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From: David Kastrup on 21 Oct 2005 14:18 Tony Orlow <aeo6(a)cornell.edu> writes: > David R Tribble said: >> Tony Orlow: >> >> I already showed you the bijection between binary *N and P(*N). >> >> What didn't you like about it? It is valid. >> > >> >> David R Tribble: >> >> No, you showed a mapping between *N and R, which is equivalent >> >> to a mapping between *N and P(N). That's easy. >> > >> >> Tony Orlow: >> >> What do you want me to try, anyway, and infinite mapping, >> >> element-by-element? A bijection's a bijection, right? >> > >> >> David R Tribble: >> >> Yes, that would be nice. Please show us your bijection. >> >> Tony Orlow: >> >> f(0) = ...000 = {} >> >> f(1) = ...001 = {0} >> >> f(2) = ...010 = {1} >> >> f(3) = ...011 = {0,1} >> >> f(4) = ...100 = {2} >> >> f(5) = ...101 = {0,2} >> >> f(6) = ...110 = {1,2} >> >> f(7) = ...111 = {0,1,2} >> >> etc. Any questions? >> >> David R Tribble: >> >> Where are the infinite subsets, such as the set of even numbers? >> > >> >> Tony Orlow: >> > Well, infinitely far down the list of course! >> > . >> > . >> > . >> > f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} >> > f(2N/3)= 0:101010.....101010 = {1,3,5,7,9,....} >> >> All of the sets you listed, finite and infinite, are composed of >> only finite naturals. Which means that they are all subsets of N. > Incorrect. See the digits to the left of the ellipses? Those are in infinite > positions, representing infinite evens and odds. >> >> You're using N (and N/3, 2N/3, etc.), which is a member of *N. > Yes we are talking aboout *N. >> >> So all you're doing is showing a bijection between *N and P(N), which >> is what I said you were doing in the first place. That's easy, >> and is equivalent to a bijection between R and P(N). > No, it's P(*N). The sets include elements of infinite value. Then you should be able to tell us whether e, the index of the set of even numbers E such that f(e)=E, is an element of E. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: imaginatorium on 21 Oct 2005 14:24 Tony Orlow 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. > > 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. 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? > 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 I "assign a set size"? Sorry, Tony, my brain is getting old and tired. Kindly fish out for me the place in my proof where I "assign a set size"... Just for the hell of it, here's the proof, with boxes [ ], so you can just put a tick where I "assign the size". Note that the proof does not include the words "finite" or "infinite": it depends only on the simplest properties of all sets - basically that they have members, and something either is or is not a member. Consider an arbitrary set A. [ ] (Thus any entity in the universe [proper class] of all entities either is or is not a member of A [ ].) Consider the power set of A. [ ] (The set of all subsets of A, including A itself; thus for any particular one of these subsets B, any element of A either is or is not a member of B [ ].) Now consider any mapping f from A into P(A) [ ]. (At least one such mapping exists [ ]: the canonical map that takes any element x e A [e means 'element of' OK?] to the subset {x} e P(A). So no worries about existence.) We want to show that whatever mapping f is chosen, there will be at least one element q of P(A) (q is a subset of A, right?) which the mapping f does not map anything onto. [ ] Consider an arbitrary element x e A [ ], and suppose f(x) = y e P(A). [ ] Well, y is a subset of A, so x either is or is not an element of y. [ ] Thus we can divide the elements of A into two classes: [ ] CLASS I: those which map to a subset of which they are a member [ ] CLASS II: those which map to a subset of which they are not a member [ ] Call w the subset of A which includes exactly those elements in CLASS II. [ ] [-----end prologue-----] Now I assert that the particular mapping f we are considering at the moment does not map any element of A to the subset w in P(A). [ ] Every element p inside w is mapped to a subset that does not include p, so it cannot be mapped to w. [ ] Every element p' outside w is mapped to a subset that _does_ include p', which thus includes an element outside w, and thus p' cannot be mapped to w. [ ] Therefore the mapping f is not a surjection. (We called that "onto" when I was doing this stuff.) Thus there is no bijection. But anyway... > 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. Crapola. You are at least getting to the point of reiterating what would have been the natural view before Cantor pointed out it was wrong. But again you imply that I would go around "finally declaring" sets. I deny this very strongly. I have never wittingly "finally declared" a set in my life, and I demand an apology. Sorry - deleting the digital fixation babble. Brian Chandler http://imaginatorium.org
From: imaginatorium on 21 Oct 2005 14:34 Tony Orlow wrote: > Daryl McCullough said: > > Tony Orlow says... > > > > >stevendaryl3016(a)yahoo.com said: > > >> Here it is once again: .... ah, it's that proof again... > > >> Let A be any set whatsoever, finite or infinite, it doesn't matter. Good. I choose the set of all topologically distinct polyhedra. > > >> 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. > > > > You mean A. Yes, that's what it means to have a > > surjection f between two sets A and P(A) (a bijection > > is a kind of surjection). It means that for every w in P(A) > > there exists a y in A such that w = f(y). If y is not in > > A, then the fact that w = f(y) is irrelevant to whether > > there is a bijection between A and P(A). > But, is S the COMPLETE set, and if so, how many bits does it have? One, old chap. Any polyhedron has exactly one bit. The bit inside the polyhedron, as opposed to the bit outside the polyhedron. This is set theory, not junior computing, you know. <snip> <babble babble last natural declare victory viagrra cheap spam> > And what does that prove? That any attempt to place a range on an infinite set > and declare it finished causes a contradiction. Yep. Absolutely. That's why people who have outgrown your mediaeval imponderables never "place ranges" nor "declare finished"... Brian Chandler http://imaginatorium.org
From: Ross A. Finlayson on 21 Oct 2005 14:40 The expansion of the real number in the unit interval is semi-infinite. It has a beginning, and no end, where infinite means no beginning and no end, like a circle. Consider if there could be base one, or base infinity. In base one, integers, are represented by tally marks, eg IIII is four. The Romans introduced a method of tally marks with various replacements and positional systems, eg V is five, IV is four, and MMV is 2005, Roman numerals. Those are sufficient for representing positive integers. The unary, base one number however are not so great in representing numeric values in the expansion, which is basically the most well-known contemporary method of representing those number beside fractions, generally "decimal" with the "decimal point". Returning to four, IIII, that's basically four marks out of the totality of marks there could be, in a semi-infinite sense, all the marks, 4/oo, 4/infinity. The radix in the expansion is basically the separation between integral values greater than or equal to one, and less than one and greater than or zero, with zero on both sides of the radix. It's more than that. So, a similar notion can be considered, lightly, in representing the values between zero and one with only unary, only succession. Then, IIII is basically four, four, divided by infinity, some kind of "infinite value", an infinity, the infinity of the natural numbers, where the order type of ordinals is an ordinal. A similar thing happens with the infinite alphabet, base infinity, .4 is ... 4/oo. This is basically leading towards the antidiagonal argument, in this unary or infinital notion, the only way to form a list with a rule that constructs for each list, including a finite list, a new element is to introduce the successor onto the list. If the successor is on the list, and inductively each of its successors, then the list is complete. The expansion in base two has no end. So, very differently than base one or base infinity, if that's even the correct word to call those things, having something along the lines of: ..000 ... (infinitely many zeroes) ... 001 Does not, per se, exist. ..000 ... (finitely many zeroes, more than can be counted incrementally by the fastest computer running for a billion years) ... 001 Does. Basically, that sequence is so absurdly long that for any medio-scale process, for many purposes it is infinite. Ross
From: Randy Poe on 21 Oct 2005 14:42
imaginator...(a)despammed.com wrote: > Tony Orlow wrote: > > imaginatorium(a)despammed.com said: > > > 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.) > > Yes, your reiterations of the standard nonsense don't go very far with me. > > 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? I think, though he is unable to articulate it, that what Tony is rejecting is the Axiom of Infinity. So he's grasping toward a system without one, and replacing "the set of natural numbers" with some vague, ever shifting thing whose membership is never the same twice and always has just enough members for whatever you're using it for at the moment. At any rate, since we work with a system in which AofI is included and Tony rejects it, that seems to be pretty much the end of the story. Of course there's no hope of convincing Tony of anything, least of all that he contradicts himself sometimes within the space of a single sentence. For at least 4000 posts the game instead has been: - see how long this thread can go, - try to goad TO into ever more ridiculous proclamations - try to figure out what TO's axioms are - Randy |