An uncountable countable set
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From: Dik T. Winter on 13 Sep 2006 22:06 In article <1158159038.615731.156100(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > Why don't you accept the axiom of > > > infinity? There is an infinite set of natural numbers. All they are in > > > the list (in unary representatition). By the axiom this is accomlished. > > > And there is nothing else. > > > > Indeed, and 0.111... is *not* a natural number, so why should it be in > > the list? Why do you think 0.111... is a natural number? > > There is a set which contains { }, and if it contains A then it > contains A U {A}. This gives an inductive set which we can interpreted > as the set of all natural numbers or the set of all indexes. Yes. And 0.111... is not in there. Because (1) it is not a natural number and (2) it can not index any digit. > > > > If it does not hold you should be able to give an index position > > > > such that it is false. > > > > > > It is not in the list, although we cannot give an index position which > > > is responsible for that fact. All possible indexes are in the list by > > > the axiom of infinity. > > > > It is not in the list because it is not a natural number. > > All indexes are in the list. All numbers which can be indexed are in > the list. The first is true, the second is false. > > > Because it is not in the list which, by definition, contains all > > > numbers which can be indexed because there are all numbers which can > > > index. > > > > This is nonsense. By the definition of your list the list contains all > > finite natural numbers. That has nothing to do with indexing. > > Index = natural number. There is no infinite index, because indexing is > identifying. That does *not* explain why the list contains all numbers that can be indexed. > > If not all digits can be indexed, there should be a digit that can not be > > indexed. But *by the definition* I gave above, all digits can be indexed. > > Why is that wrong? > > Because all digits which can be indexed are in my infinite list. Could > all digits of 0.111... be indexed then all its digits were represented > in the list. But every sequence of 1's all digits of which are > represented in the list, is completely in the list. A proof of the last statement, please. Notably the 'every'. > > > Indexing and covering "by all list numbers" is equivalent. Both > > > is true (can be done) or both is false (cannot be done). > > > > You again do not clearly state what you mean. But the statement as it > > stands is *false*. > > If a sequence can be completely indexed using any list number required Note that you change again from "all" to "any". The sequence can not be completely indexed by "any" list member, as it can not be covered by "any" list member. Nevertheless it can be indexed by "all" list members. I have no idea what covering by "all" list members would mean. > for that purpose, then this sequence can be covered by a list number, > and vice versa, if a sequence can be covered by a list number, then and > only then can it be indexed completely by list numbers. Yes. But there is not a single list number that is able to index K. > > > "Not to terminate" is not a property which can serve to distinguish a > > > number from others, because we can never observe the end (because it > > > does not exist) neither the non-end (because it is nothing but a > > > negation of an unobservable property). What can serve to distinguish > > > two numbers in unary representation is a 1 at a digit position. > > > > You are not talking nonsense. > > No? A typo, I meant "now". > > As 0.111... is different from al sequences > > 0.111...1 because it is infinite in length, there is for each 0.111...1 > > an digit position where 0.111... is different from that specific 0.111...1. > > 0.111... is different from all unary numbers which can be indexed > (because they are in the list by the axiom of infinity). Yes, utter nonsense. I have not yet seen a proper proof of the "because". > Hence there > must be a digit which cannot be indexed. "Not to terminate" is not a > property which can serve to distinguish a number from others. Why not? > > > > And if you think that that definition is wrong, please *prove* that. > > > > > > It has been proven by showing that 0.111... is not in the list. > > > > Proof that it should be in the list. But remember: 0.111... is *not* > > a natural number. > > That means: It is not a number which can be completely indexed. > Therefore it cannot be in the list. (It is not a number at all.) Circular reasoning. Based on the assumption that a non-terminating list of naturals can not index the digits of a non-terminating decimal (or whatever base) number. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 13 Sep 2006 22:18 In article <45085df2(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Mike Kelly wrote: > > Tony Orlow wrote: > >> Mike Kelly wrote: > >>> It doesn't. I never said it does so please STOP asking that question. I > >>> say that 111..... doesn't represent a (natural) value at all. > >> If it is a whole number (no fractional component), is finite, and has > >> successor and precedessor, then ....1111 is a finite natural. > > > > It isn't. Now what? > > It's not what? A whole number? Show me the fractional component. Finite? > SHow me the bit position where an infinite sum is possible. With > successor and predecessor? There is no successor to this largest > natural, unless the number line is taken to be circular. TO is claiming numerals (digit strings) are numbers, which shows that he has no notion of what numbers are. And in order for a number to be a natural number, it must be a member of the smallest inductive set, which means that it must be representable by a digit string with both a first and last digit. Proof: Let S be the set of naturals(i.e., the smallest inductive set) representable by digit strings with a first and a last digit (which can be the same digit). (1) the first natural is a member of S (2) if x is a member of S the so is the successor of x. Therefore EVERY natural is representable by a digit string with both a first and a last digit.
From: Dik T. Winter on 13 Sep 2006 22:16 In article <1158159208.048232.260050(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <45070bcb(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > ... > > > What about an unboundedly large but finite number? > > > > Let me ask you to, before I can answer such a question. What is your > > definition of "number"? (I asked the same from Wolfgang Mueckenheim, > > but his answer was not satisfactory, also not to himself, I think, > > because he never answered to questions about it.) > > Didn't you read my paper on the physical constraints of numbers? Yes, I read it. Mathematically it makes no sense. > Here I give briefly my definition which is not obligatory but useful to > avoid the weaknesses of set theory. It is derived from the first > principle of the notion "number", i.e., a means to count. In German > this becomes even more evident by the wording "Zahl" and "zaehlen". > A number is an entity which: > 1) either is realized by a fundamental set like the following > fundamental set of 3: > {III, Dik, {a,b,c}, {father, mother, child}, {sun, moon, earth}, > ...} I complain. When I am considered as a set I would think that the set contains only a single element. > 2) or is completely determined by a series of digits. Question. A terminating or a non-terminating series? > Each of these two requirements is sufficient to guarantee that the > numbers satisfy the trichotomy. That is the most important property of > a number. Sufficient, perhaps. Necessary, no. > In this sense irrational numbers, complex numbers, hyperreals, > transfinite numbers, vectors, etc. are not numbers but ideas which > consist of numbers, like 2 + 3i, or not, like pi. Yes, that is your preoccupation with the word numbers. In mathematics that word is used much more broadly than your preoccupation allows. I would state: talk the lingo if you want to attack the concept. > > Once you have answered that question we can probe further into the > > definition of "unboundedly large but finite number". > > > > As far as I understand, for any kind of numbers, they are fixed, so > > by definition not unbounded. > > O course. But the sequence or the set or the variable may be unbounded > though always finite (because it consists of or takes on finite > entities - in contrast to 0.111...). This is (at least for me) incomprehensible. I was talking about numbers, not about sequences, sets or variables. But every definition of 0.111... takes on finite entities only. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 13 Sep 2006 22:22 In article <45085f4c(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > Tony Orlow wrote: > >>> If you remove an element, the proper subset should ALWAYS be smaller by > >>> 1. That is the case for me. For a theory to claim a proper subset is the > >>> same "size" as the proper superset is an immediate deal-breaker for me. > > > > David R Tribble wrote: > >>> If by "different size" you mean that you cannot pair up all the > >>> elements from both sets, then you're going to have a difficult > >>> time proving that for any infinite set. (You have never show this, > >>> BTW.) > >>> > >>> If by "different size" you mean something other than some way of > >>> denumerating (counting) the elements of the set (e.g., by assigning > >>> them different natural indices), then you should use a different term, > >>> because it's confusing. Oh, and you have to prove that it works > >>> (you have never shown this, either). > >>> > >>> Start with a simple proof: Take the set of naturals, N = {0,1,2,3,...} > >>> and remove one element to get set S = {1,2,3,...}. Now show that > >>> the "T-size" of N is exactly one less than the T-size of S. In other > >>> words, find a way to show that every counting of S versus every > >>> counting of N always leaves one element of N (0) left over. > > > > Tony Orlow wrote: > >> Use IFR. > > > > A.k.a. a bijection. You see where this is going? > > > > Yes, bijection with measure. > > > > >> N maps to S using f(n)=n+1. The inverse of that function is > >> g(x)=x-1. > > > > Which proves that every n in N has an x in S. Where is that > > leftover element that was removed from N? If N has more > > elements than S, shouldn't N have a member that can't be > > mapped to any member of S? > > > > One can map all sorts of sets. N contains every element of S - map those > first, to themselves. Now you have one left over. > > > > >> So, over the range of 0 to N, |S|=|N|-1. > > > > Funny how you don't define what |X| is. You're using standard > > symbols but obviously with a different meaning, since "|X|" means > > "cardinality of X" when X is a set. Your IFR bijection proves that > > |S| = |N|. > > > > |X| means size of, like the absolute value of a real. > > > > >> Map N to the Evens E using f(n)=2n. The inverse function is > >> g(x)=x/2, so over the range of N, > >> the evens have |N|/2 elements. Isn't that intuitively satisfying? And > >> gee, it works for finite sets accurately too!! > > > > Same thing as above, it proves that every n has an x. Where are > > the leftover unmapped elements of N that make S a smaller proper > > subset of N? > > > > All the odds. N contains all elements of E, plus the elements of O. IFR > works to the level of accuracy of detecting a change of 0ne element out > of an uncountable number. But in standard mathematics, it is trivial that any ordering of sets in which every proper subset of a set is 'smaller' than its superset, is no more that a partial ordering, and can never be a total ordering, so that there must be sets which cannot be compared. So that TO's requirements are self-contradictory.
From: Tony Orlow on 14 Sep 2006 11:51
Dik T. Winter wrote: > In article <4506e412$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Dik T. Winter wrote: > > > In article <45005670$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > > > Dik T. Winter wrote: > ... > > > > Well, if you set up a Turing machine such that it will take an infinite > > > > number of operations to get to any finite digit, then you have probably > > > > not designed it well. > > > > > > You do not seem to understand. If you set up a Turing machine such that > > > given a number n as input calculates the first n digits of some number > > > can be very well designed. But can you proof that it will stop? > > > > Depending on what number you are calculating, you should be able to > > prove that it will reach a certain degree of accuracy in a certain > > number of steps. I'm not sure how to answer that question in such > > general terms. > > Apparently you do not understand it. Ever heard about the halting problem? Yes, of course. > > But to get more in your field. Consider Newton-Raphson for the calculation > of the square root of a floating point number in the range [1/4, 1]. > Also consider that the arithmetic is truncating (in the direction of 0) to > the precision used. Now consider the stopping criterium to be that two > successive iterands are equal. Does that algorithm stop? The answer is: > in general it will stop. There are precisely two input numbers in the > complete range for which it will not stop. Those being 1 and oo? I am not intimately familiar with the Newton-Raphson solution, but a cursory examination of the method indicates it will not stop, in terms of actually identifying sqrt(2) in its entirety. But, it will stop once it gets within any finite distance of sqrt(2). This bijective function between iterations and precision holds in general, given certain constraints. Thanks for your input. :) Tony |