Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 25 Jul 2005 15:04 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. In this case, there is justification for the axiom of induction. Peano didn't make it up randomly. It is a statement about how a recursive proof can cover an infinite set without requiring an infinite statement. As such, it really is an axiom about an infinite recursive process, and if we keep this in mind, and note that our inductive step (f(n)->f(n+1)) involves an increment (n+ 1 is finite), then we can see that the infinite loop that this statement represents ends up incrementing the value an infinite number of times, to produce an infinite value. By keeping in mind the justifications for our axioms, we can more carefully see where they do and do not apply, and refine them as needed. In my opinion, mathematics needs to be more fully integrated, and not further decomposed into independent axiomatic systems without justification. -- Smiles, Tony
From: Virgil on 25 Jul 2005 15:03 In article <MPG.1d4eb55b5c315b4b989f67(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > The fact is that any set of whole numbers greater than or equal to 1 > MUST have at least one element with a value at least equal to the > size of the set. That is a fact and has been proven. That is not a fact, and cannot be proven, without some additional assumption, such as that the set is finite or, equivalently, that it contains a maximal member. If everything that is true about finite sets had to carry over to infinite sets, infinite sets would be finite, too.
From: Chris Menzel on 25 Jul 2005 14:55 On Mon, 25 Jul 2005 14:32:09 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > stephen(a)nomail.com said: > ... >> I can flesh it out a bit. >> >> Inductive step. Show that if n is finite, then n+1 is finite. >> Proof by contradiction. >> >> Suppose that n is finite, but that n+1 is infinite. This >> means there exists a bijection f from { 1, 2, 3, ... n+1} >> to some proper subset S of { 1, 2, 3, ... n+1}. Without >> loss of generality we can assume that S does not contain n+1. >> >> If we apply the function f to { 1, 2, 3, ... n} we >> get the set S-f(n+1). Because S does not contain n+1, >> S-f(n+1) is a proper subset of {1, 2, 3, ... n}. This >> means there exists a bijection from {1, 2, 3, .. n} >> to a proper subset of {1, 2, 3, ... n}, which means >> that n is infinite which contadicts the assumption that >> n was finite. >> >> Stephen >> > Yes, it relies, as usual, on the lack of a largest finite integer, but > that doesn't mean that there are no infinite integers. It means EXACTLY that; he just proved that the successor of a finite number is finite. Since 0 is finite, it follows by induction -- you say you understand induction, right? -- that ALL natural numbers are finite. > There is no smallest infinite integer, ... If you actually understood induction, as you claim, you'd know that it is equivalent to the principle that, for any property P, if any natural number has P, there is a smallest natural number that has P. So if there are any infinite natural numbers, there has to be a smallest one. What can we conclude (from other well known axioms of arithmetic) about infinite natural numbers? Again, your ignorance of even basic arithmetic, let alone the more advanced topics you presume to address, is evident for all to see. Haven't you suffered enough shame?
From: Chris Menzel on 25 Jul 2005 14:57 ["Followup-To:" header set to sci.logic.] On 25 Jul 2005 11:41:05 -0700, georgie <geo_cant(a)yahoo.com> said: >>Faith?? "Cantorianism" is embodied in a completely rigorous axiomatic >>theory. > > How can axioms be rigorous? Aren't they supposed to be self-evident? > Doesn't that sort of imply that you "believe" them to be true? Isn't > that faith? I am using "axiom" in a purely formal sense: A sentence in a formal language. Completely rigorous. The question of their self-evidence is a completely separate issue.
From: Tony Orlow on 25 Jul 2005 15:08
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 proprty true for n=0, then for succ(n), then succ (succ(n)), etc. This is the justification for the axiom. That is why inductive proof is said to work. > > >Your #2 above is the recursive part of the proof; it proves something > >true based on the truth of its predecessor. > > No, the only thing that you prove is the implication > Phi(x) implies Phi(x+1). Yeah that's what i said. > > >When you actually "run" the proof, > > You don't "run" proofs. That's why I put it in quotes, but it is really a recursive proof with an implied loop. > > >#2 acts like a loop, iterating its way through all the members > >of the set, with no stop condition. > > No, it's not like that, because you don't "run" proofs. Why do you think inductive proof is agreed to work? Think. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony |