An uncountable countable set
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From: Virgil on 12 Sep 2006 17:04 In article <45070bcb(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <4506d58e(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> David R Tribble wrote: > > > >>> You are obviously missing something. > >>> > >> What I left out was that the power set includes countably infinite > >> subsets of N, and such countably infinite bit strings are not considered > >> standard naturals. However, they represent finite whole values > > > > For any standard meaning of finite, TO's claim is FALSE! > > > > In order for a binary string (function from N to {0,1}) to have a > > standard correspondence with any finite natural number value it must > > have only finitely many non-zero bits, and thus be equivalent to a > > finite string with leading bit 1 or with only one bit. > > > > Any string with infinitely many non-zero bits cannot correspond to any > > finite numerical value. > > What about an unboundedly large but finite number? There are no such things in any exant system. If TO ever produces a complete system of his own, only then will he be able to produce such basilisk beasts.
From: David R Tribble on 12 Sep 2006 19:05 mueckenh wrote: >> What you are arguing is only the beginning of infinity, because you are >> unable to see more than this little realm. Try to determine the natural >> number which consists of the 10^100 digits following the first 10^1000 >> digits of pi. Your unability is not due to lack of time. This is only >> one of the infinitely many numbers which you cannot deal with (because >> it is not a number). > David R Tribble wrote: >> Why is it not a number? It has all the mathematical properties >> required of a number. Having "all its digits enumerated" is not a >> required property. > Tony Orlow wrote: > What IS a number, and what IS mathematics? > When you can answer one or both of these questions, please do. I'll borrow someone else's words: [http://en.wikipedia.org/wiki/Mathematics] Mathematics is the discipline that deals with concepts such as quantity, structure, space and change. It evolved, through the use of abstraction and logical reasoning, from counting, calculation, measurement and the study of the shapes, and motions of physical objects. Mathematicians explore such concepts, aiming to formulate new conjectures and establish their truth by rigorous deduction from appropriately chosen axioms and definitions. That last part is key. [http://en.wikipedia.org/wiki/Number] A number is an abstract entity that represents a count or measurement. In mathematics, the definition of a number has extended to include abstractions such as fractions, negative, irrational, transcendental, and complex numbers. I'd also add algebraic numbers, ordinals, cardinals, quaternions, octonions, matrices, tensors, p-adics, hyperreals, and surreals to that list. You can see my confusion about Mueckenh's definition of "number".
From: David R Tribble on 12 Sep 2006 19:20 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? > 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? > 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|. > 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?
From: Dik T. Winter on 12 Sep 2006 19:21 In article <4506dbd2(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: .... > Yes, it should, and Wolfgang's and my position is that N is unbounded > but finite, given that it only contains finite values, and only one per > unit of value range. By the definition of "finite", "unbounded" and "finite" are in contradiction with each other. So a number can not be both. > Since the value range is finite, given that no two > naturals are infinitely different, and given that only a finite number > of elements can be fit sparsely in a finite range, the set is finite, > even though it's unbounded and infinite by the Dedekind definition. Not the Dedekind definition, but the standard definition. There are models with sets that are infinite, but Dedekind finite. (But in that case you do not have AC, because AC implies the two kinds are the same.) > At which bit position can the string achieve an infinite value? None > that exists in that string. Indeed. And that is why you need to do something to give actual meaning to such strings. In the 2-adics such is done, and in that case, ....111 is sum{n = 0 -> oo} 2^n = lim{n -> oo} (1 - 2^n)/(1 - 2) = -1 because in 2-adic metric lim{n -> oo} 2^n = 0. (The metric is defined as d(a, b) = 1/2^n if the n-th digits of a and b are the highest order digits where they differ.) > > This should be a rather big clue that something's wrong with your > > concept, and that it does not, in fact, "work as a natural". It does indeed not "work as a natural", but it works as a 2-adic. > > Which presumably requires some extended Peano axioms in order > > to exist? > > No, it just requires a loosening of the requirement that we only look at > the minimal set satisfying those axioms. But it is useful to look at the minimal set satisfying those axioms. There are too many sets that satisfy those axioms so that it is difficult to state anything with sense when the sets are not minimal (unless they are special). Take the Gaussian integers. They satisfy the axioms when we state the successor function as S(k + i.l) = k + 1 + i.l Take the normal integers, they also do satisfy the axioms with the standard successor function (+1), take the rational numbers, they also do satisfy the axioms when we define the successor of p/q as p/q + 2. > I don't see that it contradicts > them as they stand. It does not contradict them, but it is the minimality that makes things provable that otherwise would not be provable. > Extensions of the naturals such as this also fit > into that model. Yes, like a host of other extensions. -- 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 12 Sep 2006 19:27
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? 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |