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From: Tony Orlow on 31 Oct 2005 13:45 William Hughes said: > > albstorz(a)gmx.de wrote: > > > You misinterpret totally when you say, I think there must be an > > infinite natural number. I don't think so. I only argue that, if there > > are infinite sets, there must be infinite natural numbers (since nat. > > numbers are sets). > > > > OK. Make the substitutions, natural numbers = Greeks, sets = mortals > and > infintite = German [1] > > Then the statment "if there are infinite sets, there must be infinite > natural numbers (since nat. numbers are sets)" becomes "if there are > German mortals, there must be German Greeks (since Greeks are > mortals)". > Aristotle must be rolling in his grave. > > -William Hughes > > [1] Deutchland, Deutchland, ueber alles > > Come on William. If Albrecht is stating that sets are natural numbers and natural numbers are sets, he is not saying a natural number is a KIND of a set, like Greeks are a type of mortal. This is disingenuous. Sorry. -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Randy Poe on 31 Oct 2005 14:00 Tony Orlow wrote: > Randy Poe said: > > > > Tony Orlow wrote: > > > David R Tribble said: > > > > We are forced to conclude that there is no natural s that maps to > > > > *N, and that therefore your mapping scheme is not a bijection > > > > between *N and P(*N). > > > You are not forced to conclude that this prevents a bijection. 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. 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. > > > > The objection above is that there exists at least one element > > of P(*N) which is not equal to f(s) for any s in *N. That > > objection says nothing about "last". > Ahem. Is the entire set *N not the last element in the ordered set P(*N)? No. I don't believe endless sets have an end. You are the only one who can't get through a sentence about infinite sets without talking about their ends. > In my > biojection it is, and it's the entire set is generally considered the last > element of the power set, "Generally" people don't talk about the "last element" of any infinite set. Generally, the entire set can be considered the largest element of the power set. But so what? A bijection proof doesn't need to worry about "last", just "every". And bijections certainly don't need to be order-preserving. Most aren't. Once more for the proof-blind: - you're not allowed to call something a bijection if it leaves something unmapped. - you left something unmapped. > the null set being the first. So, at least in the > case of my bijection, this is exactly what that missing mapping translates to, > and it is probably provably always the case. The "missing mapping" between a set and its power set depends on the map. The set is w = {x in X: x not in f(x)} Do you really not understand that there are infinitely many maps possible between a set and its power set, and so infinitely many different possible sets w? > > The same objection can not be raised about f:N->E, f(x) = 2x. > > Every element in E is mapped by some element in N. Every > > single one. Nobody but you says you have to consider "the > > last" in order to justify the claim that this statement is > > true for every single member of the set of evens. > But you are saying that for my bijection, effectively. Here's what I'm saying, not "effectively", but outright. Every element in E is mapped by some element in N in the map f:N->E, f(x) = 2x. Every element in P(*N) is NOT mapped by some element in *N in your map f:*N->P(*N). Is. Isn't. Why is this so difficult for you to see? There is a requirement for bijections which you haven't met. I don't care what your excuse is for not meeting it, the fact is you haven't met it. Guilty as charged. Pay the fine. > > The surjection part of my simple proof of the bijectivity of > > f(x)=2x had this simple structure: Let y be any element of the > > codomain. Then there exists x such that f(x) = y. > > > > If you wanted to establish surjectivity of your map f:*N->P(*N) > > you need the same structure: Let y be any element of P(*N). > > Then (TO can show) there exists x in *N such that f(*N) = y. > That bijection is established by the mutual bijection between the infinite > binary strings and these two sets. If Randy can just "say" that *N is the same > as the set of all infinite binary strings, then why can't I? You are certainly allowed to define the meaning of the symbol *N. I didn't see one, so I adopted a definition. If you have another, which is more precise, by all means put it forward. But of course I didn't say this: "the set of all infinite binary strings". I first defined "infinite binary strings" as having exactly one bit for every finite n in N. You don't get the difference between the start and the end of a proof, do you? The start is the definition of terms. The end follows after deduction. I'm "allowed" to define a symbol. I'm not "allowed" to define a mapping as bijective. Bijection requires proof. There's supposed to be some of that REASONING stuff you are so averse to before you declare a bijection. If you want to talk about a different set, feel free. But in defining your infinite binary strings, tell me what set labels the bits. Then we will have a common basis. > And, certainly, if > I am representing the power set as a set of binary strings, there is a bit for > every one of the infinite number of naturals in *N OK, you're having trouble seeing the structure of the proof. Let me outline what's happening here. We have two sets which we want to establish a bijection between, using infinite strings. If the strings are defined so that they are automatically bijected to one set, then the bijection to the other set needs to be proven. A <-> strings <-> B In my approach, I defined the infinite strings as being my set *N, and then had to go through a process of prove to establish the correspondence to P(N). You're going the other way. You define the strings in question as a representation of P(*N). That makes that side of the bijection automatic. So the side you have to prove is a bijection between YOUR strings (not the same as my strings) and *N. Either of us is allowed to define symbols so that one side of the bijection is automatic. What you aren't allowed to do is to claim that both sides of the bijection are true "by definition". You realize that your strings are not my strings, right? I defined my strings because I needed a definition of *N to work with. Do you have a definition of *N separate from the strings? What set labels the bits of a number in *N? There certainly is no proof having to do with *N, without a definition of *N. - Randy
From: William Hughes on 31 Oct 2005 14:02 Tony Orlow wrote: > William Hughes said: > > > > albstorz(a)gmx.de wrote: > > > > > You misinterpret totally when you say, I think there must be an > > > infinite natural number. I don't think so. I only argue that, if there > > > are infinite sets, there must be infinite natural numbers (since nat. > > > numbers are sets). > > > > > > > OK. Make the substitutions, natural numbers = Greeks, sets = mortals > > and > > infintite = German [1] > > > > Then the statment "if there are infinite sets, there must be infinite > > natural numbers (since nat. numbers are sets)" becomes "if there are > > German mortals, there must be German Greeks (since Greeks are > > mortals)". > > Aristotle must be rolling in his grave. > > > > -William Hughes > > > > [1] Deutchland, Deutchland, ueber alles > > > > > Come on William. If Albrecht is stating that sets are natural numbers and > natural numbers are sets, he is not saying a natural number is a KIND of a set, > like Greeks are a type of mortal. This is disingenuous. Sorry. Note that Albrecht makes the statment "since natural numbers are sets" The two statements "since natural numbers are sets" and "since natural numbers and sets are the same thing" are different. If the second statement is meant, Albrecht should say so explicitely. (Other statements by Albrecht are are contradictory on this point, at times he seems to claim that any set is also a natural number, at other times he backs away from this claim.) -William Hughes > -- > Smiles, > > Tony > http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 14:27 David R Tribble said: > David R Tribble said: > >> [Alrecht] are claiming that one of the members of the infinite set is > >> equal to the size of the set. This is not obvious to us because it > >> contradicts proven set theory. It is your responsibility to prove > >> that your claim is true, and that standard set theory is wrong. > >> Saying that your claim is "obvious" is not a proof. > > > > Tony Orlow wrote: > >> His claim is obvious based on the diagram he offered, which graphically > >> shows the element values along one side of a growing square, and the > >> element count along the other side. They are obviously, graphically, > >> inductively, always equal. One is not infinite while the other is finite. > >> It's impossible. > > > > Here's his second diagram: > > > >> # O O O O O O O O O ... 1 > >> # # O O O O O O O O ... 2 > >> # # # O O O O O O O ... 3 > >> # # # # O O O O O O ... 4 > >> # # # # # O O O O O ... 5 > >> # # # # # # O O O O ... 6 > >> # # # # # # # O O O ... 7 > >> # # # # # # # # O O ... 8 > >> # # # # # # # # # O ... 9 > >> . . . . . . . . . . . > > > > Every natural is represented by a horizontal string of #'s. The vertical > > string of zeroes directly to the right of the last # in each natural denotes > > the size of the set through that natural. With the addition of each natural, > > both the horizontal number of #'s and the verticle number of 0's are > > incremented in tandem, so this equality is preserved. The slope of -1 shows > > that there is a 1-1 correspondence between element value and set size. > > This is a good graphical illustration of what I have been trying to say > > regarding the naturals. You only have an infinite number of them when you > > allow infinite values for them. > > If that's the case, at what point do the naturals on the right side > stop being finite and start being infinite? Oh geeze. Here we go again. "No largest finite!!!" No kidding. There is no single step where that happens, as you well know. Let me ask you this. At what point does the count of naturals become infinite? That's the point at which the element values become infinite. (sigh). > > All I see is an infinite list of rows, where each row is labeled by > a finite natural k and contains k leading # marks and an infinite > number of trailing O marks. > > >From the other angle, I see an infinite list of columns, where each > column contains a finite number of O marks on top and an infinite > number of # marks on the bottom. So, you cannot tell that the finite #'s are equal always to the finite 0's? You can't see that the #'s represent the value and the 0's represent the count? Look again. > > > But why can't there be an infinite number of finite naturals? Because the count, the number of 0's in the column, cannot be infinite unless the #'s, the element value, is also infinite. Look at the picture. > Do you have some kind of proof of this? You have never offered one. I have offered several. Lett me dredge one up for you. > You've offered plenty of statements about "unidentifiable largest > naturals" and "set ranges" and "unit infinities", but you've never > actually pinpointed how you get from the finites to the infinites. Here's a nice long one, with plenty to chew on: ------------------------------------ With a set of symbols of size S, one can create a set of unique strings of length L, whose maximum size is N=S^L. Among the Axioms of Finiteness (which probably exist already, somewhere): For A and B finite (non-zero) and infinite, A^B or A*B is either infinite or finite according to the following table- A Finite Infinite ----------------------------------- B | Finite | Finite Infinite Infinite | Infinite Infinite Note that this table is equivalent to the OR truth table, when you replace "finite" with 0 and "infinite" with 1. In other words, the result is infinite if either A OR B is infinite, and is finite if A AND B are finite. When we combine these two "axioms", we see that N=S^L is only infinite for either infinite S or L. Therefore, reapplying what our symbols mean to "reality", we can only have an infinite set of strings if we either have an infinite set of symbols with which to build them, or allow infinite length strings. Now we introduce the digital number systems, defined as usual, with S equal to the radix, a finite whole number greater than 0, the symbols in the set representing the whole numbers from 0 to the radix minus 1, and each string position representing a power of the radix that increases going leftward, with a digital point directly to the right of the digit place representing the zero power of the radix. The value represented by the digital number, V, is given by V=sum(i=-oo->+oo: a_i * b^i), where a_i is the digit at digit place i, and b is the radix. That sounds about right, doesn't it? Did I miss anything? Now, if i is infinite and positive for a given digit, then it is infinitely far to the left of the point, and represents the radix to an infinite power, right? So, given our axiom of finiteness, we can say that a finite radix to an infinite power gives an infinite result, right? And, if a_i is non-zero, then we can also use our axiom of finiteness to say that this infinite b^i multiplied by the finite non-zero a_i, yields an infinite result for V, right? Finally, I think we can agree that, if all digits in infinite digit places are zero, then it is equivalent to having a finite number of digits, not an infinite number, so if we require an infinite number of digits, that means we require an infinite number of non-zero digits. So, we have three axioms (actually it needs to be about 10, right, to cover the digital numbers ?): 1. N=S^L 2. A^B and A*B are only finite for finite A and B, and otherwise infinite. 3. A digital number with nonzero digits in digit places infinitely to the left of the point represents an infinite value. So, infinite(N)->infinite(S) OR infinite(L) by 1 and 2 NOT(infinite(S))->infinite(L) by definition of radix infinite(L)->infinite(V) by formula for V Hopefully this is step-by-step enough to get SOMEBODY to understand this line of reasoning. ------------------------------- Here's another, shorter one: This proof relies on some assumptions. Let me know if you disagree with any of them: (1) All naturals are finite. (you already "proved" this one) (2) The number of strings of length L that can be constructed from a set of symbols of size S (set size, not symbol size, remember (sigh)), is S^L. (3) For finite A and B, both A^B and A+B are finite. Are we good to go? Okay. Proof that f(n), the number of strings in the set of all strings up to and including length n in N, on a finite alphabet of size S, is finite: 1. For n=1, we have S^1 strings of length 1, for a total of S strings less than or equal to 1 in length. f(1)=S is finite, as stated. 2. For a finite number of srings of length n or less, we can add S^(n+1) strings of length n+1, to get f(n+1),the number of strings of length n+1 or less. S is finite and n+1 (a natural number) is finite, so S^(n+1) is finite. S^(n+1) is finite and f(n) is finite, so f(n+1)=f(n)+S^(n+1) is finite. 3. Therefore, for all n in N, f(n), the number of strings up to and including n symbols, which are created from a finite alphabet, is finite. There is no n in N, in other words, which maximum length of string will allow you to have an infinite set of strings. > > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/
From: Tony Orlow on 31 Oct 2005 14:37
David R Tribble said: > David R Tribble said: > >> So your infinite natural that maps to the entire set *N is: > >> s = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ... > >> s = 1 + 2 + 4 + 8 + 16 + ... > >> > >> Thus s is a sum of an infinite number of finite powers of two (2^p), > >> each term being twice the previous term. So s is your 2^N-1, where > >> N is your "unit infinity": > >> N = 1 + 1 + 1 + 1 + 1 + ... > > > > Tony Orlow wrote: > > The dawn dawns on David Tribble.... > > > > David R Tribble said: > >> (We'll pretend for the moment that s and N are different, and ignore > >> the fact that they are actually the same value.) > > > > Tony Orlow wrote: > > ....and the clouds enshroud the sun. > > > > Do you really think sum(n=0->N:1) is the same value as sum(n=0->N:2^n), when > > every term of the second (except the first) is greater its corresponding term > > in the first? Oy..... > > I know you can't comprehend this little tidbit dealing with infinite > sums, but I'm glad you asked. > > Let: > n = 1 + 1 + 1 + 1 + ... > and: > s = 1 + 2 + 4 + 8 + ... > > Now let's group the partial sums of n: > n = 1 + (1 + 1) + (1 + 1 + 1 + 1) + (...) + ... > n = 1 + 2 + 4 + 8 + ... > > Obviously, we can group each set of 2^p 1's in our infinite sum so > they are identical to the terms in the series for s. So n = s. Your rearrangement violates the rules for infinite series. You are introducing implied terms of 0 in your series, by leaving blanks not correlated with any elements in the first set. If you have N terms in n, then your s (which you also labelled n) has log2(N) terms. n is log2(s). It is obvious that s>n, since all but the first term in S are greater than the corresponding term in n. So, your rearrangement is bad. One can get all sorts of ridiculous contradictory results for infinite series by just misbehavior. > > Looking at it from a TOmatics standpoint, s is an infinite binary > number containing all 1 digits (one digit for each 2^p term). Uh huh. If n=N, then s=2^N-1. > Likewise, n is the sum of an infinite number of 1's, so it must also > be an infinite number composed of all 1 digits (what else could it > be?). So n = s. No, the sum of an infinite number of 1 is not all 1's, but the unit infinity, 1:000...000. Do you somehow think the sum of any number of 1's in binary is a string of 1's? It's not. > > Isn't this fun? I don't know. These review sessions get a little tiresome, but whatever is necessary is just that. > > -- Smiles, Tony http://www.people.cornell.edu/pages/aeo6/WellOrder/ |