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From: Tony Orlow on 20 Oct 2005 16:00 Virgil said: > In article <MPG.1dc03e89763835a898a4e2(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > William Hughes said: > > > > > > albstorz(a)gmx.de wrote: > > > > David R Tribble wrote: > > > > > Albrecht S. Storz wrote: > > > > > >> Cantor proofs his wrong conclusion with the same mix of > > > > > >> potential infinity and actual infinity. But there is no > > > > > >> bijection between this two concepts. The antidiagonal is an > > > > > >> unicorn. There is no stringend concept about infinity. And > > > > > >> there is no aleph_1, aleph_2, ... or any other infinity. > > > > > > > > > > > > > > > > David R Tribble wrote: > > > > > >> For that to be true, there must be a bijection between an > > > > > >> infinite set (any infinite set) and its powerset. Bitte, > > > > > >> show us a bijection between N and P(N). > > > > > > > > > > > > > > > > Albrecht S. Storz wrote: > > > > > > At first, you should show, that bijection means something to > > > > > > notwellordered infinite sets. > > > > > > > > > > > > Bijection is a clear concept on finite sets, it also works on > > > > > > wellordered infinite sets of the same infinite concept. Aber: > > > > > > Show me a bijection between two infinite sets with the same > > > > > > cardinality, where one of the sets is still not > > > > > > wellorderable. Than I will show you a bijection between N and > > > > > > P(N) or N and R or P(N) and P(P(N)) or what you want. > > > > > > > > > > I see. I'm supposed to show you a proof before you can show me > > > > > your proof. Okay, I give up, you win, so your proof must be > > > > > correct. > > > > > > > > > > Come on, now. It's up to you to prove your own claim, > > > > > especially when it contradicts established mathematics. I know > > > > > you cannot show a bijection between N and P(N). > > > > > > > > > > > > > > > P.S. > > > > > > > > > > Let > > > > > D(n,i) = floor(n/2^i) mod 2, for all i=0,1,2,3,... > > > > > This is the i-th binary digit of natural n > > > > > L(n) = i where D(n,i) = 1 and D(n,j) = 0 for all j > i > > > > > This is the number of binary digits of natural n, or > > > > > ceil(log2(n)) > > > > > Let > > > > > M(n) = sum{i=0 to L(n)} 1/2^(i+1) where D(n,i) = 1 > > > > > This reverses the binary digits of n. > > > > > > > > > > Then M(n) is a mapping N -> N, from all n in N to M(n) in N, > > > > > but the set of all M(n) is not a well-ordered set. Happy? > > > > > > > > > > > > Sad! First of all you don't argue on my claim of the thread. > > > > Second, your above argueing is not clear to me, since both sets > > > > are well-ordered. But it's nice, so I give this: > > > > > > > > N {1},{2},{3}, ... N/{1},N/{2},N/{3}, ... > > > > {1,2},{1,3},{2,3},{1,4},... N/{1,2}, ... ... > > > > > > > > Now count in diagonal sequence. You may think of Cantor's first > > > > diagonal proof. > > > > > > > > Which subset of N is not included? > > > > > > Sigh. Try {2,4,6,8,...} Or any other infinite subset that is not > > > the complement of a finite set. > > > > > > It is common for people to note that the finite subsets of N are > > > countable and incorrectly claim that the subsets of N are > > > countable. Adding in the complements of the finite subsets does > > > not change things very much. > > > > > > Those who do not study anti-cantor cranks are doomed to repeat > > > their idiocies. > > > > > > - William Hughes > > > > > > P.S. No TO, going to TO-infinite rows is not going to help us. > > > The problem is that for any TO-finite row, the subsets listed > > > will have an bounded finite number of elements, or be the > > > complement of a subset with a finite number of elements. Yes, > > > you can claim (without a shred of motivation) that when you > > > get to TO-infinite rows the subsets will suddenly have an > > > unbounded number of elements, but all this tells us is that a > > > bijection from the TO-naturals to P(N) exists. What we need > > > is a bijection from the finite TO-naturals to P(N). > > > > > > > > Look, it's certainly not my position that the pwoer set is the same > > sie as the set. > > Since the argument is not about sizes but about lack of any > surjection/bijection from any set X to its power set P(X), size in TO's > sense, is not an issue, but bjiection is. > > > It's clearly not. I just see a bijection between them > > TO accepts the proof of no bijection then immediately turns around and > claims to see a bijection. No, I accept the result that the power set is larger, but not that a bijection is impossible. You need to focus. Try ginseng. > > This is TOmatics at its best! Which is, as usual, not very good! > -- Smiles, Tony
From: William Hughes on 20 Oct 2005 16:02 Tony Orlow wrote: > William Hughes said: > > > > Tony Orlow wrote: > > > William Hughes said: > > > > > > > > albstorz(a)gmx.de wrote: > > > > > David R Tribble wrote: > > > > > > Albrecht S. Storz wrote: > > > > > > >> Cantor proofs his wrong conclusion with the same mix of potential > > > > > > >> infinity and actual infinity. But there is no bijection between this > > > > > > >> two concepts. The antidiagonal is an unicorn. > > > > > > >> There is no stringend concept about infinity. And there is no aleph_1, > > > > > > >> aleph_2, ... or any other infinity. > > > > > > > > > > > > > > > > > > > David R Tribble wrote: > > > > > > >> For that to be true, there must be a bijection between an infinite > > > > > > >> set (any infinite set) and its powerset. Bitte, show us a bijection > > > > > > >> between N and P(N). > > > > > > > > > > > > > > > > > > > Albrecht S. Storz wrote: > > > > > > > At first, you should show, that bijection means something to > > > > > > > notwellordered infinite sets. > > > > > > > > > > > > > > Bijection is a clear concept on finite sets, it also works on > > > > > > > wellordered infinite sets of the same infinite concept. > > > > > > > Aber: Show me a bijection between two infinite sets with the same > > > > > > > cardinality, where one of the sets is still not wellorderable. > > > > > > > Than I will show you a bijection between N and P(N) or N and R or P(N) > > > > > > > and P(P(N)) or what you want. > > > > > > > > > > > > I see. I'm supposed to show you a proof before you can show me your > > > > > > proof. Okay, I give up, you win, so your proof must be correct. > > > > > > > > > > > > Come on, now. It's up to you to prove your own claim, especially > > > > > > when it contradicts established mathematics. I know you cannot show > > > > > > a bijection between N and P(N). > > > > > > > > > > > > > > > > > > P.S. > > > > > > > > > > > > Let > > > > > > D(n,i) = floor(n/2^i) mod 2, for all i=0,1,2,3,... > > > > > > This is the i-th binary digit of natural n > > > > > > L(n) = i where D(n,i) = 1 and D(n,j) = 0 for all j > i > > > > > > This is the number of binary digits of natural n, or ceil(log2(n)) > > > > > > Let > > > > > > M(n) = sum{i=0 to L(n)} 1/2^(i+1) where D(n,i) = 1 > > > > > > This reverses the binary digits of n. > > > > > > > > > > > > Then M(n) is a mapping N -> N, from all n in N to M(n) in N, > > > > > > but the set of all M(n) is not a well-ordered set. > > > > > > Happy? > > > > > > > > > > > > > > > Sad! > > > > > First of all you don't argue on my claim of the thread. > > > > > Second, your above argueing is not clear to me, since both sets are > > > > > well-ordered. But it's nice, so I give this: > > > > > > > > > > N > > > > > {1},{2},{3}, ... > > > > > N/{1},N/{2},N/{3}, ... > > > > > {1,2},{1,3},{2,3},{1,4},... > > > > > N/{1,2}, ... > > > > > ... > > > > > > > > > > Now count in diagonal sequence. You may think of Cantor's first > > > > > diagonal proof. > > > > > > > > > > Which subset of N is not included? > > > > > > > > Sigh. Try {2,4,6,8,...} Or any other infinite subset > > > > that is not the complement of a finite set. > > > > > > > > It is common for people to note that the finite subsets of N > > > > are countable and incorrectly claim that the subsets of N are > > > > countable. Adding in the complements of the finite subsets > > > > does not change things very much. > > > > > > > > Those who do not study anti-cantor cranks are doomed to > > > > repeat their idiocies. > > > > > > > > - William Hughes > > > > > > > > P.S. No TO, going to TO-infinite rows is not going to help us. > > > > The problem is that for any TO-finite row, the subsets > > > > listed will have an bounded finite number of elements, or be > > > > the complement of a subset with a finite number of elements. > > > > Yes, you can claim (without a shred of motivation) that > > > > when you get to TO-infinite rows the subsets will suddenly > > > > have an unbounded number of elements, but all this tells us > > > > is that a bijection from the TO-naturals to P(N) exists. > > > > What we need is a bijection from the finite TO-naturals > > > > to P(N). > > > > > > > > > > > Look, it's certainly not my position that the pwoer set is the same sie as the > > > set. It's clearly not. I just see a bijection between them, which only bolsters > > > my argument that bijection alone is not sufficient to equate the sizes of two > > > sets. > > > > > > When it comes to the evens (let's start with 0), the value 0:010.......1010101 > > > represents such a subset, and is essentially binary N/3. > > > > Well the binary notation is certainly convenient. Let's lay > > things out in a sort of a square > > > > Take the set X > > > > How many columns do we need. One for every element of X > > > > How many rows do we need. One for every element of X > > > > Pity about the complement of the diagonal. > > > > Lets try X = N*. > > > > How many columns do we need. One for every element of N* > > > > How many rows do we need. One for every element of N* > > > > Pity about the complement of the diagonal. > > > > > > > > - William Hughes > > > > > Ha ha ha. Nice try, William. Let's get real. You want a sqaure? It's not > square. > > We have a set with N elements. Right we will have N rows (each row represents the value of the "enumeration" at a given element and there are N elements). >We will enumerate the elements of the powerset, > represented as binary strings with 1 bit per N elements, or N bits, per line of > the table. Right there are N columns > Now, how many such subsets are there? We have a power set with 2^N > elements, And you are somehow going to fit these 2^N elements into N rows. - William Hughes
From: Tony Orlow on 20 Oct 2005 16:02 Virgil said: > In article <MPG.1dc040cc19fad5ff98a4e4(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Did you have an objection to the bijection between *N and P(*N)? What > > did I do wrong there, bijection-wise? > > You assume your result in order to prove it. > Virgil, you're being a lunkhead. It wasn't a proof, and I didn't assume any conclusion. I described a bijection between *N and P(*N) and asked why it is not a bijection. No one has answered that question yet. Discredit the counterexample with established rules or accept it as a refutation of the proof. You claim to be logical? Obey logic. -- Smiles, Tony
From: Tony Orlow on 20 Oct 2005 16:06 Virgil said: > In article <MPG.1dc0411e718b821798a4e5(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > In article <MPG.1dbf2de38cf6ec898a4da(a)newsstand.cit.cornell.edu>, > > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > > > David R Tribble said: > > > > > > But that's an incomplete mapping, because there are not enough infinite > > > > > binary strings in *N to enumerate all of the subsets of *N. Try it, > > > > > if you don't believe me. > > > > > > > > > > > > > > Not enough infinite binary strings? > > > > > > Precisely. > > > > > So, you need more than an infinite amount? > > You need "more" then you have, in the same sense that there are "more" > reals than rationals even though there is an endless supply of both. > There are also more rationals than naturals, since between any two successive naturals are an infinite number of rationals, but between any two successive rationals is at most one natural. The rationals are dense in the real, having an infinite number in any given unit of value, whereas the naturals are sparse, having precisely one element per unit of value. And yet, despite the obvious differences in density and size, you accept a bijection between them, because you have an endless supply of both. The problem with the reals is that you don't have a natural well-ordering of them. I can give you that, when you're ready, which will probably never be. -- Smiles, Tony
From: Tony Orlow on 20 Oct 2005 16:08
Virgil said: > In article <MPG.1dc042163fd02d9198a4e6(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > In article <1129684276.251366.121150(a)g47g2000cwa.googlegroups.com>, > > > albstorz(a)gmx.de wrote: > > > > > > > David R Tribble wrote: > > > > > > > > > > > > > > Because I can prove it (and it's a very old proof). A powerset of > > > > > a nonempty set contains more elements that the set. Can you prove > > > > > otherwise? > > > > > > > > This argument is stupid. Is there any magic in the powerfunction? > > > > > > "Proofs" are not stupid until they can be refuted. The proof that for an > > > arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. > > > > > Except for the obvious bijection between *N and P(*N). But, hey, what's one > > measly counterexample? > > In this case it is quite "measly", since it specifically requires that > one assume most of the result in order to "prove" the result. Assume what, specifically? > > If *N bijects to the set of all endless sequences of binary digits, what > is the mechanism by which TO maps these to the set of finite naturals so > that he can then represent sets of finite naturals by endles sequences > of binary digits. Without some such countable well ordering of the set > of endless binary sequences, no such representation of subsets is > possible. Is *N not supposed to include infinite values? Isn't that what the * is? Can a binary value be infinite without infinite bit positions? Stop being dumb. > > But any such countable well-ordering would prove that the set of endless > sequences of binary digits os TO-finite. Not I am talking about actual infinitely long bitstrings, such as would represent an actual infinite value in binary numbers. Get a clue. > -- Smiles, Tony |