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From: Tony Orlow on 31 Oct 2005 12:28 Virgil said: > In article <MPG.1dcab744b1856f9198a57e(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Randy Poe said: > > > > Since f:N->E is both injective and surjective, it is bijective. > > > Okay, thanks, how about a proof of bijection between the natural > > numbers and the binary strings? > > Which set of naturals and which set of binary strings? In TOmatics there > are at least two of each. > > For the standard bijection between the Dedekind infinite set of finite > naturals and the Dedekind infinite set of finite binary strings, map > each finite natural to its finite binary string representation and voila! What do you mean "voila!"? If that's not sufficient for my bijection, then it's not for yours. Prove it is a bijection. Show me how it's done, Stud. > > Since the Dedekind infinite set of finite natural is the only set of > naturals outside of TOmatics, that bijectin will have to suffice, at > least everywhere outside chaos. Prove it is a bijection. Show me how it's done. > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 12:34 Virgil said: > In article <MPG.1dcaba9ffcd3c6998a581(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Robert J. Kolker said: > > > Tony Orlow wrote: > > > > > > >> > > > > > > > > Perhaps. Cardinality is better than nothing. And yet, I think > > > > that noting that the size really is not a number with an absolute > > > > value but needs to have some value applied to it in order to > > > > measure it, > > > > > > You continually confuse "how many" with "how much". > > > Pah! Bob, I equate "how many" with "how much" when there is a 1-1 > > correspondence between element value and element count, as is the > > case with the naturals. I am not confused. > > That sounds like a cry for help! And TO certainly could use some. > > > > > The former is denominated by cardinal numbers. That later by some > > > form of measure. Google <measure theory>. The simplest sorts of > > > measures are length, area, volume etc. and generalizations thereof. > > > > > > Mathematicians do not confuse these two kinds of quantities, but > > > you do, therefore you are not a mathematician. Your inability to > > > learn from your betters indicates strongly that you will never > > > learn to be one. > > > You have no clue what I'm doing here, do you Bob? > > One might guess trolling, since it is certainly neither mathematics nor > logic. > > The whole point is > > to bring these various disparate measures that don't get along, and > > turn them into a cohesive team that can discriminate all types of > > sets accurately. > > So how many pounds tall are you, TO? If I told you I was 6'5" and weighed 47 pounds, would you believe me? How about 3'11" and 475 pounds? Is there no relation whatsoever? Chances are, you are a fat set of cells, at least the top portion of the set. > > > > > > > > Before you criticize well established mathematical theories you > > > should learn what they are first. You have not done this and you > > > show no indication of ever doing so. > > > I know what they are, but they don't make sense. > > That TO claims they do not make sense to TO has several possible > interpretations, one of which is that TO is too stupid to comprehend, > another is that he is merely trolling. My own opinion is that it is a > mixture of these two. The third possibility is that they do not actually make sense in the big scheme of things. > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 12:36 Virgil said: > In article <MPG.1dcabbdb7bb59b1d98a582(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > > There is no such thing as 'the' size of the set, there are many > > > sizes, depending on which proprties one is looking at. But only the > > > cardinality size can be applied to every set. all other types of > > > size are limited in what sets they can compare. > > > Do you have one tool in your tool drawer? Is it a hammer, a > > screwdriver, or a wrench? Probably just a rock. > > Cardinality works for all sets. TO's "sizes" do not. > > Better having one tool that always works than any number of broken tools. So, yes, it's a rock. And when you break it, then you have TWO tools in your cave. Good, Virgil. Very good. > > > > > > > > > I hope that's okay, even if it sounds like babbling hogwash. Does > > > > deep hogwash run still? Hmmm.... > > > > > > If TO is content to wallow in his hogwash, ... > > > Come wallow, O Virgil, for the sun is hot, and the hogwash is > > refreshing! Cool thy meaty rump in the mud, as you bask in the > > afternoon breeze. :D > > The difficulty is that I cannot cross the gap between into TOmatica, and > TO's pond of hogwash lies in TOmatica, and lies, and lies. The use lie groups and take infinite steps. The sun is going down, and your tail is crusty. > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 12:47 Randy Poe said: > > Tony Orlow wrote: > > Randy Poe said: > > > Since f:N->E is both injective and surjective, it is bijective. > > Okay, thanks, how about a proof of bijection between the natural numbers and > > the binary strings? > > Which naturals, ours or yours? Which binary strings? > > How about the one I claim elsewhere exists, the bijection > between *N (or its binary representation), and P(N)? > > I like to pin things down: > > Defn: An "infinite bit string" is a map s:N->{0,1} so that > every natural number n in N is associated with either 0 > or 1. The "n-th" bit is s(n), also written s_n. > > So I've got exactly one bit for every n in N. When talking > about "infinite strings" I want to be specific about just > what labelling set I'm using to label my bits. (Being > ambiguous about that is exactly where the fallacy in your > "bijection" lies). Ha ha. That's a good one. You are stepping into the trap to which you refer, right now. > > Do we have a definition of *N separate from the bit strings? > I'm not sure. So let me just say that *N will be the set of > all infinite bit strings under the above definition of > infinite bit string. So, you construct a bijection by declaring them the set to be the same thing, but I am not allowed to do the same? This is precisely the mapping I offered, and it was rejected. Why? > > Consider the obvious map to subsets: f:*N->P(N), where > f(s) = {n in N: s_n = 1}. So s_n = 1 if and only if > n is in f(s). Yes, that's the mapping I used. > > There is also the obvious inverse g:P(N)->*N. Let w be > any element of P(N), i.e. w \subset N. Then g(w) is > an infinite string with g(w)_n = 1 if n in w, 0 otherwise. Yes, this is obvious. > > Claim: f is surjective. Proof: g(w) exists and is a valid > infinite bit string for any w in P(N), and f(g(w)) = w. Thus, > for any w in P(N), there exists s in *N such that f(s) = w, and > f is surjective. You can't just declare your conclusion, can you? Prove that g(w) is valid for all w and that f(g(x))=w. You haven't proven this. So, I guess I can't be asked to prove any such thing. I can just say "voila!", like Virgil, or just "say" that *N is the same as the set of infinite binary strings, without proof. Is that okay? I don't diagree with this part of the proof. But, I want it done right. If you are going to complain when I say something is so by the definition I gave, then you cannot do the same. > > Claim: f is injective. Proof: Consider s and s' such that > f(s) = f(s'). That is, for every n in N, (n in f(s)) > if and only if (n in f(s')). But then (s_n = 1) > if and only (s'_n = 1). So s = s'. > Since f(s) = f(s') => s = s', f is injective. Okay. > > f is a map from *N to P(N) which is surjective and injective. > Therefore f is a bijection from *N to P(N). But you have not covered all of *N. If all n in N are finite, then |N| is finite, and therefore |P(N)| is finite, but *N is infinite. you run out of finite sets to map to your infinite numbers. Anyway, thanks for an example of how you do things, even though you would not have accepted such a proof from me, for the reasons I gave above, if not for more unreasonable reasons. > > - Randy > > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 12:53
Virgil said: > In article <MPG.1dcaca55cb16499798a584(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > If I am bijecting *N and P(*N), which was my original bijection, then *N > > contains infinite elements with infinite bit strings. *N is not limited to > > finite values. > > To keeps claiming that he can bijects *N to P(*N), but never demostrates > the validity of that claim, merely reitierates it. As if repetition will > make it eventually valid. > > > Yes, if P(S) has N members, then there must be log2(N) elements in S. Very > > good. Keep in mind, I am by no means trying to say the power set is the same > > size as the set, simply that there is a bijection. > > TO has things backwards. Sameness of "size", at least as TO defines it, > implies bijectability but not conversely. > > TO cannot present any instance of a provable bijection between sets of > unequal "size", but there are many bijections between sets of unequal > "sizes". For example x -> 2*x bijects [0,1) to [0,2), even though he > latter set is twice the size of the former. > > > You are not forced to conclude that this prevents a bijection. > > > Yes we are! > > > The entire > > infinite set would be the final subset enumerated, but the set does not end, > > nor does the set of its subsets, so the bijection will never, ever get to > > this > > point. > > Then it is not a bijection. > > In order for a mapping to be a bijection, it must "get to" every point! > > Or perhaps any old mapping can be called a bijection in the strange > world of TOmatics. > > Outside of TOmaics, a function from a set S to a set T is a bijection > ONLY IF for every s in S, f(s) is a unique member of T, AND for every t > in T, there is a unique s in S such that f(s) = t. > > TO's alleged "bijections" do not meet these criteria, so that, outside > TOmatics, they are not bijectinos at all. This bijection certainly meets those criteria, for every element you can name in P(*N), besides the very last member, the entire set *N. None of your bijections for unending sets ever name a mapping for a last element. There is no such thing. > > > You used "TOmatics" fairly well. What if we apply it to the evens? > > Whatever n you choose, you have 2n, but 2n is a natural too, so you have to > > map > > it to 4n, etc. If you were to assume any such "last element" > > TO is the only one making that false assumption of "last element". Right, as if the entire set *N is not the last element of P(*N). Wake up, Virgil. The splinter you perceive in my eye is but a reflection of the log in yours. > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/ |