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From: David R Tribble on 21 Oct 2005 11:19 Sorry for the length of this post. Normally, I would only respond to the most egregious errors, but obviously you (Tony) did not understand anything I said in my post. Tony Orlow: >> No, it was specifically a bijection between two sets of infinite binary >> strings representing, on the one hand, the whole numbers in *N starting >> from 0, both finite and infinite, in normal binary format, and on the >> other hand, the specification of each subset of whole numbers in *N, >> where each bit which, in the binary number, represents 2^n denotes >> membership of n in the subset. This is a bijection between the whole >> numbers in *N and P(*N), using an intermediate bijection with a common >> set of infinite binary strings. > David R Tribble: >> We can show a mapping (surjection) from the finite naturals in N >> to the finite subsets of N (the finite members of P(N)) using your >> mapping strategy. But then you don't have any more naturals left >> to map to the infinite subsets in P(N). > Tony Orlow: > No more FINITE naturals. Does *N include infinite values or not? Yes, *N contains all the finite and all the infinite naturals. I'm the one who starting using the "*N" notation in this thread in the first place, after all, to distinguish the Tony set of naturals from the standard set. But you obviously are not paying attention. Here I'm talking about a bijection between N and P(N). We already agree that a bijection between *N and P(N) is easy, so I don't even bother mentioning it. The next paragraph is where I talk about a bijection between *N and P(*N). David R Tribble: >> Pretty much the same thing for *N - using your binary mapping >> stragegy, we can show a mapping between the finite naturals in *N >> to the finite subsets of *N (which are the finite subsets in P(*N)), >> and a mapping between the infinite naturals in *N to some of the >> infinite subsets in P(*N). But after we've run out of members in >> *N for mapping, there are still more infinite subsets remaining >> in P(*N) that we can't map. > Tony Orlow: > Uh, when do we run out of the unending list of infinite natural numbers? > Which is the last one? For every one, there is always another after it, > right? So, when does the bijection break down? After the 2^Nth element of > the set of N naturals? We all know that there is no last member of *N; this has been mentioned hundreds of times already in this thread. Just let it go. It's obvious that you don't understand. We run out of members of *N, finite and infinite, because we use them all to map to subsets of *N. But there remain subsets of *N that still are not mapped, and we've "used up" all the members of *N in our mapping process, so there are more subsets of *N than members in *N that can map to them. You can't use a member of *N to map to more than one subset; once you've mapped a natural to a subset, it's "used up" by the mapping process, and you have to choose a different natural to map the next subset. That's my whole point, that there are subsets of *N that cannot be mapped by any members of *N, which is why your strategy is not a proper bijection. David R Tribble: >> Consider the subsets S(n) for any n in *N, using your binary index >> mapping scheme. If n is finite, the subset S(n) is also finite; >> for example, >> S(42) = S(101010(2)) = {1,3,5}. >> If n is infinite, S(n) is also infinite; for example, >> S(...999) = S(...111(2)) = {0,1,2,3,...}. >> >> But there are still more subsets left in P(*N), and there are not >> enough left over members in *N, finite or infinite, to map to them. >> These subsets contain at least one infinite natural, and that >> natural has no bit index corresponding to a member of *N. > Tony Orlow: > What? Every element of *N has an infinite number of bits, even if the string > represents a finite value. Just because an element has an infinite value, and > is represented by an equally infinite bit position, that is not a problem, > since infinite values in binary REQUIRE 1 bits in infinite positions in the > string. It's obvious that you don't understand. A bit position number must be longer than the bitstrings containing its corresponding bit value. Bit position j corresponds to the bitstring value 2^j, and 2^j > j, right? If you're going to use index j in a subset, then there must be a natural k >= 2^j, right? But how do you deal with an infinite natural j, which appears in an infinite number of subsets of *N, if you've already previously mapped all the finite and infinite naturals in *N to the other subsets containing only finite naturals? David R Tribble: >> For example, subset B = {0, ...999} has no corresponding n in *N >> that can map to it, because there is no natural bit index available >> to represent the second member. > Tony Orlow: > You are assuming a finite natural number of bits, but that is simply not the > case with *N, or it would ony contain finite values. Are we talking about the > same set? You obviously don't understand what I'm saying. You're trying to create a straw-man argument out of what you *think* I'm saying, them saying it's false, so therefore your logic wins. But it's obvious I'm not saying anything of the sort; I said nothing about a last element nor a last bit, because we all know there are no such things. We're talking about infinite sets and infinite naturals, right? Just let it go. Tony Orlow: > Lets say that, since we are using binary, you mean 0:111...111 with N digits. > This number is 2^N-1. That's the number that corresponds to the subset > containing the first N elements. It's not IN those first N elements, but it's > in the infinite set. And it's not in any of the sets that can be mapped by any member of *N, because all of those members are used to map to other subsets that do not contain any infinite naturals. Try it for yourself. List all the subset mappings using your strategy, and show us where any natural in *N, finite or infinite, maps to any subset containing an infinite natural. There are none. David R Tribble: >> This example is just one of an >> infinite number of finite and infinite subsets of *N that cannot >> be mapped with the finite and infinite members of *N. > Tony Orlow: > Ah, but they can, if you avoid assuming in some way that you have identified > the last of them. That's what infinite bijections are all about. You simply > push all differences down the line, until theyre infinitely far away and you > can't see them any more. At lease, that's how transfinite cardinality works. Tony, Tony, Tony. You obviously can't follow what I'm saying. Just let it go. David R Tribble: >> The conclusion is that there does not exist a surjection from the >> members of *N to the members of P(*N), and therefore no bijection >> between them is possible. > Tony Orlow: > But that conclusion assumes a last element, whence it draws its > contradictions. Obviously, you're just ignoring what I said. I state quite clearly and explictly that your strategy leaves some subsets of *N unmapped by any members of *N, so even though it is an injection from *N to P(*N), it is not a surjection, so it is not a bijection. It is incumbent upon you to show otherwise, and so far you have not done that. There is no red herring straw-man last element. Just let it go.
From: David R Tribble on 21 Oct 2005 11:28 Tony Orlow wrote: > The problem with the reals is that you don't have a natural well-ordering > of them. I can give you that, when you're ready, which will probably never > be. No, by all means, please do. I'm ready. Oh, but let me guess. The first point in the real line is: 0:0.000...001, and the next point is 0:0.000...002, and so on. But then how do you explain the points between them (you know, 0:0.000...000:000...001 0:0.000...000:000...002 and so on), if the real line is continuous? You do know what "continuous" and "infinitely divisible" mean, right?
From: Tony Orlow on 21 Oct 2005 11:33 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. > 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. 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. > > 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? 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? > > "Can therefore identify the element that maps to the entire set" makes > no sense. We are considering any mapping from a set A to its power set > P(A). In some mappings there is an element mapped to the complete set - > e.g. (a crank favourite) > > 0 -> {} > 1 -> {0, 1, 2, 3, ... } // the complete set of naturals > 2 -> {0} > 3 -> {1, 2, 3, 4, ... } // all naturals except 0 > 4 -> {1} > ... That looks like one of Albrecht's. It reminds me of my adjustment to the Peano axioms to include both finite and infinite. It could have some application. But, it's not what I offered. > > > > ... 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? > > > ... Despite the fact that, within any range up to > > S, you cannot map every subset to an element within that range, the lack of a > > largest element makes it so there DOES exist an element that maps to all n > > through S, but it is more than S. 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? > > 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. We can speak of the entire set, but as soon as we speak of any specific size, we will encounter contradictions. > > Brian Chandler > http://imaginatorium.org > > -- Smiles, Tony
From: Randy Poe on 21 Oct 2005 11:47 Tony Orlow wrote: > imaginatorium(a)despammed.com said: > > Right: well done!! You got something right. > So, what about that situation is not a bijection? You claim to have a bijection f from N to P(N). One element of P(N) is the set N itself. You admit that there is no y in N such that f(y) = N. Therefore f is not a bijection from N to P(N). > What is wrong with my bijection? No specific rule has been broken, has it? The definition of bijection, that it be "onto" P(N), that every element of P(N) be mapped by some element of N. - Randy
From: Daryl McCullough on 21 Oct 2005 11:55
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. 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. 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". 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. -- Daryl McCullough Ithaca, NY |