Obections to Cantor's Theory (Wikipedia article)
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From: Daryl McCullough on 25 Jul 2005 19:25 Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >Daryl McCullough said: >> No, that's no progress at all. You can prove that for finite sets >> if there is a bijection between set A and set B, then they have >> the same Bigulosity. But that fails for infinite sets. So Bigulosity >> is an inconsistent notion of "size". > >Huh? Bigulosity doesn't rely on bijections. That why it DOES work. It doesn't matter whether it relies on bijections or not. It is a fact that for finite sets, two sets have the same bigulosity if and only if there is a bijection between them. Isn't that right? But now you say that for infinite sets, two sets can have different bigulosities even if there is a bijection between them. Why are the rules for bigulosity for infinite sets inconsistent with the rules for bigulosity of finite sets? -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 25 Jul 2005 19:31 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >Daryl McCullough said: >> > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> >> 1. Phi(0). >> >> 2. for all natural numbers x, Phi(x) implies Phi(x+1). >> >> >The proof has a finite form, much like a recursive algorithm. A recursive >>>algorithm will run forever if it doesn't have some stop condition, like running >> >out of nodes in a tree path, which is bad for a computer program. >> >> But unlike an algorithm, there is no implied infinite number of >> steps. >Yes there is. You show the property true for n=0, then for succ(n), then succ >(succ(n)), etc. This is the justification for the axiom. No, there is no implication of an infinite number of steps. There are exactly *two* steps: 1. (prove Phi(0)), 2. Prove Phi(x) implies Phi(x+1). That is all that it takes to prove the statement forall x, Phi(x) -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 25 Jul 2005 19:48 Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >Okay, well that is only because you view all "countable" sets as equivalent. >In my book, the set of multiples of 1/2 is twice the size of the whole >numbers. Then I would throw away your book, since it's wrong. >It seems perfectly countable. And, the powerset of a countable set, >as far as I can see, is also countable That is provably false. >> existential quantification, >> universal quantification, >> addition, >> multiplication, >> equality, >> set membership, >I am not redefining any of the above concepts, except for equality >as regards the sizes of infinite sets. I'm asking you to *define* the concept of "size" in mathematical terms. >Gee, if I stick to standard concepts, then don't I just come up >with standard analysis? I'm not asking you to keep the Cantorian definition of cardinality. Throw that out. What I'm asking is that you replace it by something that is as mathematically clear---your replacement should be expressible in mathematical terms. >You don't know what an infinite number, set, string, tree, or >process are? I know what *I* mean by those terms. But you are rejecting the standard definitions. In that case, you need to substitute your own definitions. [Your definition of "infinite"] >I mean that there is no end to the set. What does that mean? Think about the set of real numbers between 0 and 1 (inclusive). One "end" is 0. The other "end" is 1. So the set has an end (it has *two* ends). But most people would say that there are infinitely many real numbers between 0 and 1. [Your definition of "smaller"] >Yes it does. It is a proper subset. That's wrong. Look at these two sets: A = { 0, 1, 2, 3, 4 } B = { a, b, c } B is smaller than A, but it isn't a proper subset. Your definitions don't work. -- Daryl McCullough Ithaca, NY
From: malbrain on 25 Jul 2005 20:21 Virgil wrote: > In article <MPG.1d4f082a49cdb89d989f7b(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > > But there is no 1-1 correspondence between naturals and INFINITE > > > bit strings (only with finite bit strings). This is just another > > > instance of that same delusion that TO has that there exist > > > naturals with more than finitely many naturals as predecessors. > > > > > If they have all 1's in finite positions, then there is indeed a > > bijection between infinite bit strings and finite naturals. > > Not in any standard representation, in which the leading digit of an > n-ary representation of a natural must be non-zero. That's why Tony is using the "value" of the string representation, not the representation as a thing-in-itself. > > If they have any 1's in positions infinitely to the left of the > > point, then they represent sets that include infinite integers, and > > also have infinite values as binary numbers. > > If one includes all such strings of infinitely many 0's and 1's, there > are as many such strings as there are subsets of N, which is a > cardinality greater than that of N itself. And also the size of the set of real numbers in base 2. karl m
From: malbrain on 25 Jul 2005 20:30
Virgil wrote: > In article <MPG.1d4f01ca1b87261c989f76(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > hale(a)tulane.edu said: > > > > > > > > > Tony Orlow (aeo6) wrote: > > > > malbrain(a)yahoo.com said: > > > [cut] > > > > > That's not what our axiom says. It says that induction covers > > > > > all the natural numbers in a single step, a single > > > > > leap-of-faith. > > > > That's really not the case. It is a recursive proof where the > > > > property is proven true for each element depending on its truth > > > > for the preceding element. f(n)->f(n+1), for n=1 to oo. > > > > Otherwise, how do you think it proves things for each and every n > > > > in N? > > > > > > Do you agree that one can prove the statement "If n is even, then > > > n+1 is odd" without using induction? > > > > > > I would say that one could. > > > > > > In such a proof, one would start off with something like: "Suppose > > > n is even. Then, there is an m such that n = 2 * n. etc" > > > > > > In such a proof, you would not be running through elements n = 1 to > > > oo. Or, do you disagree with this? If you disagree, then we can > > > reduce the problem to this case and eliminate the discussion about > > > how induction works. > > Are you offering a different type of proof that all whole numbers are > > finite? I have seen only the inductive form, so that might be > > interesting. > > > > > > Then, to answer your last question, I prove that f(n) is true for > > > each and every n in N by invoking "axiom of induction" after I have > > > proved f(1) is true and "f(n)->f(n+1)" is true (this without > > > invoking induction). > > Okay, so you ARE using induction, invoking it after not invoking it. > > Right? > > > > > > Induction is an axiom, not a recursive proof. The idea of recursing > > > from n = 1 to oo is an intuitive justification of the axiom of > > > induction. But, an axiom doesn't need a justification (in a certain > > > sense). > > > > > > -- Bill Hale > > > > > > > > Well, Bill, I have come to realize that this is one of my most > > central complaints about mathematics as it is today. There is great > > emphasis put on axiomatic systems, and axioms are given this status > > as unquestionable atomic fact without justification, when really > > facts NEED to be justified somehow, from outside of the axiomatic > > system. Axioms are taken as fact within an axiomatic system and used > > for proofs and arguments, but that doesn't mean every axiom is > > universally true as stated, or true at all outside of the system > > where they reside. > > No one says that axioms are true outside of their system, any more that > anyone claims that 4 balls and is a walk or three strikes is an out hold > outside of baseball. They are true inside of baseball because they are > the "axioms" of baseball. But EVERYTHING exists as part of a SYSTEM. Three-strikes holds lots of other places. It's part of the common language. See your local District Attorney for example. karl m |