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From: Daryl McCullough on 21 Oct 2005 14:46 Tony Orlow says... > >Daryl McCullough said: >> Notice that this is not the case with *finite* subsets. >> If S is any finite subset of naturals, then we can associate >> it with the natural number >> >> n = sum of S_i 2^i >> >> where S_i = 1 if i is an element of S, and S_i = 0 >> otherwise. This is a rule for associating every finite >> set of naturals with a natural, and there are no >> finite subsets that are left out. Therefore, there >> is a bijection between the set N of naturals and the >> set FS(N) the finite subsets of N. > >But, if there are N finite naturals, there are still 2^N finite subsets. No, the number of finite sets of finite naturals is exactly the same as the number of finite naturals. 2^N is the number of *all* subsets, including infinite and finite subsets. Yes, I know you think that N is finite, but you are an idiot. >This power set relation holds for all power sets. The power set of the finite naturals includes both finite sets and infinite sets. >> If you give me any rule for mapping sets of naturals >> to naturals, then I will give you a rule for creating >> a set of naturals that is not in the range of your >> mapping. >And what does that prove? That there can be no definable bijection between any set and its power set. >> Even though you can never complete the process of generating >> an infinite set, you *can* complete the process of giving >> a rule for one. For example, the rule "x is in the set if and >> only if x is even" defines the infinite set of even natural >> numbers. The rule "x --> 2*x" is a rule for mapping the set >> N of natural numbers to the set E of even numbers. > >Right, and the rule I offered gives a mapping to a unique >subset for each natural in N. No, you offered something that you mistakenly *believed* was such a mapping. The mapping you described was this: Let n_i be the ith bit of the binary representation for n. Then f(n) = { i | n_i = 1 } But that map is not a bijection between *N and P(*N). This has been pointed out to you many times. In particular, there is no x such that f(x) = *N. What properties would such an x have? Well, in particular, in its binary representation it would have to have a 1 in *every* position, where "position" ranges over the *entire* set *N, including x itself. So your x would have to be a number with a 1 in position number x. That implies that x > 2^{x-1}. That is impossible. What you are doing is chasing your own tail. If you start with the finite naturals, then you can represent all finite sets of naturals, but you can't represent any infinite sets of naturals. So you introduce the concept of an infinite natural. Fine, your infinite naturals can represent *every* set of finite naturals, including infinite sets. But they don't represent infinite sets of *infinite* naturals. They can't. That would be a contradiction. You could, if you like, introduce a whole hierarchy of bigger and bigger infinite naturals: Level 0: The finite naturals. These can be used to represent finite sets of naturals, but cannot be used to represent infinite sets of naturals. Level 1: These can be used to represent infinite sets of Level 0 naturals. Level 2: These can be used to represent infinite sets of Level 1 naturals. Level 3: These can be used to represent infinite sets of Level 3 naturals. etc. You will never find a level that can be used to represent infinite sets of elements of that level. That's *provably* the case. The fact that you don't believe the proof is not because of any flaws or loopholes in the proof, but because you are basically an idiot. You are mathematically incompetent. -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 21 Oct 2005 15:09 David R Tribble said: > William Hughes said: > >> Well, as you have not even defined 2^N for infinite TO naturals, you > >> have some work to do. > > > > Tony Orlow wrote: > > Defined 2^N? It's 2 to the Nth power. What do you need to know? In binary, > > it's a 1 followed by N 0's. Please state your complaint in the form of a > > question. > > It is entirely reasonable to ask you to define 2^N. > > In the past, you've defined 'N' as the sum of an infinite series: > N = 1 + 1 + 1 + 1 + ... That is standard N. > or perhaps: > N = 1 + 2 + 3 + 4 +... That is (N^2+N)/2 > or perhaps: > N = 1 + 2 + 4 + 8 + ... That is 2^N-1 > > Most of the time, you simply say that 'N' is an arbitrary infinite > number. Yes, or it can be the standard unit infinity. It's the unit we happen to be using for comaprison. > > For 2^N, "a 1 followed by N zeros" doesn't cut it, if for no other > reason than because your 'N' is arbitrary. Perhaps you should define > N a bit more precisely before you go throwing it around so freely. That definition holds for finite or infinite N. It's fine. > > Defining your "infinite naturals" in more concrete terms, perhaps > sets and successors, would be a step forward in making sense of your > number theories. Yes. I wopuld like to be able to express the sum of the harmonic series, or of the primes, for one thing. Things like log2(N), however, seem only describable in terms of some number of zeroes. > > -- Smiles, Tony
From: Daryl McCullough on 21 Oct 2005 14:58 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. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 21 Oct 2005 14:55 imaginatorium(a)despammed.com says... >Right. Fairly soon, I think we should all give in. Accept that we just >can't understand what Teacher Tony is Telling us. (OK guys? Let's give >in. Tony's right. Anything that's been repeated at least 4000 times >just has to be true, no?) Do share with us what your next step is? Are >you going to write a text book? At least a web page? Convert a >university department? Or just tell your grandchildren that the people >on Usenet said you were right, so all the universities just must be >wrong? You're right. People are only arguing with Tony because they have jobs that depend on the blind acceptance of "establishment" mathematics. Or in some cases, because they have spent a lot of money and time learning material that is nonsense, and they can't admit to themselves that they fell so hard for such a scam. After Tony mathematics has replaced the standard mathematics in schools and textbooks, will we be able to practice pre-revolution mathematics in our own homes, in our spare time, or will that be outlawed? -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 21 Oct 2005 15:16
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. > > Standard mathematics *has* gone through this process. Your junk > has not. That's why standard mathematics is much higher quality > than your bullshit. It's not because mathematicians have a higher > IQ, but because they have a much stricter quality control process > than you do. You have no quality control AT ALL. Even when you > are confronted with an out-and-out contradiction in your ideas, > it hardly fazes you---you don't change your ideas in the face of > evidence that they are wrong, you just make up new weasel words > like "tenuous existence", "unidentifiable naturals". I express things the way I see them. From your axiomatic standpoint, your arguments are airtight. From my perspective, they all rely on contorting a very few paradoxical facts into general rules that defy reason. I don't expect you to see that. I also won't call you an idiot for not seeing what is obvious to people like Martin, Zuhair, Ross, myself and others. > > The fact that you are not stupid makes your behavior all the > worse: your idiocy is a *choice* on your part. You would rather > be lazy and pretend to be doing mathematics than to work hard > and *actually* do mathematics. Do you think it is lazy to try to construct a better foundation for mathematics? Well, it is a lot easier than regurgitating what I learned in books. It's so much easier to paint than go to the museum. Only lazy people make anything new. The real work is in defending the status quo against the crackpots. Oh, the hordes at the gate! Woe to those lazy barbarians! > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony |