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From: David R Tribble on 21 Oct 2005 15:23 William Hughes wrote: >> 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. > Albrecht Storz wrote: >> 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. Albrecht, if every number is a set, which set is 3? What number is the set {a, {b}, {{c}}, {{{d}}}, ...}? If sets are numbers, then adding sets should obey the same rules as adding numbers, right? What is {1,2,3} + {0,2,4}? Is it 3+3 = 6?
From: David R Tribble on 21 Oct 2005 15:23 Albrecht Storz wrote: > You can biject all constructable subsets of P(N) to N, but not the > unconstructable. That's an argument against the diagonal proof, since, > if a set will be constructable in form of an antidiagonal of a list, > it's a constructable set and therefore it is element of the set of the > sets which are bijectable with N. Then that constructable set will be in the original list, and the diagonal set will be different from it, and it can't be a set constructed from the diagonal. But that just proves that you can't construct a set from the diagonal of the list of sets while allowing that set to also be in the list. No such set exists in the list. Therefore, the constructed set must be a set that does not exist in the mapping, and therefore the mapping does not map every possible set in the list.
From: stephen on 21 Oct 2005 15:27 Tony Orlow <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: >> Tony Orlow <aeo6(a)cornell.edu> wrote: >> > stephen(a)nomail.com said: >> >> Tony Orlow <aeo6(a)cornell.edu> wrote: >> >> > 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. >> >> >> >> So which number does that first 1 represent? >> >> > log2(N-1). Perhaps we should say we have 2^N subsets and strings of N digits. >> > That's what I should have said. Then that number is 2^N/3, and the first 1 in >> > it represents N-2, since the bits are numbered from 0 through N-1. >> >> So this is not the set of all even numbers then, because >> according to you there are even numbers larger than log2(N-1). > This is the imagined completed set. The number of digits is variable according > to what infinity you are working in. You have to choose a unit to measure. Do > you have a ruler with no marks on it? Excuse me while I notch my infinite > number line. These numbers are relative. If it does not contain all the even numbers, it is not the set of even numbers. <snip> >> >> So where is the string that represents all the even numbers? > ....101010101. Is that better? oo/3. So where is the string that represents all the even numbers plus ....101010101? ....101010101 is a member of your set after all, and so it must also appear in the elements of the power set. So what is y such that f(y) = { x: x is even or x is ....101010101 }? >> >> Your arguments would be a lot more consistent and sensible >> if you just dropped the whole notion of "infinity". It >> is clear that you do not actually believe in infinite sets. >> For you all sets must have a first and last element, and >> all your arguments require all sets to be finite. > No, all measurements require two endpoints. The endpoint is variable, and can > be finite or infinite. Sets are not variable, and their "endpoints" are not variable. Face it Tony, you do not believe in infinity. Stephen
From: imaginatorium on 21 Oct 2005 15:32 (Yes, Randy is right...) Tony Orlow wrote: <snip> > It assumes a completed set with a completed string and a natural that equals > that string, corresponding to the completed set. For any completion you > consider, than number is beyond it. So, don't assume a completion to the set. Pretty muddled, yet pretty clearly a rejection of the axiom of infinity. So you don't have infinite sets in the sense meant by mathematicians. Back to the middle ages, Tony... Brian Chandler http://imaginatorium.org
From: Virgil on 21 Oct 2005 15:42
In article <MPG.1dc2c70fefb2cc0198a523(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dc078e83d0081bc98a4f6(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > I don't understand why an inductive proof of an equality relationship > > > does not extend to infinite numbers. > > > > Informally, the Peano axioms may be stated as follows: > > 1 There is a natural number 0. > > 1 Every natural number a has a successor, denoted by S(a). > > 3 There is no natural number whose successor is 0. > > 4 Distinct natural numbers have distinct successors: > > if not(a = b), then not(S(a) <> S(b)). > > 5 If a property is possessed by 0 and also by the successor of > > every natural number which possesses it, then it is possessed > > by all natural numbers. (This axiom ensures that the proof > > technique of mathematical induction is valid.) > > Says nothing about finiteness there...... > > > > > More formally, we define a Dedekind-Peano structure to be an ordered > > triple (X, x, f), satisfying the following properties: > > 1 X is a set, x is an element of X, and f is a map from X to itself. > > 2 x is not in the range of f. > > 3 f is injective. > > 4 If A is a subset of X satisfying: > > 5 If [x is in A, and > > (If a is in A, then f(a) is in A)] > > then A = X. > > Nope, still no finiteness...... But when one constrins one self to finiteness, one loses absolutely nothing of what Peano provides, so no infinitenes is needed, or possible. > > > > > A member of the set of such objects so defined is defined as finite if > > the set of such objects of which it is ultimately a successor is a > > finite set by the Dedekind definition of finite sets. > So, all finite numbers are successors to finite sets. That is not what was said, TO. Learn to read. > Then, which finite is a > successor to any infinite portion of the infinite set of finite naturals? > None? > Hmmmm..... That doesn't seem to work very well. It works absolutly wonderfully everywhere outside TOmatics. And what goes on inside TOmatics is irrelevant to that. > > > > With that definition of finiteness of such objects, all such objects are > > provably finite. > Only the ones that are successors to a finite set. The ones that are > successors > to infinite-size sets are infinite in value. The way things are defined, sets need not have successors, only (finite) naturals need have them. > So, you assume there are only finite naturals NO! We define things so that all Peano naturals correspond to Dedekind finite sets, and are thus definably Dedekind finite. |