virgil on 6 Jul 2006 14:29 In article <1152191557.066779.80730(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152133409.564175.47720(a)p79g2000cwp.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > > > 0.1 > > > > > > > 0.11 > > > > > > > 0.111 > > > > > > > ... > > > > > > > 0.111...1 > > > > > > > ... > > > > > > > > > > > > > > But in this list the number 0.111... is not contained. Hence not > > > > > > > all > > > > > > > of > > > > > > > its digits can be identified. > > > > > > > > > > > > You have just proved that your example supports Cantor's theorem > > > > > > by > > > > > > producing a number, 0.111..., not in your own list. > > > > > > > > > > All digits of a number must be indexed by natural numbers. Otherwise > > > > > they cannot be identified. All digits which can be identified are > > > > > pesent in the list which contains all unary representations of > > > > > naturlal > > > > > numbers. > > > > Those are not natural numbers at all in the list, but decimal, or other > > base, fractions. > > These are unary representations of natural numbers: 1 = 0.1, 2 = 0.11, > 3 = 0.111, ... What is your point in having only a sequence of natural numbers? No one has claimed that the set of naturals is uncountable and no one has claimed that a sequence of distinct natural numbers converges to a natural number limit, so that a Cantor "anti-diagonal", or a limit value for a sequence of distinct naturals is irrelevant. ..
From: Virgil on 6 Jul 2006 14:30
In article <1152191614.643204.79420(a)s26g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1152134107.848676.131670(a)75g2000cwc.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > David Hartley schrieb: > > ... > > > > So you acknowledge that your process doesn't give, in the limit, a > > > > bijection N -> Q+ ? > > > > > > Such a bijection does not exist, because there is no smallest positive > > > rational. > > > > There is no need of such for a bijection. Consider the set of positive > > numbers P and the set of negative numbers N. Clearly there is a bijection > > from P to N: f(x) = -x. But N has no smallest element. A bijection is > > not necessarily order preserving. > > You are right. But the transpositions which I defined are order > preserving. Then none of them disrupt the ordering. |