Obections to Cantor's Theory (Wikipedia article)
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From: David Kastrup on 25 Jul 2005 09:58 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Virgil said: >> >> It is not me who keeps requiring a largest finite, but TO, by his >> delusional insistence that there must be non-finite naturals within >> the Peano system. >> > While it may seem that counting ala Peano can never produce any > infinite numbers, It does not seem so. It is provable from the axioms quite easily. > the insistence on the set being infinite requires the existence of > infinite values in the set. No. Sulking won't make it so. > Perhaps you should call your set "boundless" or "bigfinite" or > something, but it's not "infinite" It is infinite if it can be bijected to a proper subset of it, and the successor relation does that. > if you only count a finite number of times You can only count finite numbers, but there is no finite limit to how large they may be, even though every single one of them is finite. > and only have finite values. You know, the Peano axioms can easily > be inverted Different axioms, different game. Irrelevant. If you invert the Peano axioms, you get a rule "every number has a predecessor", and of course this means that the set we get can't have 0 as a member, since it then would also need -1. > to produce infinite whole numbers that count down, with exact > symmetry compared to the finite end of the number circle. Problem is that then you'll never reach 0, since there is no predecessor operation that will magically cross from the infinite to the finite. > If one uses such a set of axioms, does that mean that finite whole > numbers cannot exist, because that set of axioms doesn't seem to > allow us to count down that far? Of course not. But the axioms for natural numbers don't count down, so your silliness does not apply. > This insistence on determining the dividing point between finite and > infinite is clearly a waste of time, and your repetitions of this > non-point don't make it any more important. You can't count through > the divide in any finite number of steps. So what? That doesn't mean > that one side exists and the other doesn't. It means that the other side is not part of natural numbers. > Infinite set sizes ARE infinite whole numbers. Whatever you want to call them: they are not natural numbers as determined by the Peano axioms. > I can't understand why this isn't clear to everyone. Because the natural numbers are defined in a way that does not admit the size of the set of natural numbers into the set itself. That does not mean that the set size "does not exist" or is "not a number" or whatever else. But it means that it is not a natural number. Wherever you want to go looking for it, it is useless looking _in_ the natural numbers for it. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Han de Bruijn on 25 Jul 2005 10:28 Robert Low wrote: > Han de Bruijn wrote: > >> Robert Low wrote: >> >>> Han de Bruijn wrote: >>> >>>> Robert Low wrote: >>>> >>>>> Han de Bruijn wrote: >>>>> >>>>>> Clearly, it has never crossed their minds that such a nice >>>>>> relationship >>>>>> between topology and calculus could possibly esists. >>>>> >>>>> Apart from trivialities like de Rham cohomology and >>>>> the Atiyah-Singer Index theorem, anyway. >> >> But, as I suspected, these don't compare with my remarkably _simple_ >> result. Which nevertheless went unnoticed by mainstream mathematics. > > I was referring you to large bodies of mathematics which > relate topology and calculus. If you want a simple one, > then the *usual* integral to find the winding number > is already well-known to compute the crossing number. But my "awkward" integral to find the crossing number is NOT well known. I find this very strange, if not unbelievable. > But the complaint that something is a large body of > well-developed mathematics is hardly strengthening the > claim that mainstream mathematics is ignorant of > nice relationships between calculus and topology. Somewhat overdone, I admit. But nevertheless ... Han de Bruijn
From: Daryl McCullough on 25 Jul 2005 10:43 Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >Daryl McCullough said: >> >How do we get an infinite set, then, if m<=n is finite for any finite >> >n in N? >> >> You get an infinite set by (1) Pick some starting number a. >> (2) Pick an operation f(x) that, given a number n, returns a new >> number that is greater than n. (3) Then form the set >> >> { a, f(a), f(f(a)), ... } >> >> That's guaranteed to be infinite. >Meaning it goes on forever? Meaning that if you start enumerating the elements of the set (say one per second), you will never finish. >If it goes on forever, for an infinite number of >iterations, each time incrementing the value of the >next element (assuming your f() is successor/increment), >then the value of the next element will become >infinite. I'm telling you that a certain process never completes. You are saying that *after* it completes, the next element will be infinite. That's completely nonsensical. You seem to be thinking that *first* you generate all the finite elements, and *then* you start on the infinite elements. But the first step (generate the finite elements) *never* completes. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 25 Jul 2005 10:59 Tony Orlow (aeo6) 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. No, it's not a fact, and no, it hasn't been proven. You are assuming that the size of every set of naturals must be a natural. What is the basis for believing that? What definition of "size" are you using? -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 25 Jul 2005 11:27
David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> > >> > Barb Knox said: > >> >> In article <MPG.1d4726e11766660c989f2f(a)newsstand.cit.cornell.edu>, > >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >> >> [snip] > >> >> > >> >> >Infinite whole numbers are required for an infinite set of whole numbers. > >> >> > >> >> Good grief -- shake the anti-Cantorian tree a little and out drops a > >> >> Phillite. Here's a clue: ALL whole numbers are finite. Here's a > >> >> (2nd-order) proof outline, using mathematical induction (which I > >> >> assume/hope you accept): > >> >> 0 is finite. > >> >> If k is finite then k+1 is finite. > >> >> Therefore all natural numbers are finite. > >> >> > >> >> > >> > That's the standard inductive proof that is always used, and in > >> > fact, the ONLY proof I have ever seen of this "fact". Is there any > >> > other? I have three proofs that contradict this one. Do you have any > >> > others that support it? > >> > > >> > Inductive proof proves properties true for the entire set of > >> > naturals, right? > >> > >> For each of its members. > >> > >> > That entire set is infinite right? Therfore, the number of times you > >> > are adding 1 and saying, "yep, still finite", is infinite, right? > >> > >> No. He is not adding 1 more than a single time, just to check that > >> for n finite (which means that the set 0..n obeys the pigeon-hole > >> principle) n+1 is still finite (the case of one additional pigeon-hole > >> can be reduced to the case n if you check for the hole in position > >> n+1 and in 0..n both before and after permutation). > > > How do you know 6598367 is finite? Because you got it by adding 1 to > > 6598366, the 6598366th number. > > Certainly not. Don't tell me that you have spent the effort of > counting to 6000000 manually just to be able to talk about that > number. > > I know it is finite because it has a form (a finite number of digits) > that can be shown to always refer to a finite number. I see that the > number has 7 digits, and I know it is finite. > > That is, I know it is finite because it obeys certain laws, and those > laws can be shown to hold for all natural numbers, by induction. I > don't need to carry out any individual steps to make use of that > knowledge. Then why do you ignore those laws of representation, which are based on N=S^L, and pretend you can have an infinite set of such finite strings? Why do you feel entitled to violate established laws of math? You cannot form an infinite number of strings with a finite alphabet, without strings of infinite length. This should be obvious. > > > So how to you get the aleph_0th number? What IS that number? it's > > aleph_0. > > But that number does not obey a form I can recognize as belonging to a > finite number. And indeed, taking a look at its properties, it > becomes clear that it can't be a finite number. Well, no, it's infinite. Aleph_0 is basically 111...111+1. This is a clear contradiction within set theory, this assumption of all finite values. Try my new axioms. You can count from both sides at once. > > >> > So, you have some way of adding an infinite number of 1's and > >> > getting a finite result? > >> > >> No, he is just checking that the conditions for the fifth Peano axiom > >> hold. The whole point of the axioms is not to have to check an > >> infinite number of steps in order to get a statement about the members > >> of a particular infinite set, the naturals. > > > And yet, one cannot apply an increment an infinite number of time > > without adding infinity. > > It is neither necessary nor feasible to increment "an infinite number > of time" or "add infinity". The Peano axioms do just that. they start at 0 (or 1) and define successors, which are interpreted to be the previous value plus 1. That's called an increment, and if you do it x times, your value increases by x, wether x is finite or infinite. > > -- Smiles, Tony |