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From: albstorz on 24 Oct 2005 11:24 David Kastrup wrote: > albstorz(a)gmx.de writes: > > > You are not able to respond to my concept. You prefer to correct all > > over again the same mistakes (if you are shure to accord with the > > majority in this aspect). > > They don't go away by ignoring them. > > -- > David Kastrup, Kriemhildstr. 15, 44793 Bochum You are right. But you are not able to give respond to my argumentation. You correct mistakes which have no connection to my argumentation. You just ignore it. You speak with yourself. Why do you post in this thread if you are not willing to debate my argumentation? This is not only adressed to you, David, but also to all the other participants of this thread. Regards AS
From: David R Tribble on 24 Oct 2005 15:04 Albrecht S. Storz wrote: > But you are not able to give respond to my argumentation. You correct > mistakes which have no connection to my argumentation. You just ignore > it. You speak with yourself. Why do you post in this thread if you are > not willing to debate my argumentation? > > This is not only adressed to you, David, but also to all the other > participants of this thread. Okay, I'll respond to your original post, one more time. > Let's start with a representation of the natural numbers in unitary > (1-adic) system as follows: > > O O O O O O O O O ... > O O O O O O O O ... > O O O O O O O ... > O O O O O O ... > O O O O O ... > O O O O ... > O O O ... > O O ... > O ... > . > . > . > > 1 2 3 4 5 6 7 8 9 ... > > Each vertical row shows a natural number. Horizontally, it is the > sequence of the natural numbers. Since the Os, the elements, are local > distinguished from each other, we can also look at the rows as sets. A > set may contain the coordinates of the elements as their representation > or may look like this: S3 = {O1, O2, O3} e.g. . > > From the Peano axiomes follows that the set of all naturals is > infinite. So, the set of the elements of the first horizontal row is > infinite. Actually the set of the elements of every horizontal row is > infinite. And the set of all the elements in this representation is > also infinite. > > But there is no vertical row with infinite many elements since there is > no infinite natural. That is correct. The set of natural numbers (N) derived from the Peano axioms is an infinite set containing only finite naturals. > Now let's fill the horizontal rows or lines with other symbols. We have > to take into account that only lines should be filled with #s which > containes Os. > > # O O O O O O O O O ... 1 > # # O O O O O O O O ... 2 > # # # O O O O O O O ... 3 > # # # # O O O O O O ... 4 > # # # # # O O O O O ... 5 > # # # # # # O O O O ... 6 > # # # # # # # O O O ... 7 > # # # # # # # # O O ... 8 > # # # # # # # # # O ... 9 > . . . . . . . . . . . > . . . . . . . . . . . > . . . . . . . . . . . > > The vertical sequence of sets of #s fullfill the peano axiomes exactly > as the horizontal sequence of the sets of Os does. > But there is a slight difference. Since there is no infinite natural in > form of a set of Os and since after every set of #s there should be a > O, the size of the set of the naturals as sets of #s could not extend > the "biggest" number of the naturals in form of sets of Os. > Since there is no biggest number and since there is no infinite number, > the size of the set of numbers in form of sets of #s is undefined as > the biggest natural number is undefined. > > But the sequence of the sets of # fullfill the peano axiomes. So this > set must be infinite. > > The cardinality of a set is not able to be infinite and "not defined" > at the same time. > > This is the contradiction. There is no contradiction. The set contains only finite elements, and the set is infinite (i.e., contains an infinite number of elements). The size (cardinality) of the set is infinite, but this set measure is not a natural number, and is not a member of the set itself. Infinite cardinal numbers are not natural numbers, so they are not members of sets of naturals. The "largest natural number" is undefined because it does not exist; there is no such number. On the other hand, the size of the set is well-defined as the smallest infinite set size (Aleph_0). But the size is not a natural number. It is reasonable to say that the number of elements in the set is greater than any member in the set. Therefore, the size cannot be a member of the set, and it cannot be a natural number. > Or let's say it in another form: The first vertical row of #s could not > exceed the biggest vertical row of Os (and could not be smaller). So, > the cardinality of this set is undefined like the biggest natural > number. But the set of the elements of the first vertical row of #s has > the same cardinality like the set of the natural numbers. > --> Contradiction. > > Or did I construct a monster set which cardinality is subtransfinite? Neither. If you consider the second set to be a set of subsets (where each subset is represented by a row of #s and Os), the argument is the same - the set is infinite, but no member of the set is infinite. Your contradiction is wrong. The contradiction comes from saying that the size of the set is also a member of the set. There is not much else to say about it.
From: David R Tribble on 24 Oct 2005 15:10 David R Tribble wrote: >> I've got a set S = {0, 2^0, 2^2^0, 2^2^2^0, ...}, which contains >> all the powers of 2 of the form 2^p, where p=0 or 2^q. >> 1) If it is not an infinite set, tell me how many members it has. >> 2) If it is an infinite set, tell me what the smallest (first) >> infinite number is a member of it. > Albrecht Storz wrote: > A short view upon this makes me think that you are writing sensless > symbols. S don't contain all numbers of the form 2^2^q if you think > about the sequence 0, 2^0, 2^2^0, 2^2^2^0,... . If this is your > intention you may have a infinite sequence 0, 1, 1, 1, ..., and a set > {0,1}, so > 1) 2 > 2) ? Obviously, you do not understand. My set is: S = {0, 2^0, 2^2^0, 2^2^2^0, 2^2^2^2^0, ...} S = {0, 1, 2, 4, 16, 65536, ...} All of the members of S are 0 or finite powers of 2. 1) If S is not an infinite set, tell me how many members it has (or what the largest member is). 2) If S is an infinite set, tell me what the smallest infinite number is a member of S. I know how many members are in S. Do you?
From: Tony Orlow on 24 Oct 2005 16:30 Virgil said: > In article <MPG.1dc1c0b757ad81cd98a518(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > > For what set is TO assuming that each member of its power set can be > > > represented by a single bit in an infinite sequence of bits? > > I never said that. > A member of the power set is a subset of the entire set. You are saying that I am saying that each of these subsets can be represented by a single bit in an infinite sequence? If you take the power set of the power set, then you have sets of subsets, each of which can be assigned a bit corresponding to its order in the enumeration of the power set. But, that is not what you seem to be saying here now. > If each subset of *N is to be represented by an infinite binary sequence > of digits with 1 in some position representing the presence of a member > *N and 0 representing its absence, then one element sets must be > represented by strings with one 1 in them. This makes sense, but doesn't jibe with what you said before, as far as I can tell. Yes, each singleton set in P(*N) will map to a natural whose binary representation has a single bit. This only includes N out of 2^N subsets. > > So that, while he may not have been aware of it, TO was saying precisely > that. That TO is often unaware of the menning of what he is saying has > long been apparent. Precisely what? You are being particularly opaque, even for Virgil. Is this Virgil's little sister on his computer again? > > > > > > > If that set is the set of all Dedekind infinite binary strings, which is > > > uncountably infinite, or any set bijectable with it, then TO's > > > assumption is false. Unless we are operating in TOmatics where an > > > "infinite string of bits" can contain an uncountable number of bits. > > > What is the limit on the number of bits? One bit for every real number in the > > entire real interval. How's that? > > To do that in a string, TO must not only well-order the reals, but order > them isomorphically to the set of finite naturals. But the finite naturals have finite numbers of bits, so what do you mean? They can be enumerated in linear order, but like *N, require an infinite number of bits. > > > > > > > > > > > > > I got a little confused, > > > > but you still haven't proven anything like > > > > the impossibility of a bijection with the power set. > > > > > > We have to those who are not so permanently confused. > > > > > > > > > > In other cases > > > > bijections are performed without regard to such discrepancies. > > > > > > The 'discrepancy' is that when mapping a set to its power set, there > > > must always be a member of the codomain which is not the image of > > > anything in the domain. > > > > > > When that happens, whatever one does have, it is not a bijection. > > > So, which element of the power set does not have a natural mapped to it? > > {x in S:x not in f(x)} Oh yeah, the entire set, last element and all. What element was that again? > -- Smiles, Tony
From: David R Tribble on 24 Oct 2005 20:03
Virgil said: >> If each subset of *N is to be represented by an infinite binary sequence >> of digits with 1 in some position representing the presence of a member >> *N and 0 representing its absence, then one element sets must be >> represented by strings with one 1 in them. > Tony Orlow wrote: > This makes sense, but doesn't jibe with what you said before, as far as I can > tell. Yes, each singleton set in P(*N) will map to a natural whose binary > representation has a single bit. This only includes N out of 2^N subsets. > Your scheme listed these mappings: f(0) = {} f(1) = {0} f(2) = {1} f(3) = {0,1} f(4) = {2} f(5) = {0,2} f(6) = {1,2} f(7) = {0,1,2} f(8) = {3} ... This list defines only the mappings of finite naturals in *N to the finite subsets in P(*N). Your list does not show any infinite naturals, nor does it show any infinite subsets, nor does it show any subsets containing infinite naturals. Your list shows an infinite number of finite naturals mapping to an equally infinite set of finite subsets. That's easy. But it's not a bijection between *N and P(*N). It's not a bijection between *N and P(N), or even between N and P(N). It is a bijection between N and some of the members of P(N). Obviously that's not good enough. Even if we were to grant you that the list somehow includes infinite naturals in *N mapping to subsets, you would still be missing a lot of subsets in P(*N). So fill in the blanks, and show us those other mappings. Otherwise your "bijection" is incomplete and is not, in fact, a bijection between *N and P(*N). |