An uncountable countable set
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From: Dik T. Winter on 21 Aug 2006 07:42 In article <ecb927$ce6$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Dik T. Winter wrote: > > In article <ec8if0$rtd$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: .... > > > > > > Indeed, I agree with WM's logic concerning the identity relationship > > > between element count and value in the naturals. He's quite correct in > > > that regard. > > > > Well, you and he are not. The logic is flawed. > > How so, precisely? Because you by the same logic the ordinal number of the set of naturals is *larger* than the largest natural, or *smaller*. > > > I'm a devoted post-Cantorian delver into > > > infinity. While he takes the argument put forth as proving that the set > > > of naturals is finite, he does so with the assumption that infinite > > > naturals cannot exist. > > > > And his proof is not a proof. > > When any positive infinite value is considered greater than any finite > value, and you prove inductively the p(x) is true for x > finite y, > well, yes, it's a proof. You can not prove it inductively based on properties for finite x. The induction axiom only is about finite x. > > > For my part, I agree that the set of finite > > > naturals is finite, though unbounded, > > > > In that case you are not using standard mathematical terminology. I > > have no idea what a finite but unbounded set is. > > Yes, that's contradiction in terms given the Dedekind definition of an > infinite set. WM's point is very subtle. A *subtle* contradiction in terms? > I know that because I have > raised it a bunch of times over the last year+, as I think has he. But, > where there is a constant finite distance on a line between points, and > no point is infinitely distant from any other, there is a finite range, > and only a finite number of such disjoint intervals can occupy that > space. Yes, you are arguing that, and WM is arguing that, but offering no proof. > > > but that there exists an infinite > > > set of naturals, which includes infinite values. > > > > That is alright with me, only, do not call them naturals, because that is > > extremely confusing. > > Yes, I think "hypernaturals" as a superset of the naturals is best, > though that name has been used, and may carry some unwanted ssumptions > with it. Still, for now, "hypernaturals"? That is probably less confusion, although it also has been used. > Dik - have you given much thought to infinite values? I have studied the surreals quite thoroughly. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 21 Aug 2006 07:47 In article <1156148541.934474.46690(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1155998763.083248.291940(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > > > Where can I find that quote? I have problems with reading utf-8. > > > > > > Works, p. 278, first paragraph (his famous article on the diagonal > > > proof). > > > > I will look it up when I am back at work. > > > > > > Well, let me be clear. Constructable numbers is used in two senses in > > > > mathematics. The sense I meant is "computable numbers". So let me > > > > refrase with this word: > > > > The set of computable numbers is countable, but the diagonal is > > > > not computable. > > > > > > No diagonal is computable, except in simple cases like my list, because > > > no infinite list is computable, again with the exception of closed > > > formulas and simple cases like my list. But that is completely > > > irrelevant. > > > > You apparently do not know what the word "computable" does mean. To > > be precise: all algebraic numbers are computable in the mathematical > > sense. > > To be precise: Each algebraic number is computable. But not all. > Because then the list of all algebraic numbers was computable. But it is. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 21 Aug 2006 07:53 In article <1156148773.420187.122140(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: .... > > > I *defined* that all digit positions which can be indexed belong to > > > numbers of my list. The indexes are natural numbers and all natural > > > numbers are in my list. > > > > > > And I showed that your number is not in my list. > > > > > > Hence one of our definitions must be wrong. > > > > No. *Both* definitions are correct, there is no contradiction between > > them. It is your later assertion that is incorrect: > > "If all digit positions can be indexed, the number is in the list". > > It does not follow from your definition. > > You are dreaming. It does follow from the form of the numbers of my > list. > Try to come back to reality. Simply show how a digit position n can be > indexed the complete number of which, i.e., the complete sequence of > digit positions 1 to n is not in the list. Come off your high horse. Tell me why K: K[p] = 1 for all p in N can not be indexed. ANd do not come back with "because it is not in the list". That would be circular reasoning. > > > Either there are not all natural numbers in my list and nowhere else, > > > or your number cannot be indexed completely. > > > > Nothing of that. Give me an index position in my number that is not in > > the list. (By definition, there are none.) But all natural numbers > > are in your list, as are all index positions of my number. But the > > number itself is not in the list, because it is not a natural number. > > In my list there are all numbers which can index the position n+1 if > they can index the position n. But they cannot index a position larger > than all naturals. > All positions which are not larger than all naturals are in my list. Yes. And my number does not have positions larger than all naturals. > You assert that a set of small numbers can surpass a large number if > there are enough small numbers. That is wrong. 0.111...1 < 0.111... > such that any attempt to index digit positions of 0.111... leaves > unindexed positions. Makes no sense. True, if we index position n, higher positions are not indexed. That does not mean that they *cannot* be indexed. > It is the same with the staircase: If the total height is H, then there > must be at least one stair of height H. Wrong. Think asymptote. > > All digit positions are indeed in the list, and it is itself not in the > > list. Why must it be in the list itself? Consider the digit positions > > as bricks. I state the number I have is built of bricks, but I do not > > state that the number I have is a brick itelf. That is what *you* are > > claiming. > > You forget that the numbers of my list leave no other outcome. But they do. > > > > That is not a proof. Indeed, the sum of infinitely many differences > > > > of 1 make up an infinite number, > > > > > > are you sure? Indeed, even in set theory a sum of infinitely many 1's > > > is infinite? > > > > I would think so, yes. > > We have infinitely many differences between natural numbers. Hence they > sum up to a non-natural number. Contradiction. What is the contradiction? There is nothing in mathematics from which you can find that "an infinite sum" of natural numbers is a natural numbers. (And I put that in quotes because that is not really defined in mathematics.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 21 Aug 2006 08:09 In article <1156149324.184708.191200(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > Now, if we complete the tree (and the list above) the number > > of edges is still countable, but *none* of the edges terminates at 1/3. > > You want "a last line" for completion, but there is none. > > I don't want a last line, nevertheless the set of all edges is > countable. So 1/3 is not in your tree. > > You simply > > do not accept completion without there being a last one. And that is > > precisely what the axiom of infinity asserts, the set of natural numbers > > exist (that is, it is possible to talk about it), but there is no > > largest natural number. > > Nevertheless they are all countable, just like the edes of my infinite > tree. But 1/3 is not in the tree. > Have you got it now??? Have you got it now??? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 21 Aug 2006 08:07
In article <1156148970.634196.169840(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > Dik T. Winter schrieb: > > > > Again a conclusion without proof. Use my definition of K: K[p] = 1 > > > > for all p in N, and there are no other digits. > > > > > > Why should I? > > > > I would think because we are talking about that number? You keep stating > > that that number is not indexable, without proof. When you want to show > > that you have to start with that definition. .... > > > > Show that that number can not be indexed or show that that number can > > > > be covered by a natural number, give a *proof* of either. > > > > Again, I ask you that. > > Indexing is covering due to the form of the numbers of my list. > If position n can be indexed, then 1 to n are covered. In repeatal mode again? That is no proof for either of my statements. Each position (say n) of my number can be indexed, so each segment from positions 1 through n can be covered. This does *not* mean either that my number can not be indexed or that my number can be covered. > > > I define the list such that your definition is wrong. if all natural > > > numbers do exist, my list is complete. > > > > What is the wrongness of the definition? Let me refrase: > > Let L be the list of all natural numbers > > Let K be the sequence of digits such that for each p in L, K[p] = 1. > > What is *wrong* with that definition? > > It cannot be satisfed. Why not? > > > The required digit positions are not contained in the list in form of > > > this infinite number but only disperged over many finite numbers (which > > > is impossible, of course). > > > > Why is that impossible? There is no "of course" here, because that "of > > course" is just stating "because the set of all naturals does not exist". > > Because all digit positions up to a given one are all contained in one > single number. > > In my list holds the following law: > > If a digit position n is covered then there is at least one number not > less than n in the list. Right. > If an infinite number of digit positions is covered than there is at > least one number with infinitely many digit positions in the list. Right. > But > in my list there is no number with infinitely many digit positions. Right. > If you would understand that, then all you question were answered. No. Because my number can not be covered. It is your lack of a proper proof that if each digit of a number can be indexed that number can be covered. And such a proof does not exist. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |