Obections to Cantor's Theory (Wikipedia article)
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From: MoeBlee on 25 Jul 2005 16:17 >From a post by georgie: > Please provide an effective method by which we may determine > whether a formula is an axiom. Each step in the definition of 'phi is an axiom of T' is given recursively, therefore, there is an effective procedure to determine whether phi is an axiom of T. At the very least, to determine whether a formula is an axiom, one needs only to consult the recursive definition. To see proofs that the defintions are recursive, consult any book on the subject, such as Herbert B. Enderton's [i]A Mathematical Introduction To Logic[/i], or, for example, consult whatever online postings there are of Godel's incompleteness paper. Since it is well known, and not controversial, that formal theories have effective methods for determining whether a formula is an axiom, I don't know the point of your question. Do you have difficulties following recursive definitions to determine whether formulas are axioms?
From: Robert Low on 25 Jul 2005 16:26 Tony Orlow (aeo6) wrote: > Robert Low said: >>The connection is fundamental. Set A is bigger than set B if there >>is no surjection from B to A. If set B is the integers, then >>saying that set A is bigger than set B is saying that there is >>no surjection from the integers to A, which implies that there >>is no bijection between them, and hence that there is no enumeration >>of A (by definition of enumeration). >>Now, until you give a definition of 'bigger', and explain >>why my first paragraph above is rendered irrelevant, we'll >>all be very grateful. Or at the very least, very >>surprised. > Your definition of "bigger" works fine for finite sets. And so far, you have given no evidence of even having any hint of a definition of what it means for one infinite set to be bigger than, or the same size as, another. All you have is a vague intuition which you insist we all agree with. (In spite of the fact that it leads you to patently wrong conclusions.) When you can actually come up with a better definition of what it means for two sets to be the same size (heck, even one that's no worse) then there'll be some point in listening to you. At the moment, I see no evidence whatever that you're capable of rational discourse.
From: Tony Orlow on 25 Jul 2005 16:56 Dik T. Winter said: > In article <MPG.1d483583ff4dfb97989f41(a)newsstand.cit.cornell.edu> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > ... > > > Now this case is quite dissimilar. Cantor's proof (about the cardinality > > > of powersets vs. the cardinality of the base sets) is indeed simple and > > > can be written in very few steps. > ... > > You know, if the only conclusion drawn from Cantor's proofs was that a power > > set is necessarily larger than its base set, I would have absolutely no > > problem. > > Actually that is the only conclusion. But that conclusion can be shown to > be equivalent to other conclusions. > > > That fact is always the case and I don't dispute it. Infinite sets of > > finite natural numbers, on the other hand, are self contradictory, > > Yes, you have told that already numerous times without actually showing > that is true. But let us go from the binary to the dyadic notation. > In that case the digits used are 1 and 2. 0 can not be represented, but > each natural number can be represented in only *one* way by a string of > those digits. A short table to show the idea: > 1 = 1 > 2 = 2 > 3 = 11 > 4 = 12 > 5 = 21 > 6 = 22 > 7 = 111 > 8 = 112 > 9 = 121 > 10 = 122 > etc. > In this representation each finite natural number is represented by a single > finite string. Now how many of such finite strings are there, given that the > stringlength is unbounded? This is precisely the kind of question which cannot be answered. There is no bound to the lengths of the strings, but you claim they are finite. What number am I supposed to use to calculate this set size? Still, the question is interesting if infinite digits are allowed. If you allow infinite strings and values, then I say there are N numbers, by definition. Now, this is a little different from normal digital number systems. At first I said "two symbols, S=2", but the problem here is that the strings have different lengths and cannot be simply extended using leading zeroes and using S=3, because the zeroes are ONLY allowed to the the left of the 1's and 2's, so we cannot have some constant infinite L. So, we can say there are 2^L strings for each L, and therefore sum(x=1->n: 2^x) strings less than or equal to length x, or 2^(n+1)-2. Therefore, if 2^(n+1)-2 =N, then 2^(n+1)=N+2, n+1=log2(N+2), and n=log2(N+2)-1 digits required for N numbers. In other words, it has one fewer digits than would be required to make a binary number that is 2 greater in value. > How would your "number" "N" be represented in > that representation? The largest number would be 2222...222 obviously. Since the digit places represent powers of two, as in binary, this would appear to be twice as large as 111...111 in binary, but as noted above, with one less digit we can make a number that is two greater. When we have one less binary digit, we are essentially dividing by 2, which we can do by shifting each digit to the right, or by dividing each digit by 2, since each digit is a 2 in this case. Then we get 111...111 in normal binary, which is what we expect. This is the normal number of digits, which is one less than would be required to represent a number that is 2 greater, since that would require the additon of another digit. That was a hard one, but not impossible. > > > and the > > conflation of "larger and infinite" with "uncountable" is arbitrary. > > You are missing the point. Each set "larger" in some sense than the naturals > as mathematicians think about them is (by definition) uncountable. Sure, unless you consider sets like the halves or square roots to be larger than N, which I do. besides, the reals are only "uncountable", without bijection with the naturals, only because the naturals are not allowed to be infinite. > > > The > > diagonal proof has hidden assumptions that mathematicians seem to > > acknowledge and not question, > > Wrong, the diagonal proof proves the same thing as the proof that the > powerset of a set is larger than the base set. The reals can be put in > a 1-1 relation with the powerset of the natural numbers. (But only when > you consider natural numbers in the sense of the mathematicians.) > And if you allow infinite naturals, then the naturals can also be put in 1-1 correspondence with the power set of the naturals via binary strings. -- Smiles, Tony
From: Tony Orlow on 25 Jul 2005 16:59 Virgil said: > In article <MPG.1d485d5b56de5b6989f50(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > And yet, one cannot apply an increment an infinite number of time without > > adding infinity. > > > That TO says it cannot be done does not prove that it can't be done. > > Anyone but TO, and a few others of equally limited capabilities, can do > it quite easily. > You can add an infinite number of 1's and get a finite result? Wow, you're good! -- Smiles, Tony
From: David Kastrup on 25 Jul 2005 17:02
Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > This is precisely the kind of question which cannot be > answered. There is no bound to the lengths of the strings, but you > claim they are finite. Can you tell us the difference between "arbitrarily large" and "infinite"? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |