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From: imaginatorium on 21 Oct 2005 15:50 Daryl McCullough wrote: > imaginatorium(a)despammed.com says... > > >Yep. Absolutely. That's why people who have outgrown your mediaeval > >imponderables never "place ranges" nor "declare finished"... > > Is Tony an idiot, or is he pulling our legs? It's hard for > me to figure it out. He ends his post with "Smiles", which > might be a clue that it's all a big joke with him. It is hard to know. Trouble is that endlessly repeating the same confusions, even when they are wrenched into daylight, isn't exactly high in the humour rankings. I think the bottom line is that Tony has some ideas - basically he has studied some finite combinatorics and stuff, but simply assumes he knows what "infinity" means. He is much more coherent than most of the cranks, often lasting several sentences without collapsing (I mean, Ross Finlayson can't even do verbs and their objects yet). The problem is basically a combination of a massive misconception (that mathematics is a branch of experimental computing, roughly) with a strangely inflated confidence level. I think if I had never met another crank I would have assumed that Tony is just trying to pull our legs, but now I'm happy to accept his (presumed) assurance that he isn't. I also wasted a bit of time the other day reading some of the "Phil" threads - I had forgotten quite how similar the arguments were. There's always eventually this wriggling bit while sets expand to fill the requirements of the latest objection. Also, originally Tony appeared with this "bigulosity" idea, trying to assign "sizes" to infinite sets. One big problem IMO is that many people write, carelessly, as though cardinality (under the standard definition) is The Only Correct Way to Assign a Size to a set. And this is plainly not true: just the simple subset relation as a partial ordering is - at least in some sense - a sort of sizing. It looked as though if the Cardinality Good Anything Else Bad droners could be kept out of the way, Tony might have been guided to seeing how to formalise what he was thinking of, and even (!!) possibly to have seen why it isn't as satisfactory as he first thought. But then things went downhill, and we've been through several ridiculous episodes where he simply argues blindly against some elementary result, which he plainly doesn't properly understand. (I really _don't_ believe anyone could properly understand the A->P(A) bijection proof and still argue against it.) Brian Chandler http://imaginatorium.org
From: David R Tribble on 21 Oct 2005 15:58 Brian Chandler: >> Yes, we have the complete set of naturals. >> No, we have not "identified the last", because there isn't one. > Tony Orlow wrote: > Then how do you know the size, or how many bits you need in y? You have, in > effect, declared N as the size of the set, and noted that y=2^N and is thus > outside the set. The set doesn't end. There is no natural outside of it. If y > is a natural, 2^y is a natural. When is that not the case? It's always the case that for any finite natural y in N, 2^y also exists as a finite natural in N. (You just said so.) Which proves that there is no last (or "largest") finite member in N, because for any finite natural y in N, you can always find a larger finite 2^y in N. So you just proved that there is no largest finite natural, and that N has no last finite member, and that N is an infinite set. Or do you still maintain that N is a finite set, contradicting what you just said above?
From: Virgil on 21 Oct 2005 15:59 In article <MPG.1dc2d44c87ee12e398a524(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > imaginatorium(a)despammed.com said: > > Tony Orlow wrote: > > > stevendaryl3016(a)yahoo.com said: > > <snip> > > > > Let A be any set whatsoever, finite or infinite, it doesn't matter. > > > > Let f be any function from A to P(A). > > > > Let w = { x in A | x is not an element of f(x) }. > > > > Let x = any set in A. > > > > Let u = f(x). We prove that u is not equal to w. > > > > > > > > By definition of w, we have x in w <-> x is not an element of f(x). > > > > So x in w <-> x is not an element of u. That means that there are > > > > two cases: Case 1: x in w, and x is not in u. In that case, u cannot > > > > equal w. Case 2: x is not in w, and x is in u. In that case, u > > > > cannot > > > > equal w. > > > > > > > > So what we have proved is that forall x, w is not equal to f(x). So > > > > w is not in the image of f. So f is not a bijection between A and P(A). > > > > > > > > There's no induction. There's no assumption that A is finite. > > > > > But there is an assumption that y is in S. If you are assuming you have > > > the > > > complete set of naturals, that you have identified the last, and can > > > therefore > > > identify the element that maps to the entire set, then you indeed run > > > into a > > > contradiction. ... > > > > Goodness you are dim. The axiom of infinity says (in effect) "The > > naturals are a set". This means we talk about the set of naturals, > > letting all the other axioms apply to it. Why do we need to have an > > _axiom_ to let us talk about the naturals? Because it is an infinite > > set. It goes on forever, and never ends. There is no last natural. Then why does TO keep making assumptions that require a last naturals? TO seems to want to have things both ways, no largest and having a largest simultaneously. Fortunately for his delusion, in his world of TOmatics both can be true and both false simultaneously. > > There is no end to them. The "end" is not merely "unspecified", > > "unidentified", "tenuous", or any such, it is *nonexistent*. > > (Remembering that nonexistence, like existence, is not a predicate.) > > > > Having proved that there is no last natural, we nonetheless talk about > > the complete, entire, total, set of all naturals. All naturals. There > > is no natural anywhere in any of your nonsensical "doubling" and > > "thinning" operations that does not already belong to the mathematical > > set of all naturals. That's what "all" means. > > > > Sorry, mustn't go on - it's pointless anyway, since after thousands of > > posts it's pretty unlikely you will ever grasp any of it. But anyway, > > to go back to your paragraph: > Yes, your reiterations of the standard nonsense don't go very far with me. TO has his own brand of nonsense which he much prefers, since in his one can be on both sides of every fence simultaneoulsy. > Sorry. Having noted that there is no last natural, and that the size > of the set to which any natural is successor is itself, you > nonetheless assign a set size while declaring the nonexistence of a > largest element. Here, you again declare you have the entire set, > derive a contradiction from that, and deflect it at bijections. The > bijection will fail with any finally declared set. Where the sets > don't end, it never fails. Then let TO produce any bijection between the endless set of finite naturals and its power set. Or does TO still claim that the Dedekind infinite set of finite naturals ends? > > > > Yes, we have the complete set of naturals. > > No, we have not "identified the last", because there isn't one. > Then how do you know the size, or how many bits you need in y? We do not need to have a "size" in any TOish sense to do all that needs doing. Cardinality serves us well, and we do not need any of TO's artificial additions to standard math. > You have, in effect, declared N as the size of the set We have, in effect, not decalred a "size" at all, other than a cardinality. > > > ... However, both the infinite set of naturals and the infinite > > > power set go on forever, so you never run out of naturals to map to > > > subsets, > > > nor subsets to map to naturals. > > > > Right: well done!! You got something right. > So, what about that situation is not a bijection? NO! > > > This is a prime example of where the value range matters in the > > > bijection. > > > > No it isn't. There is no "value range" in a bijection. "Value ranges" > > are only used in TOmatics, remember. But you are clawing your way > > towards grasping what was the natural assumption (that if two sets go > > on forever it is never possible to say there can't be a bijection > > between them) before Cantor pointed out that it is wrong. > > What is wrong with my bijection? No specific rule has been broken, has it? Yes, the one that says that every member of the codomain must be in the image. > > > > > Here's a question for you. Obviously if the TOnats (T) are a set, > > within the framework of conventional set theory, the proof applies to > > them, and there can be no bijection T <-> P(T). But are they a set? Is > > it possible to have the complete, final, total, can never be extended > > in response to the next question, set of TOnats? > It is never possible to place any exact number on it. Infinite sets > can only be compared with each other over any given range. At least inside TOmatics, though not ouside its idiocy. > We can > speak of the entire set, but as soon as we speak of any specific > size, we will encounter contradictions. So outside of TOmatics, one merely avoids speaking of "size" and uses only cardinality. > > > > Brian Chandler http://imaginatorium.org > > > >
From: David Kastrup on 21 Oct 2005 16:03 Tony Orlow <aeo6(a)cornell.edu> writes: > Daryl McCullough said: >> Tony Orlow says... >> >> >Yes, your reiterations of the standard nonsense don't go very far with me. >> >> That's because you are an idiot. You are incapable of formulating >> or understanding a mathematical argument. Worse, you believe yourself >> to be much more competent than you actually are. > So, you wanna go out for a beer, or what? LOL >> >> That's the point, really, of rigor and formalization. If you are >> speaking loosely, it's very easy to fool yourself into thinking >> you are making sense. But if you are forced to write down your >> reasoning in a careful, detailed, rigorous way, you will discover >> where your ideas are full of hot air. > Then perhaps they will fly. Sure. But pray open the window instead of blowing them here. They stink. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Dave Rusin on 21 Oct 2005 16:19
In article <1129901285.084347.34810(a)g49g2000cwa.googlegroups.com>, William Hughes <wpihughes(a)hotmail.com> wrote: >Yes all sets have a cardinality. That's your Choice. |