An uncountable countable set
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From: Tony Orlow on 15 Sep 2006 20:53 Mike Kelly wrote: > Tony Orlow wrote: >> Randy Poe wrote: >>> Tony Orlow wrote: >>>> Haha!! Good one. "There are the same number of primes as there are >>>> naturals, a proper superset." Good answer. >>> Obviously you don't believe that. >> Good guess. >> >>> Here's an equivalent statement: If n is a natural, >>> there is an n-th prime. >>> >>> Do you think that's false? Do you think that there >>> is a natural n such that there is an (n-1)th prime, >>> but no n-th prime? >> No, I don't think that, but I do recognize that the set of primes is a >> subset of the naturals which proportion thereof has a limit of 0 as the >> range of the naturals approaches oo. > > So, uh, do you think that there is always an nth natural? And always an > nth prime? So arne't there, like, the same number of both, given the > equivalence between the two ideas? Why do you *always* avoid answering > the most pertinent questions directly? I have answered that, directly, several times. No, I do not consider simple bijection to be grounds for considering two sets equal, when the value range is not determined. If you speak of all reals or all naturals, that covers the same real range called "all", so if one set is denser than another, and covers the same range, it's a bigger set. It's that simple. If you claim the naturals and evens, for instance, have the same count, that in essence says the evens have twice the range, since every even is twice its corresponding natural. That's wrong. > >> Do you honestly think bijection works as an infinite analog to equality? >> Can a dense set like the rationals, with an infinite number of them >> between any two naturals, really be no greater a set than the naturals, >> which are an infinitesimal portion of the rationals? > >> That's just poppycock. > > Doesn't matter if you feel that way. All cardinality says in set theory > is "sets with this cardinality are bijectable with one another". Yes, it's the claim of mathematicians that this is a valid extension of the finite meaning of "set size" which is so irksome. Go ahead and have your broad equivalence classes based on nothing other than bijections, but stop pretending that hocus pocus applies to anything outside of Cantor's Garden O' Tricks, because it doesn't. Until the concept of measure is applied to infinite sets of reals, they cannot be properly compared. > > Your personal feelings on whether this really means they have "the same > number" of elements seems highly irrelevant, given that all set theory > "cares about" is that "these sets are bijectable", something which you > don't seem to dispute (give or take one really stupid argument about > the powerset of the naturals being countable). Sure, and set theory plays Banach-Tarski Ball, and Omega the Littlest Giant, and The Elements that Didn't Matter, and other such nonsense. > > Most people intuitively think that if two sets can have all their > elements paired up then they have "the same number of" elements. You > don't. So what? Set theory doesn't "care" if that's what you think > "bijectable" means or not. > Some people do, and many are turned off by the hat tricks of transfinitology. It's not correct that you can cut a ball into five pieces and reassemble it into two solid balls of the original size, and it's not true that you can put ten balls in a vase and take one out, over and over, and ever get an empty vase. If you don't think people "care" whether what they think makes sense or not, then I guess you don't care whether what you think makes sense. Tony
From: Tony Orlow on 15 Sep 2006 20:58 Virgil wrote: > In article <450b0042(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> David R Tribble wrote: >>> Tony Orlow wrote: >>>>> Yes, I am including infinite values on the number line, since it's >>>>> "infinitely long". >>> David R Tribble wrote: >>>>> Yet another thing you have to define or prove. Where do these >>>>> "infinite values" appear on your real number line? >>> Tony Orlow wrote: >>>> Further from 0 than any finite number. >>> Then how are they "on" the same "line"? >>> >> By trichotomy. For all x ad y on the line, either x>y, x=y or x<y. >> That's what makes a line in concept. > > But until TO proves trichotomy for all his imaginings, they are not on > any line. > > And he has not done so. > > Note that using either the Dedekind cut or Cauchy sequence definition of > reals, one can prove trichotomy for them >>> David R Tribble wrote: >>>>> Are these infinite values connected (in the mathematical sense) to >>>>> the finite values on the line? If not, how is it a "line"? >>> Tony Orlow wrote: >>>> Yes, they are on the line, with all finite values between them and 0. >>> That's not what "connected" means. If they are not connected, >>> they are simply two separate sets, which means it is more correct >>> to visualize them as two unconnected (but "ordered") "lines". >>> >> That is like saying the starting point of a line segment is disconnected >> from the rest of the line. > > Until TO proves his trichotomy, and connectivity, he has neither. Note > that one can prove the density and "continuity" of the reals (contunuity > being the LUB and GLB properties. > > Accordingly TO must show that any non-empty set of his extended reals > which is bounded above must have a least upper bound. > > The set of standard reals is bounded above if there are any infinite > reals, so must have a LUB, so what is it TO? I have already said I have no least infinity. You're the one with an LUB of aleph_0, even though the same logic applies to that fallacy as to the largest finite, unless one disregards that removing elements makes a set smaller. What you are saying applies only within any finite neighborhood on the real line. >>> Tony Orlow wrote: >>>>> You are mistakenly applying >>>>> the standard cardinalistic fact that the number or reals in [0,1] is the >>>>> same as the number of all reals. That is false in my system. > > What "system" is that? Bigulosity.
From: Tony Orlow on 15 Sep 2006 21:00 Virgil wrote: > In article <49edf$450aacfc$82a1e228$14539(a)news2.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >> Mike Kelly wrote: >> >>> Han de Bruijn wrote: >>> >>> Plagiarism? I don't get it. Who is plagiarising what? >> "Your" would-be arguments against mine are not really yours. They are >> just a _plagiary_ of well-known "arguments" employed by the mainstream >> mathematics community. >> >> Han de Bruijn > > What is common knowledge can be used by anyone without plagiarizing. > > Otherwise only its original author could use "2 + 2 = 4". Yes, and Virgil would be collecting the royalties from every first-grade class.
From: Tony Orlow on 15 Sep 2006 21:02 Virgil wrote: > In article <450b06ce(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> 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. >> You dingbat. I never said that the subset relation was the only method >> of ordering by any means. > > But anything that includes it must fail. You mean anything that doesn't contradict that obvious fact for finite sets would have to fail for infinite sets? That's a baseless statement. Why do you think such nonsense?
From: Mike Kelly on 15 Sep 2006 22:27
Tony Orlow wrote: > Mike Kelly wrote: > > Tony Orlow wrote: > >> Mike Kelly wrote: > >>> Tony Orlow wrote: > >>>> Mike Kelly wrote: > >>>>>>> 1) A finite string of 1s represents a (finite) natural number. > >>>>>>> 2) An infinite string of 1s represents a (finite) natural number. > >>>>>>> > >>>>>>> 1) doesn't imply 2). > >>>>>>> > >>>>>> If the string is unbounded but finite, then 2) follows. > >>>>> What's a finite but unbounded infinite string? > >> I missed that you slipped "infinite" in here. Unbounded but finite may > >> be considered potentially, but not actually, infinite. > > > > By whom? Nobody knows what you mean by those terms. > > > >>>> One with all finite bit positions but no greatest. > >>> So... why does 2) follow from 1? > >> Because then the string is considered finite. It's not finite AND infinite. > > > > It's not finite if it's unbounded, except for very stupid and > > unituitive meanings of "finite". And wasn't the goal of all this to > > satisfy intuition? > > > >>> 1) A finite string of 1s represents a finite natural number. For > >>> example, 101 represents > >>> > >>> 1* (2^0) + 0* (2^1) + 1*(2^2) > >>> = 1 + 0 + 4 > >>> = 5 > >>> > >>> this representation bijects finite bit strings of 1s and 0s and natural > >>> number. > >>> > >>> 2) doesn't work however... Take ...11111. Then the natural this > >>> represents would be > >>> > >>> 1 + 2 + 4 + 8 + 16 + 32 + .... > >>> > >>> but there is no such natural. This sum is divergent. > >>> > >> The sum diverges to an infinite value as n APPROACHES oo. It is finite > >> for all finite n. If all n are finite, it's finite. It's not infinite > >> until n is infinite, since 2^n is finite for all finite n. > >> > >> Is n ever actually infinite? You sum diverges ****in the limit****. > > > > Well, I'd more properly say that it diverges at every step. No matter. > > The sum doesn't sum to a finite value? Right? > > That depends whether it ever gets to infinite n. The sum is finite for > all finite n. But it doesn't SUM to a finite value? Do you even understand what I mean by that? The sum of a series (the limit of the sequence of partial sums) is a fixed value. It either exists or it doesn't. Does it make sense to you that the series 1/2 + 1/4 + 1/8 + 1/16 + .... sums to 1? And that the series 1 + 2 + 3 + 4 + 5 +6 ... doesn't sum to anything? Are these ideas familiar to you from high school, perhaps? > > You seem to have completely missed my point about a dozen times now. > > Either I'm not explaining things well or you're unintentionally(?) > > obtuse. I have never claimed that the sum > > > > 1 + 2 + 4 + 8 + 16 + 32 + .... > > > > "gets to" an "infinite value". I don't even know what that would mean, > > because I don't know what an "infinite number" is. > > Apparently, then, you agree with Wolfgang that there is no infinite > number. Is that correct? I agree that there are no infinite natural numbers. Indeed, I don't even know what it would mean for a NATURAL number to be infinite. I don't know the definition of "number" but I can think of some examples of types of number that may be infinite (cardinal, ordinal, surreal). >Or, are you just regurgitating what you've been > told and managed to remember? Oh please. Your debate tactic of spreading effluent is offensive and not even efficacious. I'm not saying "Tony, you're wrong because you don't have a mathematical education" so please cut the arguments against the man. Your attempt to paint the mathematically educated as brainwashed parrots is rather amusing. Mathematics is a discipline that absolutely requires a highly critical mind. > > I've claimed that the sum doesn't sum to a finite value. So it doesn't > > "represent" a finite value. Is this really so hard to follow? > > Your claims are not substantiated, especially when you also claim > "I have never claimed that the sum > > > > 1 + 2 + 4 + 8 + 16 + 32 + .... > > > > "gets to" an "infinite value". " > > > > > NB - If you argue that "it doesn't ever get to an infinite value" again > > I am going to have a hard time continuing to assume good faith on your > > part, given how many times I have specifically said that I am not > > claiming that. > > > > Then on what grounds do you claim that the string represents an infinite > value? Either it does or it doesn't. It's hard to tell what you're > saying when you contradict yourself. Oh for the love of... Read this carefully, Tony : I do not think that 1 + 2 + 4 + 8 + 16 + 32 ... sums to an infinite value. I think it diverges i.e. it has no value that it sums to. I do not think that when interpeted as a binary natural "111111..." represents an infinite value (there are no infinite naturals). I do not think it represents a finite value (what finite natural could it possibly be?). Therefore, I think it doesn't represent a value at all. This is because the sum it represents is divergent. i.e. it does not sum to any value. Capisci? -- mike. |