An uncountable countable set
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From: Randy Poe on 15 Sep 2006 16:54 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. > > Do you honestly think bijection works as an infinite analog to equality? What does "as many as" if not "there's one of these for every one of those"? That's a long-winded way of saying: "yes". Besides, it follows by T-induction. The set of the first n naturals is the same size as the set of the first n primes for every n. Hence it's true for the whole sets. > 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? Yes. There's one of these for every one of those. > That's just poppycock. Well, but that's a statement based on emotion rather than reason. Reason says "there's one of these for every one of those". - Randy
From: MoeBlee on 15 Sep 2006 17:00 Tony Orlow wrote: > 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. A set is not dense onto itself. A set is dense under an ordering. And the set of natural numbers is dense under certain orderings. MoeBlee
From: Mike Kelly on 15 Sep 2006 18:01 Tony Orlow wrote: > Mike Kelly wrote: > > Han de Bruijn wrote: > >> Han de Bruijn wrote: > >>> Mike Kelly wrote: > >>>> Sorry for jumping in so late. But VM is quite right, of course. We have > >>>> encoutered utterly absurd consequences of thinking otherwise, like the > >>>> mainstream "theorem" that the probability of a natural being a multiple > >> of 3 doesn't exist. While the obvious truth is that it is equal to 1/3 . > >>>> This topic has been discussed at length in a thread called "Calculus XOR > >>>> Probability". Let Google be your friend, eventually. > > > > Please don't snip this necessary context. > > > >>> So you still don't know what "probability" means. > >> On the contrary. Very much better than you. > > > > Interesting. What do you base this claim on? Unabashed and unjustified > > egotism? > > I think Han brought up a good point in Calculus XOR Probability. I think Han demonstrated some gross ignorance of basic probability theory (along with limits, infinity, the difference between the physical sciences and math etc. etc.) and drew a very stupid analogy with calculus. "My poor understanding of informal descriptions of calculus and my poor understanding of informal descriptions of probability do not jive well together. We must unite against the Cantorian fascists for decieving us so!". > > Perhaps to demonstrate your firm grasp of these matters you could > > define "probability" and then explain how one determines the > > "probability" that "a natural" has some property P? > > Does the set of naturals with property P have a mapping function from > the naturals? :) Explain what it means for a property P to have a mapping function from the naturals xD Pretend it does. Then explain how one determines the probability that a natural has property P. Go wild. For bonus credit explain what happens when property P doesn't have a mapping function from the naturals =D! Here are a few questions for you to practice your new theory on : What is the probabilty that a number is n? What is the probability that a number is even? What is the probability that a number is greater than n? What is the probability that a number is greater than another number (also randomly chosen)? What is the probability that a number is a perfect square? What is the probability that a number is prime? What is the probability that the gcd((n^17)+9, ((n+1)17)+9) is not equal to 1 for "a number" n? Don't forget I asked for a definition of "probability", too. I'm dreadfully ignorant on these matters o_O Oh and please stop with the emoticons at least when replying to me. They do not belong in prose. Adding a smiley emoticon to the end of a paragraph adds no meaning and is intensely irritating. :| > >>> How predictable. > >> Same to you. No? > > > > Sure, posting rubbish about a subject one knows too little about is > > liable to get one called out by someone or other. > > > > Yes, you might want to watch that, Mike. Interesting. Both you and Han have implied that I don't know anything about probability. Based on what? That I think your ideas about mathematics are very, very stupid and will never amount to anything of any value whatsoever? Cute. -- mike.
From: Mike Kelly on 15 Sep 2006 18:10 Tony Orlow wrote: > Mike Kelly wrote: > > Tony Orlow wrote: > >> Mike Kelly wrote: > >>> Tony Orlow wrote: > >>>> Mike Kelly wrote: > >>>>> Tony Orlow wrote: > >>>>>> Mike Kelly wrote: > >>>>>>> Tony Orlow wrote: > >>>>>>>> Virgil wrote: > >>>>>>>>> In article <44fe2642(a)news2.lightlink.com>, > >>>>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>>>>>> > >>>>>>>>>> Virgil wrote: > >>>>>>>>>>> In article <44fd9eba(a)news2.lightlink.com>, > >>>>>>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>>>>>>>> > >>>>>>>>>>>> Dik T. Winter wrote: > >>>>>>>>>>>>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > >>>>>>>>>>>>> writes: > >>>>>>>>>>>>> Your axiom uses things that are not defined. What is the *meaning* of > >>>>>>>>>>>>> "x<z"? > >>>>>>>>>>>> Geometrically it means that x is left of z on the number line. > >>>>>>>>>>> And for someone standing on the other side of the number line would x be > >>>>>>>>>>> on the right of z? > >>>>>>>>>>> > >>>>>>>>>>> And does the line stay horizontal as one moves around earth? Which way > >>>>>>>>>>> is larger if the line ever goes vertical. And how does the "larger" work > >>>>>>>>>>> at antipodes? > >>>>>>>>>>> > >>>>>>>>>> Silly questions. > >>>>>>>>> In response to a silly definition. > >>>>>>>>>>>> It means > >>>>>>>>>>>> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it > >>>>>>>>>>>> needs to, wouldn't you say? > >>>>>>>>>>> Not hardly. > >>>>>>>>>>> A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) > >>>>>>>>>>> is a bit better but still insufficient. > >>>>>>>>>> True, I should have specified y<>x and y<>z. I guess it's usually done > >>>>>>>>>> using <= for this reason, eh? > >>>>>>>>>> > >>>>>>>>>>>>> > > That is not a definition, because it makes no sense. "The set of > >>>>>>>>>>>>> > > naturals > >>>>>>>>>>>>> > > is as large as every natural"? > >>>>>>>>>>>>> > > >>>>>>>>>>>>> > It is not larger than all naturals > >>>>>>>>>>>>> > >>>>>>>>>>>>> That is something completely different again. > >>>>>>>>>>>> It's not LARGER than every finite. > >>>>>>>>>>> Which natural(s) is it "not larger" than", in the sense of not being a > >>>>>>>>>>> proper superset of that natural or having that natural as a member? > >>>>>>>>>> ....11111 binary (all bit positions finite) > >>>>>>>>> Unless that string has only finitely many bit positions as well as only > >>>>>>>>> finite bit positions, it is not a natural at all, as it is then neither > >>>>>>>>> the first natural nor the successor of any natural, and every natural > >>>>>>>>> has to be one or the other. > >>>>>>>> It is the successor to ....11110. Duh. I've already proven that this is > >>>>>>>> a finite value, given that all bit positions are finite, and that > >>>>>>>> therefore no place in that string can achieve an infinite value, and > >>>>>>>> that any such number has predecessor and successor. The cute thing is > >>>>>>>> that the successor to ...1111 is 0, and that ...1111 is essentially -1. :) > >>>>>>> Does it not bother you that nobody else agrees with, or even > >>>>>>> understands, your proof? > >>>>>>> > >>>>>> I find it disappointing, but not surprising, that you don't understand > >>>>>> such a simple proof, since it's contradictory to your education. I do > >>>>>> find it annoying that you feel the right to disagree with it without > >>>>>> understanding it. If you feel there is a problem with the proof, please > >>>>>> state the logical error I made. If the string is all finite bits, and > >>>>>> none of them ever can possibly achieve an infinite value, then how can > >>>>>> the string have an infinite value? There's nowhere in the string where > >>>>>> that can occur. It's that simple. Grok it. > >>>>> 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? 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. 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? 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. -- mike.
From: Mike Kelly on 15 Sep 2006 18:21
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? > 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". 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). 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. -- mike. |