Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 28 Jul 2005 14:00 David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > No, I understand what you mean by "countable". My issue with your > > "uncountability of the powerset of the naturals" is that you can > > easily form a bijection between the members of the powerset and the > > set of infinite bit strings denoting set membership in the subsets, > > Yes. > > > and between those bit strings and whole numbers, > > No. > > > and the only reason this bijection is rejected is because of the > > refusal to allow infinite whole numbers in the set of whole numbers, > > They don't fit the axioms. But let's, for the sake of the argument, > take a look at the set of infinite binary bit strings (which is a > superset of the naturals). All positions of those bit strings _are_ > already completely used up in the powerset mapping for just the > naturals. Let's call the bit stream picking out all numbers dividable > by k B_k, so B_1 = ...111111111, B_2=...10101010101, B_3=...1001001 > and so on. > > Now tell us the number of the set {B_2, B_3}. It has 2 bits set. > What is the distances between those 2 bits, and where is the rightmost > bit? Okay, so now you want to take this up a level, and assign numbers to subsets of the power set. I would note that B_1 is equal to B_2+2*B_2, or 3*B_2, and also equal to B_3+2*B_3+4*B_3, or 7*B_3. So, we have B_2=N/3 and B_3=N/7, and the distance between them being 4N/21. Can we express these numbers? Well, not very easily in terms of bitstrings. Nevertheless, conceptually we can imagine that we have an infinite string of bits, with two 1's, 1/7 and 1/3 the way from the beginning of the bitstream. It's somewhat irrelevant whether we can work with these values easily or not. The concept is still valid. > > The infinite bitstreams you dragged in to be able to number the power > set have exacerbated the problem: you can't indicate subsets > containing them. The only subsets you can now index are subsets > containing only finite integers. In a sense, and yet, that really doesn't matter when considering the question as to whether such infinite strings can represent a power set. > > Since the additional members have not solved the problem of the > powerset mapping, it is less complicated to not admit them in the > first place. Then we only have a single level of problems. You are taking the power set of the power set, which complicates matters some, but doesn't destroy the basic concept. Just because things get sticky doesn't mean they don't make sense. > > > despite the fact that an infinite set of whole numbers requires > > infinite whole numbers, > > Whining does not make it so. It requires arbitrarily large numbers, > but none of them need to be infinite. I am not whining and you are not correct. > > > so you don't consider the infinite bitstrings to represent whole > > numbers. It's like all the wrong choices have been made, almost on > > purpose, in order to make some grand distinction which really isn't > > there. > > Well, it's like you try to fix something that you perceive as a > problem by making it much worse. > > > If you allow infinite whole numbers, as is required, > > Whining won't make it so. And making statements like that doesn't make it any less so. > > > suddenly this whole "uncountability of the power set" vanishes in a > > puff of smoke. > > But then you still can't account for the resulting powerset: you can > only cover the powerset of the finite integers with your infinite bit > strings. > > -- Smiles, Tony
From: Tony Orlow on 28 Jul 2005 14:04 Daryl McCullough said: > Tony Orlow (aeo6) wrote: > > >Daryl McCullough said: > > >> So, you agree that for *finite* sets, two sets have the same > > >> bigulosity if and only if there is a bijection between the two? > >> But that no longer holds for infinite sets? > >> > >> Then how is bigulosity an improvement over cardinality? > >Because Bigulosity takes into account the nature of the > >bijection in order to determine a precise relative size of infinity. > > In other words, bigulosity is whatever you want it to be, and > so you have a lot more flexibility. Just make it up as you go > along. > > A = the set of natural numbers { 0, 1, 2, ... } > B = the set of base ten numerals { "0", "1", "2", ... } > C = the set of base two numerals { "0", "1", "10", "11", "100", ... } > > A and B have the same bigulosity. A and C have the same bigulosity. > But B and C do *not* have the same bigulosity (clearly C has a smaller > bigulosity than B). That's bigulosity for you... > > -- > Daryl McCullough > Ithaca, NY > > If you consider your numerals to have N digits, then base ten has 10^N elements and base 2 has 2^N elements, but we consider the decimal system to only require log10(N) digits and the binary system to only require log2(N) digits, thereby making each set the same size, as one might epect, since they represent the same concepts. That's Bigulosity for you. And, please make sure you capitalize it. Big B for Bigulosity. :) -- Smiles, Tony
From: malbrain on 28 Jul 2005 14:27 Poker Joker wrote: > "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message > news:dc9f0t03cj(a)drn.newsguy.com... > >>> >Robert Low said: > >>> > > >>> >> OK, so how many elements are there in the set of all finite > >>> >> natural numbers? > > > > Tony replied. > > > >>>>Some finite, indeterminate number. > > > > That is an out-and-out contradiction. Let FN be the > > collection of all finite natural numbers. You say that > > FN is finite. You say that that means that its size is > > equal to some finite natural number. > > He never said that. He said "Some finite, indeterminate > number." He didn't say "Some finite, > indeterminate natural number." But even if he did, > mathematicians have their own meaning of words > and therefore his natural numbers might be different than > mathematicians. Words have an agreed meaning. That's why we have dictionaries. Mathematicians learn to use words like everyone else. karl m
From: David Kastrup on 28 Jul 2005 14:27 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > David Kastrup said: >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> > and the only reason this bijection is rejected is because of the >> > refusal to allow infinite whole numbers in the set of whole >> > numbers, >> >> They don't fit the axioms. But let's, for the sake of the >> argument, take a look at the set of infinite binary bit strings >> (which is a superset of the naturals). All positions of those bit >> strings _are_ already completely used up in the powerset mapping >> for just the naturals. Let's call the bit stream picking out all >> numbers dividable by k B_k, so B_1 = ...111111111, >> B_2=...10101010101, B_3=...1001001 and so on. >> >> Now tell us the number of the set {B_2, B_3}. It has 2 bits set. >> What is the distances between those 2 bits, and where is the rightmost >> bit? > Okay, so now you want to take this up a level, and assign numbers to > subsets of the power set. _I_ don't want to take this up a level. You want. If you claim that the cure for the set to be mappable to its powerset is to include additional members, then _of_ _course_ you have to be able to map the powerset of the _resulting_ set. Everything else would be cheating. You can't include those members in the set itself, and ignore them when doing the powerset. > I would note that B_1 is equal to B_2+2*B_2, or 3*B_2, Ok. > and also equal to B_3+2*B_3+4*B_3, or 7*B_3. So, we have B_2=N/3 What is "N" supposed to be here? > and B_3=N/7, and the distance between them being 4N/21. > Can we express these numbers? Well, not very easily in terms of > bitstrings. Too bad, since that is your representation. > Nevertheless, conceptually we can imagine that we have an infinite > string of bits, with two 1's, 1/7 and 1/3 the way from the beginning > of the bitstream. Uh, there is no beginning of the bitstream. We are talking about infinite strings here, remember? 1/7 times infinite certainly still is infinite. > It's somewhat irrelevant whether we can work with these values > easily or not. The concept is still valid. The concept is bunk. >> The infinite bitstreams you dragged in to be able to number the >> power set have exacerbated the problem: you can't indicate subsets >> containing them. The only subsets you can now index are subsets >> containing only finite integers. > In a sense, and yet, that really doesn't matter when considering the > question as to whether such infinite strings can represent a power > set. Nobody ever disputed that infinite binary strings are perfectly capable of representing the power set of naturals, the standard, finite variant of naturals, that is. But you are claiming that those strings ought to be included into the naturals for some reason. And once you do that, the infinite binary strings are _no_ _longer_ sufficient for mapping to the power set of this _new_ set. >> Since the additional members have not solved the problem of the >> powerset mapping, it is less complicated to not admit them in the >> first place. Then we only have a single level of problems. > > You are taking the power set of the power set, which complicates > matters some, but doesn't destroy the basic concept. It is not _I_ that is taking the "power set of the power set". I am taking the power set of what _you_ claim should be the naturals, namely some set including infinite bit strings. And it is _your_ claim that this Orlow-natural set is now able to map to its own power set. > Just because things get sticky doesn't mean they don't make sense. Too bad that one can rigidly _prove_ that your augmented set is impossible to provide a mapping onto its own power set. This is not sticky. It's fatal. >> > despite the fact that an infinite set of whole numbers requires >> > infinite whole numbers, >> >> Whining does not make it so. It requires arbitrarily large >> numbers, but none of them need to be infinite. > > I am not whining and you are not correct. Whining does not make it so. >> > so you don't consider the infinite bitstrings to represent whole >> > numbers. It's like all the wrong choices have been made, almost >> > on purpose, in order to make some grand distinction which really >> > isn't there. >> >> Well, it's like you try to fix something that you perceive as a >> problem by making it much worse. >> >> > If you allow infinite whole numbers, as is required, >> >> Whining won't make it so. > > And making statements like that doesn't make it any less so. Well, let's suppose we have a mapping of the Orlow-integers O to their powerset P(O), as o->f(o). Then take a look at the set U={o in O: o not in f(o)}. This set U is different from f(o) for _all_ o in O, since U does not contain o in case f(o) contains it, and it contains o in case f(o) does not contain it, so it is different from f(o) for _any_ o. Nevertheless, it only contains elements of O, so is a member of P(O). That's not mere whining. That is a proof. You would not recognize one if it hit you in the face, but others will. And that's what makes the difference to mere whining of how you would like things to be. >> > suddenly this whole "uncountability of the power set" vanishes in a >> > puff of smoke. >> >> But then you still can't account for the resulting powerset: you >> can only cover the powerset of the finite integers with your >> infinite bit strings. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Daryl McCullough on 28 Jul 2005 15:01
Tony Orlow (aeo6) wrote: >> A = the set of natural numbers { 0, 1, 2, ... } >> B = the set of base ten numerals { "0", "1", "2", ... } >> C = the set of base two numerals { "0", "1", "10", "11", "100", ... } >> >> A and B have the same bigulosity. A and C have the same bigulosity. >> But B and C do *not* have the same bigulosity (clearly C has a smaller >> bigulosity than B). That's bigulosity for you... >If you consider your numerals to have N digits I consider that there are just enough base-ten numbers so that there is one for each natural. And there are just enough base-two numbers so that there is one for each natural. Any more is overkill. -- Daryl McCullough Ithaca, NY |