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From: Randy Poe on 27 Oct 2005 12:58 Tony Orlow wrote: > David R Tribble said: > > We are forced to conclude that there is no natural s that maps to > > *N, and that therefore your mapping scheme is not a bijection > > between *N and P(*N). > You are not forced to conclude that this prevents a bijection. Um, yes you are. If a bijection requires that there exist such an s, and no such s exists, you are forced to conclude that you do not have such a bijection. > The entire > infinite set would be the final subset enumerated, Your usual red herring. But it's nonsense. This can clearly be seen since your scheme CAN be used to establish a bijection between *N and P(N), and no recourse to "last subsets enumerated" is needed. It is provable that EVERY subset of N is mapped, just as it is provable that SOME subset of *N is not mapped, ever, by anything, in this mapping. Actually, in a sane world, the fact that your mapping produces a bijection from *N to P(N), using up all the elements of *N, would be enough to convince you that the enumeration doesn't go on to map the rest of P(*N). But no, somehow you can believe both that your list is bijection from *N to P(N), and the VERY SAME LIST magically stretches when you want it to, to become a bijection from *N to P(*N). - Randy
From: Robert J. Kolker on 27 Oct 2005 13:08 Tony Orlow wrote: > > LOL that's a good one. You declare aleph_0 to be some smallest infinity, > applying this "value" to a set with no size so as to measure it. Please! Don't > make me wet my pants laughing. I'm the only one....heh! That was good. Once again you conflate count with measure. You are a total mathematical incompetent. Bob Kolker
From: Robert J. Kolker on 27 Oct 2005 13:09 Tony Orlow wrote: >>betters indicates strongly that you will never learn to be one. > > You have no clue what I'm doing here, do you Bob? Yes I do. You are a troll. > > I know what they are, but they don't make sense. They don't make sense to YOU. But what does? Bob Kolker
From: William Hughes on 27 Oct 2005 13:30 Tony Orlow wrote: > William Hughes said: > > > > Tony Orlow wrote: > > > William Hughes said: > > > > > > > > Tony Orlow wrote: > > > > > David R Tribble said: > > > > > > Tony Orlow wrote: > > > > > > >> So, which element of the power set does not have a natural mapped to it? > > > > > > > > > > > > > > > > > > > Virgil said: > > > > > > >> {x in S:x not in f(x)} > > > > > > > > > > > > > > > > > > > Tony Orlow wrote: > > > > > > > Oh yeah, the entire set, last element and all. What element was that again? > > > > > > > > > > > > Your mapping does not include any natural that maps to the entire set > > > > > > *N, which it must do in order to be called a bijection, since *N is one > > > > > > of the members of P(*N). Show us that one single mapping, please. > > > > > It would obviously be the natural with an infinite unending strings of 1's, > > > > > right? Ah, but you want to know, how MANY 1's? > > > > > > > > No, this is irrelevent. No natural has a binary representation that > > > > consists only of 1's. > > > I assume you mean in the infinite string, otherwise 1, 11, 111, etc would fit > > > the bill. Why do you say I cannot declare that the 1's go on indefinitely? We > > > are talking about *N, the set including infinite naturals, with infinite > > > numbers of bits. When will I run out of bits for my subset elements? Remember, > > > there is no last element, as you remind me again, below. > > > > > > You do not "run out of bits". > > The reason that no natural number can > > have a binary representation composed only of ones > > is that every natural has a successor which also > > has a binary representation. > > > > Suppose that a natural n > > had a binary representation that is composed only of ones. > > What could the representation of n+1 be? It cannot consist > > only of ones (this is the binary representation of n). It cannot > > have a 0, as a binary representation containing a 0 is the binary > > representation of a number smaller than the number with a binary > > representation consisting only of ones. Thus n+1 cannot have a binary > > representation. Contradiction. > Well, golly gee. A contradiction. And all you asked me for was a natural with a > 1 bit for every natural in the set. What did you expect? Of course there are > problems with the answer. There are problems with the question. That's why you > get a contradiction. You want the complete mapping to every element of an > endless set. I want a flying pink unicorn that pees single-malt scotch. Oh > well. Guess we're both SOL. > > I am a bit confused as to exactly what you are claiming. Do you agree with the statment? Given any natural number n, the binary representation of n does not consist only of ones. Do you agree with the statement? If we specify a natural number n, the binary representation of n does not consist only of ones. Do you agree with the statement? For all natural numbers n, the binary the binary representation of n does not consist only of ones. Do you agree with the statement? The natural number n which maps to the entire set of natural numbers cannot be specified. Do you agree with the statement? The natural number n which maps to the entire set of natural numbers does not exist. -William Hughes
From: Virgil on 27 Oct 2005 15:02
In article <1130425879.936863.187760(a)g43g2000cwa.googlegroups.com>, albstorz(a)gmx.de wrote: > > > You misinterpret totally when you say, I think there must be an > > > infinite natural number. I don't think so. I only argue that, if there > > > are infinite sets, there must be infinite natural numbers (since nat. > > > numbers are sets). Does albstorz(a)gmx.de require the set of finite natural numbers to contain one or more of his alledgedly infinite natural numbers before the set of finite natural numbers can be Dedekind infinite (have an injection to a proper subset)? |