Cantor Confusion
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From: William Hughes on 11 Jan 2007 18:56 mueck...(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > The "tallest man" is something that can change. L_D is a line. > > > > A line cannot change. The analogy is not valid. > > > > > > L_D is the name of a line, like a championship title. > > > > A championship title is not the name of a team. > > No. It is an attribute of a team. And the team it is an attribute of can change. > > > "The Chicago Bulls" is the name of a team. > > "The NBA champions" is not the name of a team. > > A championship title refers to a team, but the > > team it refers to can change. The name of a > > team refers to a team, but the team it refers to > > cannot change. L_D is the name of a > > line. The line it refers to cannot change. > > Then use the name "largest line". The line referred to by "largest line" can change. The line referred to by L_D cannot change. The "largest line" is not L_D. > > > > > > > > > > > It is possible to find L_D > > > > > > > > > > It is not assumed, but it is obvious that for every given set of > > > > > natural numbers there is one line containing it. > > > > > > > > But since the set of natural numbers can change > > > > this "one line" can change. It is not L_D. > > > > > > > > > > > > > > > > If you assume actual infinity then L_D > > > > > > does not exist. > > > > > > > > > > It does exist, in potential infinity. But it is not fixed. > > > > > > > > L_D is a line. A line is fixed. L_D does not > > > > exist. > > > > > > It is the line containing the whole set. > > > > Depends what you mean by "the whole set". > > If you mean every element of the diagonal > > that can be shown to exist, then "the line" does not exist. > > If you mean "all elements > > of the diagonal that have been shown to exist", > > then "the whole set" is something that can > > change and "the line" must be something > > that can change. > > So it is. L_D cannot change. Therefore "the line" is not L_D. > > > > In either case "the line" is not L_D. > > In both cases there is a largest line, namely the largest line which > can be shown to exist or the largest line which has been shown to > exist. The first one is obviously a contradiction, it refers to actual > infinity. Hence, our only choice is the second Note that the line referred to by "the largest line which has been shown to exist" can change. Therefore L_D is not "the largest line which has been shown to exist". ..> > > > > Let L_D go from 1 to oo. > > > > This statement is meaningless. > > L_D is a line. Lines do not "go from 1 to oo". > > Not individual lines, but largest lines. > L_D is a line. L_D is not lines. Every putative description you have given for L_D is something that can change. Since L_D is not something that can change, none of these descriptions are valid. L_D does not exist. - William Hughes
From: Dik T. Winter on 11 Jan 2007 20:32 In article <1168445677.447997.154510(a)p59g2000hsd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1168351968.885524.129360(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > No, that does AC (although the statement is indirectly). V=L is the axiom > > > > of constructibility and states that every set is constructible. Most > > > > mathematicians think it is false, but it can not be disproven and is > > > > "consistent" with ZF. It *gives* a construction. > > > > > > But, alas, this construction cannot be communicated. It is and remains > > > a top secret. Otherwise most mathematicians could easily be proved > > > wrong by simply constructing a well-ordering of R. > > > > It can be communicated. > > Why does nobody dare to do so? Serious punishment by the high priests > to be expected? > > > But you are not even willing to look at the > > axiom and its consequences. > > I am not interested in consequences but in the *definition or > construction* of a well-ordering of the reals. I would say, have a look at it. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 11 Jan 2007 21:00 In article <1168511880.370120.180940(a)p59g2000hsd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1168444071.169167.287510(a)i39g2000hsf.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > That still does *not* show that the union of the sets of paths in the > > finite trees contains infinite paths. You still fail to provide a > > proper definition of the union of trees. So I can only guess. > > Here is the formal proof: Oh. A formal proof. > Theorem. The set of real numbers in [0, 1] is countable. > > Lemma. > Each digit a_n of a real number r of the real interval [0, 1] in binary > representation has a finite index n. > r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. Yes, we already know that. No proof is required here. > Proof. > A natural number n can be represented in a special unary notation: n = > 0.111...1 with n digits 1 (the leading 0. playing no role). Example: 1 > = 0.1, 2 = 0.11, 3 = 0.111, ... > In this notation the definition of the set of natural numbers, (1, 2, > 3, ...} = N, reads > > {0.1, 0.11, 0.111, ...} = 0.111.... (*) So you are now defining here 0.111... as the set of natural numbers. A pretty amzing notation, but let it stand. > Note that also the union of all finite initial segments of N, {1, 2, 3, > ..., n}, is N = {1, 2, 3, ...}. Therefore (*) can also be interpreted > as union of initial seqments of the real number 0.111.... But 0.111... is not a real number, you just defined it as the set of natural numbers! But I think I understand what you are doing. You are creating sets of natural numbers based on binary expansions. And so 0.010101... would map to the set of even natural numbers. I have two objections to this. (1) There is no difference between the mapping of 0.1 and 0.10. (2) A single real number can map to two different sets of natural numbers. Consider 1/2 = 0.100000... = 0.0111... . > Taking the uninion of all finite binary trees, we get the complete > infinite binary tree with all levels. All infinite paths representing > real numbers r of the real interval [0, 1] are in this union. We can > see this by the path always turning right, 0.111..., which is present > in the tree, according to (*). But the *set* of paths in the union is not the union of the *sets* of paths in the finite trees. > Conclusion: Every finite binary tree contains a finite set of path. Right. > The > countable union of finite sets is countable. RIght again. > The set of paths is > countable. Wrong. The union of the set of paths of the finite trees is *not* the set of paths in the union of the finite trees. You can not get the union of a sequence of sets of paths by taking the union of individual paths. Consider the following finite trees (I see now that in finite trees you terminate paths with a node). Look: 0 0 / \ / \ / \ / \ 1 2 1 2 / \ / \ 3 4 5 6 On the left-hand side we have two paths, so the set of paths is: {0-1, 0-2} and on the right-hand side we have: {0-1-3, 0-1-4, 0-2-5, 0-2-6} the union of these two sets of paths is: {0-1, 0-2, 0-1-3, 0-1-4, 0-2-5, 0-2-6}. And so is the set of paths in the tree on the right-hand side not equal to the union of the sets of paths in both trees. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 11 Jan 2007 21:21 In article <1168533543.556066.104940(a)p59g2000hsd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > That still does *not* show that the union of the sets of paths in the > > finite trees contains infinite paths. You still fail to provide a > > proper definition of the union of trees. So I can only guess. Let us > > assume two finite trees, where I number the nodes: > > 1 1 > > / \ / \ > > / \ / \ > > / \ / \ > > 2 3 2 3 > > / \ / \ / \ / \ > > 4 5 6 7 > > / \ / \ / \ / \ > > We can easily identify like numbered nodes from both trees with each other. > > And so, with respect to sets of nodes, the union of the two is the tree on > > the right hand side. With a bit of imagination we can identify the edge > > going to the left from 2 in the left-hand tree with the edge going from > > 2 to 4 on the right-hand tree, and in that way see that also with respect > > to edges, the union of the sets of edges is the set of edges of the tree > > on the right hand side. If we accept finite paths (and assume that they > > all start at node 1), there are 6 paths in the tree on the left > > There are 4 paths, because the finite trees have leaves only at the > end. Ok, we have on the left hand side the four paths (shown as a set): {1-2-L, 1-2-R, 1-3-L, 1-3-R} > > and 14 > > paths in the tree on the right. > > There are 8 paths. Again, shown as a set: {1-2-4-L, 1-2-4-R, 1-2-5-L, 1-2-5-R, 1-3-6-L, 1-3-6-R, 1-3-7-L, 1-3-7-R}. > > And in this way we accept that the union > > of the sets of paths is the set of paths on the right. > > > > Do you agree with it so far? > > Yes, except that you mistake edges for paths (or invent leaves). But > that is not important. But with this interpretation the union of the sets of paths is *not* the set of paths in the tree on the right. The set of paths in the tree on the right is missing the path 1-2-L for example, which should be in the union. Actually the set of paths on the right misses *all* paths from the tree on the left. So when you state that there are only 4 paths in the tree on the left and 8 in the tree on the right, I retract my remark. In that case the union of the sets of paths has 12 elements. > > If you agree, there is however one problem with the complete infinite tree. > > *None* of the constituent threes contains an infinite path, and so the > > union of the sets of paths also does not contain an infinite path. > > None of the finite initial segments of N contains N. Is there a problem > with N? If no element of the constituent sets in a union has property A, also there is no element in the union has property A. No path in the sets of paths is infinite, so also no path in the union is infinite. No number in the initial segments is infinite, so also no number in the union is infinite. > > You are wrong. The union of sets of paths is a set of paths were each of > > the paths is element of at least one of the sets that define the union. > > If there is a constituent of the union that contains a particular path, > > that path is element of the union set. If there is none, that path is > > not an element of the union. > > That's not required. The set of infinite paths is the union. Pray revise how union is defined. I think I quoted the definition quite proper. It *is* required. If an element a is in the union of a collection of sets, it *must* be in at least one set from the collection. > > > The relevance is that the uninon {1,2,3,...} is *not* an infinite set N > > > with cardinal number aleph_0, as long as there is not an "infinite > > > number" *in* the set. > > > > But that is just what you can find when you allow the axiom of infinity. > > Why do you refuse to allow it for the segments of paths? Why do you keep confusing "segments of paths" with "sets of paths"? > > Suppose the set X consisting of all the finite natural numbers plus a > > single infinite number. That set does not satisfy the Peano axioms, > > as a set with a largest element *can not* satisfy the Peano axioms. > > Therefore it cannot actually exist. Are you stating that the Peano axioms are nonsense because there is no set that satisfies the axioms? Or what else? > > Simply because there is no successor of the largest element. Now remove > > that largest element. If there is still a largest element, the Peano > > axioms can not be satisfied. However, the axiom of infinity states that > > there is at least one set that satisfies the Peano axioms, and that N is > > the smallest of those sets. You are claiming that there is no such set. > > You claim it for the set of paths. You are in confusion again. I do not claim that there is a largest set of paths. > > > But you do not see this small difference. You > > > claim that the union of all finite numbers or of all finite initial > > > segments {1,2,3,...,n} is the infinite set {1,2,3,...}. > > > > Elaborate on the difference. > > Why do you see it different in case of the union of all finite trees or > paths? You are again in confusion. I do claim nothing about the union of paths, I claim something about the union of *sets* of paths. > > > Therefore you > > > must add, in addition to all finite segments of that very infinite > > > path, something which you cannot describe. Something, however, which is > > > certainly not a finite segment of that path. > > > > I think I was talking about sets of paths. > > representing finite and infinite sequences of numerals of real numbers. What is the relevance? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Andy Smith on 11 Jan 2007 19:45
> > So you can have an infinite set with a first and > last member? > > Sure. Consider the following sets of real numbers: > > {x| 0 <= x <= 1} > {x| 0 < x <= 1} > {x| 0 < x < 1} > Perhaps I should have asked, can you have a "countably infinite" ordered set with a first and last member? Re my suggested example, perhaps zero crossings of sin(pi/x) with -1<=x<=1 is better. That maps the zero crossings of sin(pi.y) at -1, -2, -3 ... to -1,-1/2,-1/3 ... and +1,2,3 ... to +1, +1/2, +1/3 ... It is also antisymmetric so there is an additional zero crossing at 0.0... Each element in the ordered set generated by the zero crossings of sin(pi/x) is > than its predecessor, and < its successor, but the set has a start and an end and can be placed in 1:1 correspondence with the natural numbers (and loosely speaking corresponds to counting backwards to -inf, then the inf point itself (the extra crossing at 0 implied by asymmetry), then from +inf back to 1?). For any zero crossing at a point e<0 there exist (an infinite) number of crossings in <=e,<0 , so it is not possible to count the crossings sequentially up to 0. But there exists a point in the set at 0 (as well as the corresponding positive zero crossings). Is that still OK? -- Andy |