Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 27 Jul 2005 15:05 malbrain(a)yahoo.com said: > Tony Orlow (aeo6) wrote: > > > In any case, sure, the program will spit out finite numbers, snce it is a > > finite machine running in finite time. If the machine had infinite capacity and > > infinite funtime, it could conceivably produce infinite results. > > In order to finish generating the natural numbers in 1 second, it would > need to spit out natural number n with a calculation time of 1/2^n. I > believe George first noted this fact in a post to Daryl a few years > ago. > > Unfortunately, I've never found a way to configure my > java-virtual-machine to accelerate with the number of steps taken, so > I've not found much use for his observation. karl m > > I think it might help to perform the calculation while plummeting into a black hole, but I guess we would never get the results after the fact, eh? ;) -- Smiles, Tony
From: Daryl McCullough on 27 Jul 2005 14:49 Tony Orlow (aeo6) wrote: >Robert Low said: > >> OK, so how many elements are there in the set of all finite >> natural numbers? >> >Some finite, indeterminate number. >You tell me the largest finite number, and >that's the set size. So you really think that there is some number n such that n is finite, but if you add 1 you get an infinite number? Maybe it's 7? Maybe 7 is the largest finite number, and 8 is actually infinite? >It doesn't exist? Well, then, I can't help you. In fact, a set is finite if and only if the number of elements is equal to a natural number. There is no largest natural number, and there is no largest finite set. The collection of all finite natural numbers is an infinite set. -- Daryl McCullough Ithaca, NY
From: David Kastrup on 27 Jul 2005 15:05 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Daryl McCullough said: >> Tony Orlow writes: >> >> >Barb, you're not saying anything new. I have heard it all before. I am not >> >drawing my conclusions in any such confused way >> >> Uh, yes you are. Why don't you write down what you consider to be >> valid axioms for working with infinite sets, and then try to write >> a formal proof of your claim >> >> If a set S of strings is infinite, then S contains some infinite >> strings. >> > I guess you can't understand a proof if it has natural > language. I guess he can't accept a proof that is wrong. > Your statement is not quite right. This is what I have proven: Whining won't make it so. > > Given an alphabet of finite size S, an infinite set of unique > strings made from this alphabet must contain infinite strings. Wrong. > Here's my two axioms: Those are not axioms. They are starting points, but not axiomatic. Let's see where you go wrong, anyhow. > (1) N=S^L, where N is the size of the set of ALL strings of length > L, constructed from an alphabet of size S. N is the upper limit on > the number of strings from a set of S symbols and a maximum length > of L. Well, this is actually wrong to start with, since S^L is the number of strings of an _exact_ length of L. The number of strings with a length up to and including L would be (S^(L+1)-1) / (S-1). But let's, for the sake of the argument, assume that you have not already goofed here. > (2) A^B is finite if and only if A and B are finite. Fine so far. > Given (2), S^L is finite iff S and L are finite. Fine so far. > Given finite S, S^L is finite iff L is finite. Fine so far. > Given (1), N is finite iff L is finite. Fine so far. > Conversely, N is infinite iff L is infinite. Still fine. Unfortunately, it does not prove what you are out to prove. You have now proven (assuming for the sake of the argument that all your blunders above did not happen) that all strings with a _finite_ maximum length L form a finite set. But the set of all strings from S does not have a finite maximum length. > Well, I just got back from getting lunch, during which it occurred > to me that it would be good to put this proof in inductive format, > and watch you try to discredit it while hanging on to your proof > that all naturals are finite. This proof relies on some > assumptions. Let me know if you disagree with any of them: > > (1) All naturals are finite. (you already "proved" this one) > (2) The number of strings of length L that can be constructed from a set of > symbols of size S (set size, not symbol size, remember (sigh)), is S^L. > (3) For finite A and B, both A^B and A+B are finite. > > Are we good to go? Okay. > > Proof that f(n), the number of strings in the set of all strings up > to and including length n in N, on a finite alphabet of size S, is > finite: > > 1. For n=1, we have S^1 strings of length 1, for a total of S > strings less than or equal to 1 in length. f(1)=S is finite, as > stated. > > 2. For a finite number of srings of length n or less, we can add > S^(n+1) strings of length n+1, to get f(n+1),the number of strings > of length n+1 or less. S is finite and n+1 (a natural number) is > finite, so S^(n+1) is finite. S^(n+1) is finite and f(n) is finite, > so f(n+1)=f(n)+S^(n+1) is finite. > > 3. Therefore, for all n in N, f(n), the number of strings up to and > including n symbols, which are created from a finite alphabet, is > finite. There is no n in N, in other words, which maximum length of > string will allow you to have an infinite set of strings. Totally correct. As soon as you limit the maximum length n of the strings to a finite number, the resulting set of strings is finite. But there is no maximum length of strings: you can always make your finite string with k positions another position longer. So you have just proven that once you limit the string length, the resulting set is finite. Too bad the statement is supposed to be about strings with _arbitrarily_ large (but finite) length. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: malbrain on 27 Jul 2005 15:08 aeo6 Tony Orlow wrote: > malbrain(a)yahoo.com said: > > Tony Orlow (aeo6) wrote: > > > > > In any case, sure, the program will spit out finite numbers, snce it is a > > > finite machine running in finite time. If the machine had infinite capacity and > > > infinite funtime, it could conceivably produce infinite results. > > > > In order to finish generating the natural numbers in 1 second, it would > > need to spit out natural number n with a calculation time of 1/2^n. I > > believe George first noted this fact in a post to Daryl a few years > > ago. > > (...) > > > I think it might help to perform the calculation while plummeting into a black > hole, but I guess we would never get the results after the fact, eh? ;) Mathematics does not require physical models to be consistent. If we can mathematically specify an "accelerating" java-virtual-machine that doubles its performance with every iteration, the program will generate the natural numbers in finite time, without needing "black-holes". As such, the program running on the java-virtual-machine it becomes a valid model for the set of natural numbers. karl m
From: Tony Orlow on 27 Jul 2005 15:12
Virgil said: > In article <MPG.1d506ab0de98a030989f9e(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > Either you have an upper bound or you do not, and if there is no > > upper bound on the values of the members, then they may be infinite. > > Or they may not. > > > > how do you have an infinite set of strings with only finite lengths? > > The usual way, by not having any finite bound on their lengths. Only an infinite bound? Then they CAN be infinite? You make no sense. > > > > > > > >I am not assuming anything except for this fact. > > > > > > You are assuming that every set of strings has a natural number L > > > such that every string has length L or less. That's false. > > > > I am saying that if L CANNOT be infinite, then S^L CANNOT be > > infinite > > No one is requiring any S^L to be infinite, we merely deny that there is > any finite lid on the size the S^L can attain, as S and L are allowed to > increase without finite limit. Oh, so now you are not requiring the set of naturals to be infinite? I could have sworn..... > > > For finite S, > > S^L can ONLY be infinite with infinite L. Why is this so hard to > > understand? If S and L are both finite, then S^L is finite, isn't it? > > But no one is claiming that any of S, L or S^L is infinite, we just say > that any of them can be larger that any finite natural you choose to > name. that is, there is no finite limit on their sizes. So S^L, the size of the set, is finite but unbounded? Okay. That at least sounds like how you describe the natural numbers. > > Why TO rails against something that no one is claiming is because he has > no valid argument by which to refute what we actually are claiming. I believe, and correct me if I am wrong, that you were ALL claiming the set of finite natural numbers was infinite. I COULD be mistaken.... > > That form of argument is called the fallacy of the STRAW MAN. When TO > has no counter to an actual argument, he pretends that there was a > different argument, which he then attacks. Virgil seems to get himeslf mixed up with me. He doesn't even know what the argument is any more, and doesn't know what side he taking. This must be an indication of something.... > > TO has been doing a lot of STRAW MAN arguing recently, to no avail. > Really? That's interesting, because you keeep proving my points for me. Are you a scarecrow? What was that song he used to sing, in the Wizard of Oz? That must be your favorite. -- Smiles, Tony |