An uncountable countable set
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From: Virgil on 3 Oct 2006 03:47 In article <452211ed(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <4521fc40$1(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > >>> Does "Mueckenh" claim that there is only one function mapping > >>> {1,2,3,...} to {2,4,6,...} or vice versa? > >> There is only one natural order-isomorphic relation, defined by y=2x. > >> The inverse clearly indicates the second set is half the first, whether > >> subset or no. > > > > That presumes, among other things, that a proper subset must be in > > every sense smaller than its superset, but that is not true, as the > > bijection y = 2x proves.. > > How presumptuous of me! TO presumes lots of things contrary to fact, which does make him, among other things, presumptuous.
From: Virgil on 3 Oct 2006 03:52 In article <45221245(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <4521fcd5(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > >>> If we alter the problem to start with all the balls in the vase and > >>> remove them according to the original schedule then every ball spends at > >>> least as much time in the vase as before, but not everyone will see that > >>> the vase is empty at noon. > >>> > >>> Those who argue that having the balls spend less time in the vase leaves > >>> more of them in the vase as noon, have some explaining to do. > >> No, if you start with all the balls in the vase, it clearly becomes > >> empty, but when you are adding more balls per iteration than you are > >> removing, the eventual emptying of the vase is impossible. > > > > Thus TO claims that by putting balls into the vase earlier, but removing > > them as before, one will find fewer of them in the vase at noon. > > > > I do not see the logic of that argument. > > Because, in your version of a mind, you have decoupled the addition and > subtraction of balls, so that they no longer form the same sequence of > events as originally stated. But how does putting balls into the vase earlier produce fewer balls in the vase at noon? Until TO can explain how adding balls earlier ends up with fewer balls, he has a serious logical contradiction in his theories which his analysis has not dealt with successfully.
From: Virgil on 3 Oct 2006 03:53 In article <d8ef7$45220ece$82a1e228$19883(a)news1.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Virgil wrote: > > > History is rife with mathematical developments that, at the time of > > their development, had no uses in physics or any other science, but > > which later turned out to be essential to some science's further > > development. > > This is both true and misleading. > > Han de Bruijn It may mislead HdB, but it has no mislead me.
From: Virgil on 3 Oct 2006 03:57 In article <b3dc0$45221476$82a1e228$24601(a)news1.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Dik T. Winter wrote: > > > In article <1159726829.184763.303470(a)i3g2000cwc.googlegroups.com> > > Han.deBruijn(a)DTO.TUDelft.NL writes: > > > stephen(a)nomail.com schreef: > > > > > > > I suppose I should clarify this. You can approach the infinite > > > > using the the limit concept, but you always have to be careful > > > > when using limits, and you have to be precise about what you > > > > mean by the limit. > > > > > > Okay. But the point is whether there exist infinities that can _not_ > > > be approached using the limit concept. > > > > Yes. In ordinals they are called the "limit ordinals". The name is not > > very appropriate, I think, because they are the ordinals that defy the > > limit concept, but that is just naming. > > > > > Obviously they exist, because > > > how can we approach e.g. the Continuum Hypothesis by employing limts? > > > > I have no idea about the meaning of this statement, but off-hand I ould > > say that there is no way. > > Now we are getting somewhere. In applicable mathematics, approaching the > infinite via the limit concept is the _only_ viable way. All the rest is > nonsense (= has no "sense": can never be measured with a sensor). > > Han de Bruijn As the continuum hypothesis is independent of the other axioms of ZFC or NBG, there is no need to approach it at all. But one can approach it as one approaches anything similarly independent, by choosing whichever version sits one best.
From: Tony Orlow on 3 Oct 2006 10:55
Virgil wrote: > In article <4522007f(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> David R Tribble wrote: >>> Tony Orlow wrote: >>>>> For the sake of this argument, we can talk about infinite reals, of >>>>> which infinite whole numbers are a subset. >>> David R Tribble wrote: >>>>> What are these "infinite reals" and "infinite whole numbers" that you >>>>> speak of so much? >>> Tony Orlow wrote: >>>>> The very same, with no restriction of finiteness. Any T-riffic number >>>>> has successor. :) >>> David R Tribble wrote: >>>>> It's sporting of you to drop the requirement that all the naturals in N >>>>> have to be finite, but since all of them are, it's meaningless to say >>>>> "with no restriction of finiteness". That's kind of like saying N >>>>> contains all naturals "with no restriction of non-integer values". >>>>> I can say that, but it does not change the fact that all the members >>>>> of N are integers. >>> Tony Orlow wrote: >>>> Is the successor to ...11110000 not equal to ...11110001? >>> I don't know - how are you defining those numbers? >>> >>> Even if it is (or is not), what does it have to do with the finite >>> naturals? Are you saying that those "infinite naturals" are >>> somehow successors to the finite naturals? How? >>> >>> >>> David R Tribble wrote: >>>>> So I ask again, where are those infinite naturals and reals you keep >>>>> talking about? It's obvious they are not in N. >>> Tony Orlow wrote: >>>> [No] it's not. >>> Every member of N has a finite successor. Can you prove that your >>> "infinite naturals" are members of N? >>> >> Yes, if "finite successor" is the only criterion. > > But it is not. In addition, N must be minimal with respect to closure > under successorship, which excludes TO's "infinite naturals". Which of Peano's axioms states such a requirement? >> To prove finiteness of such a string: >> >> The bits over each sequence are indexed by natural numbers, which are >> all finite, yes? >> >> For any finite bit position, the string up to and including that bit >> position can only represent a finite value, yes? >> >> Therefore, there is no bit position where the string can have >> represented anything but a finite value, see? If the length is >> potentially, but not actually, infinite, so with the value. > > That is true of the members, but not of the set itself. The sets consists of nothing but the members. If it is true for all members of the ordered set, then it is true for every point in the set. >> >> To prove successorship of such a string: >> >> The rule for successorship for finite values is >> 1. Find the rightmost (least significant) 0 >> 2. Invert from that 0 rightwards >> >> This works for all values where there is a rightmost 0. That excludes >> ...111, which can only have successor given ignored overflow, allowable >> in some cases. > > But not in N. Ah but t'is. >> You don't really question why the successor to ...11110000 is equal to >> ...11110001, do you? > > We question why TO thinks that has anything to do with natural numbers. > > Von Neumann's N is such that > (1) {} is a member of N {} is the empty set, not a natural number. > (2) If x is a member of N then so is x \union {x} > (3) If S is an subset of N such that > (3.1) {} is a member of S, and > (3.2) if x is a member of S then so is x \union {x} > then S = N. > > And that is, by general consent, a satisfactory model for N. You may be satisfied with it. > > But that model cannot contain any of TO's allegedly infinite naturals. Well, it can. > > (1) there cannot be a first infinite natural, and, No, there can't, but you have a fallacious first infinite ordinal, so why complain? > (2) since this N can be proved to be well ordered, > Therefore, every subset of N with no first element must be empty. My N is not "well ordered" any more than the reals. |