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From: Virgil on 20 Oct 2005 23:30 In article <MPG.1dc1c76cc8869ca998a51f(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > William Hughes said: > > Because we are discussing standard mathematics here and there > > are no infinite natural numbers (only infinite ordinals and cardinals > > which are not the same thing). Yes it is true that the semi-mythical > > TO naturals have infinite members and that induction sometimes > > holds, but that is irrelevent here. > No it's not at all. The bijection I offered is between *N, which has infinite > values as members, and its power set. Except that one must assume that result to obtain it. To starts with the set of all endless binary strings and wants to represents sets of them by similar strings. thus one member sets would be strings with one 1 and the rest all zeros. If this could really be done, one could get by with representing all such infinite strings using only finite strings, the one element set strings, with all the zeros beyond the one 1 omitted. Thus if TO could do what he claims, he would only need a set of finite strings each with a single 1 to represent all his infinite strings.
From: Virgil on 20 Oct 2005 23:38 In article <MPG.1dc1cb9ec77d1dcc98a520(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > stevendaryl3016(a)yahoo.com said: > > >> albstorz(a)gmx.de says... > > >> > > >> >I don't know what you are talking about. The proof for finite sets > > >> >needs just a complete induction. This will not hold for infinity I > > >> >think. > > >> > > >> There is no induction involved in the finite case, and the > > >> infinite case is *exactly* the same proof as the finite case. > > >> > > >> Here it is once again: > > >> > > >> Let A be any set whatsoever, finite or infinite, it doesn't matter. > > >> Let f be any function from A to P(A). > > >> Let w = { x in A | x is not an element of f(x) }. > > >> Let x = any set in A. > > >> Let u = f(x). We prove that u is not equal to w. > > >> > > >> By definition of w, we have x in w <-> x is not an element of f(x). > > >> So x in w <-> x is not an element of u. That means that there are > > >> two cases: Case 1: x in w, and x is not in u. In that case, u cannot > > >> equal w. Case 2: x is not in w, and x is in u. In that case, u cannot > > >> equal w. > > >> > > >> So what we have proved is that forall x, w is not equal to f(x). So > > >> w is not in the image of f. So f is not a bijection between A and P(A). > > >> > > >> There's no induction. There's no assumption that A is finite. > > > But there is an assumption that y is in S. > > > > Of course there is an assumption that y is in S. The > > goal is to find a bijection f from S to P(S). That > > means that for every element x in P(S), there must be > > an element y in S such that f(y)=x. w is an element of P(S). > > In order for f to be a bijection, there must exist an > > element y in S such that f(y)=w. > > > > If y is not in S, then it is irrelevant to the question > > of whether or not f is a bijection from S to P(S). > > > > Do you consider the following a bijection from S={a,b,c} to its > > power set? > > > > f(a) = {} > > f(b) = {a} > > f(c) = {b} > > f(d) = {c} > > f(e) = {a,b} > > f(g) = {a,c} > > f(h) = {b,c} > > f(i) = {a,b,c} > > > > If w= { x : x in S and x not in f(x) } > > then w= {a,b,c}. > > > > Is there a y in S such that f(y) = {a,b,c}? No. > > > > Is there a y such that f(y) = {a,b,c}? Yes, f(i)={a,b,c}, but > > i is not in S, and is not part of a bijection from S to P(S). > > > > The above function is a bijection from {a,b,c,d,e,g,h,i} > > to P({a,b,c}). It is not a bijection from {a,b,c} to P({a,b,c}). > > > > Stephen > > > But, if this set went on forever, then i WOULD be in the set, and for any > subset, you could identify an x such that it mapped to that set. It is true > that no bijection is possible in the finite case. Then by TO-induction, if it is true for all the finite cases, requires that it must be true for the infinite cases also. > So tell me, what rule of bijection did I break, and at what point does the > mapping through the infinite binary strings break down. For lack of answers > to > thses questions, my bijection stands. That lack of answers only occurs in TOmatics. Outside TOmatics, there are a surplus of answers which show no such bijection can exist.
From: Ross A. Finlayson on 21 Oct 2005 00:33 sci.math_20040413: About the leading zeros, maybe it's one way to illustrate that the antidiagonal process as applied to an infinite list that is supposed to be a list of all real numbers is a contradiction in terms, because each element of the set of reals is distinct from each other, because a set is not a multiset. Here we are seeing that an antidiagonal different in representation than each element of an infinite list is not an element of that list, nor of the range of the function. sci.math_20040415: It's kind of interesting that all rational numbers have at least dual representation as simple continued fractions that are finite. As well, it appears that an element of a continued fraction might be an arbitrarily large number. Is it not so that a rational number can be represented with an infinite continued fraction, in a similar way to how a root of a power series might be an integer? sci.math_20040415: Consider other ways to represent each of the set of real numbers as a sequence, besides trivially representing each real number as the sequence (0), representations that give each real number at least one unique identifying sequence. Congratulations. Most of the sequences are infinite and most of the arguments about binary integer modulus representation apply. sci.math_20040415: I model only binary numbers where there is exactly and only one antidiagonal of the matrix of elements thus constructed, all elements of the unit interval of reals as infinite bitstrings with a beginning, and dual representation allows the existence of an antidiagonal, another extension. sci.math_20040415: The Equivalency Function is defined in a way thus that all values of EF(n) except for n=0 are indefinite, and greater than zero, and less than or equal to one. This is where the set of integers is an infinite set. sci.math_20040415: I might suggest a completely different tack on the antidiagonal: not that it shows that there is no mapping between a set and its powerset, but that infinity is dually represented as zero. sci.math_20040417: I think more about the antidiagonal of an infinite list of particularly all possible sequences of binary elements. sci.math_20040417: Thus, either a does not exist or a is dually represented. Why would that not be so? Alternatively, as the reals are uncountable, you might say, or infinite, x_n, is never the last element of the reals to affect the antidiagonal, yet x_n is always itself. sci.math_20040418: I consider the Equivalency Function, and its list of values, and what its binary antidiagonal would be. It would start with infinitely many 1's, then, at some point there might be some values that are not 1's, for around EF^-1(1/2), then, I'm not quite sure what happens. An alternative consideration is that the antidiagonal would be all ones, then perhaps some ones and zeros, and eventually all zeros. Another consideration is that it is ill-defined as all the elements are indefinite, or that the antidiagonal is .111.... sci.math_20040418: So say you want to add that to the list and generate a new antidiagonal. The list elements are sorted in ascending order, so you have to put any list element you consider in the correct location. If it's .111..., that's greater than or equal to any element of the range, and the infinite list has no end, so there's no place to put it.
From: albstorz on 21 Oct 2005 04:44 William Hughes wrote: > albstorz(a)gmx.de wrote: > > William Hughes wrote: > > > > > > > > > Coincidently natural numbers and cardinalities are undistinguishable in > > > > finity. > > > > > > They are very similar, but they are not quite "undistinguishable". > > > A natural number is a set, a cardinality is an equivalence class. > > > > You make me hopefull. Some experts make "äääh", "hömm" and > > "üüüh" if I said "A natural number is a set." One sees a > > correspondence between natural numbers and von Neumann sets after all. > > You are free to say "A natural number is a set." without "äääh-", > > "hömm-" and "üüüh-" comments. Be lucky. You are right. > > And natural numbers don't behave in any other way than sets. So, if > > there is an infinite set there is an infinite number. If there is no > > infinite number there is no infinite set. And vic versa. > > > > You have made exactly this mistake before. What mistake? > Yes every number > is a set. No, not every set is a number. For example > {peach, apple, plum, fiddle} is a set but not a number. Wrong. It's a number. The number is this aspect of the set which it have in common with the set {diddle, daddle, doddle, duddle}. To say, a set is a number or a set has a number is a slight difference which has no effect in this considerations. The important aspect is, that every set has a number of elements, since a set, totally independent of the kind of the members of the set, behaves like a set of numbers with same cardinality in all math aspects. But it's sysiphosiousely done. > Just because you have a set does not mean you have a number. > So yes, there is an infinite set. But this does not mean > that this set is a number. Indeed, no infinite set is a number. > > - William Hughes > > > P.S Actually it is not true that natural numbers must be sets, but > they can be. As you insist on using a model in which the natural > numbers are sets, I am playing along to be polite. Oh, how polite. Go play with sand. You are not able to grasp the overview to see the inconsistency of math. You hang on your indoctrination. You are unable to think independently. Regards AS
From: albstorz on 21 Oct 2005 04:57
David R Tribble wrote: > William Hughes wrote: > >> A natural number is a set, a cardinality is an equivalence class. > > > > Albrecht Storz wrote: > >> You make me hopefull. One sees a > >> correspondence between natural numbers and von Neumann sets after all. > >> And natural numbers don't behave in any other way than sets. So, if > >> there is an infinite set there is an infinite number. If there is no > >> infinite number there is no infinite set. And vic versa. > > > > William Hughes wrote: > > You have made exactly this mistake before. Yes every number is a set. > > I agree that we can assign a correspondence between natural numbers > and certain sets. But which sets? > > von Neumann sets are only one such kind of set, where each one has > a successor: > 0 = {} = { } > 1 = {0} = { {} } > 2 = {0,1} = { {}, {{}} } > 3 = {0,1,2} = { {}, {{}, {{}} } } > etc. > > But there are other, equally valid sets that can be associated with > natural numbers. E.g.: > 0 = {} > 1 = {a} u {{0}} = { a, {{{}}} } > 2 = {a} u {{1}} = { a, {{{a, {{{}}} }}} } > 3 = {a} u {{2}} = { a, {{{a, {{{a, {{{}}} }}} }}} } > etc. > > This is just one example of the many kinds of sets that can be > associated with natural numbers. > > http://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory > > So when Albrecht says that natural numbers are sets, I ask the > question, which sets? Any kind of set in particular? Yes. The natural numbers are sets which build a kind of stamp form or placeholder of every set with the same cardinality. The natural numbers are the equivalent classes of the sets with the same number of elements. Natural numbers counts sets. What else? Regards AS |