Cantor Confusion
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From: mueckenh on 14 Jan 2007 08:05 Dik T. Winter schrieb: > 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. I know that you cannot but. Regards, WM
From: mueckenh on 14 Jan 2007 08:15 Dik T. Winter schrieb: > > 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. No proof is given for the Lemma. > > > 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! 0.111... and its initial segments are *representations* (according to you, they cannot be numbers). They can represent natural numbers as well as certain real numbers. Obviously both sets of numbers are isomorphic. > 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. This is not required. Only initial segments of N and of 0.111... are required. > I have two > objections to this. > (1) There is no difference between the mapping of 0.1 and 0.10. No problem. > (2) A single real number can map to two different sets of natural > numbers. Consider 1/2 = 0.100000... = 0.0111... . Yes, some real numbers can do. Therefore there are more paths in the tree than real numbers in [0, 1]. > > > 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. When all nodes are there , then all paths are there. Or the set of digits of Cantor's diagonal is not an infinite number. > > > 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. I take the union of all nodes. That guarantees the presence of all paths. Or the set of digits of Cantor's diagonal is not an infinite number. > > 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. But every path of the left tree is contained as an initial segment in a path of the right tree. Regards, WM
From: David Marcus on 14 Jan 2007 13:48 mueckenh(a)rz.fh-augsburg.de wrote: > Here is the 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}. > > 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.... (*) > > 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.... > > A real number r of the real interval [0, 1] can be represented as one > (ore two) path in the infinite binary tree. The set of all real numbers > r of the real interval [0, 1] is then given by the infinite binary > tree: > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > ................... > > A finite binary tree is the infinite binary tree, cut off below a level > n with n in N. > Here is a tree with two levels: > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > > namely level 1 and level 2. (The root at level 0 is conveniently not > counted, because 0 is not a natural number.) > The union of binary trees is defined as the union of levels. > The union of two or finitely many different finite binary trees simply > is the largest on. > 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. If by "in this union", you mean in the infinite tree, then OK. If you mean each infinite path is in one of the finite trees, then you haven't proved this. I'll assume you mean the former. > We can > see this by the path always turning right, 0.111..., which is present > in the tree, according to (*). > > Conclusion: Every finite binary tree contains a finite set of path. The > countable union of finite sets is countable. OK, I'm with you up to here. > The set of paths is countable. Do you mean the set of infinite paths in the infinite tree? If so, you haven't proved this. Please provide the missing steps in your proof. > The set of real numbers in [0, 1] is countable. QED. -- David Marcus
From: Franziska Neugebauer on 14 Jan 2007 14:23 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> >> > This and your further questions are answerd by the following >> >> >> > texts. >> >> >> >> >> >> I did not pose any questions. I have informed you about the >> >> >> fact that there is no time and hence no temporal process _in_ >> >> >> math. >> >> > >> >> > Be informed then that mathematics is in time. >> >> >> >> From that it does not follow that there is a notion of time in >> >> mathematics. >> > >> > A failure >> >> Thanks again for confirming that there is no time im mathematics >> _now_. > > There is no time in what you may erronesously understand by > mathematics. It is not mathematics what you erroneously assign to mathematics as can be easily seen from the current http://de.wikipedia.org/wiki/Benutzer_Diskussion:W._Mueckenheim http://de.wikipedia.org/wiki/Spezial:Beitr%C3%A4ge/W._Mueckenheim discussions. [...] > ==================================================== Aha you cut and paste from different sources. "But an infinite ordinal ..."-part is from ,----[ <45a6c3de$0$97239$892e7fe2(a)authen.yellow.readfreenews.net> ] | > mueckenh(a)rz.fh-augsburg.de wrote: | > | > Franziska Neugebauer schrieb: | >> Virgil wrote: | >> | >> > In article <1168511743.391940.58070(a)77g2000hsv.googlegroups.com>, | >> > mueckenh(a)rz.fh-augsburg.de wrote: | >> [...] | >> >> Conclusion: Every finite binary tree contains a finite set of | >> >> path. The countable union of finite sets is countable. The set | >> >> of paths is countable. | >> > | >> > The set of /finite/ paths in the union is countable. | >> > But when one takes the union of sets of finite paths one only | >> > gets finite paths in that union. There are no infinite paths in | >> > that union. | >> > | >> > The same thing happens with ordinals. When one takes the union of | >> > all finite ordinals (like unary trees), there is no infinite | >> > ordinal IN that union | >> | >> Absolutely right. | > | > But an infinite ordinal is that union! | | There is no infinite number (ordinal) _in_ that union. Rephrased: | There is no infinite _member_ in that union. `---- >>> But an infinite ordinal is that union! > >> There is no infinite number (ordinal) _in_ that union. Rephrased: >> There is no infinite _member_ in that union. > > The infinite ordinal is that union. So what? > The infinite diagonal of Cantor's list is the unioin of all its finite > digits. So what? "But that object ..." is originally from <45a6c417$0$97239$892e7fe2(a)authen.yellow.readfreenews.net>. You have replied to an answer to that post: >>> An infinite union of finite sets is an infinite object. > >> But that object does *not* have an infinite _member_. > > Why should it? Complete it reads: ,----[<45a6c417$0$97239$892e7fe2(a)authen.yellow.readfreenews.net> ] | > mueckenh(a)rz.fh-augsburg.de wrote: | > | > Virgil schrieb: | > | >> In article <1168511880.370120.180940(a)p59g2000hsd.googlegroups.com>, | >> mueckenh(a)rz.fh-augsburg.de wrote: | >> | >> > Here is the formal proof: | >> > | >> > Theorem. The set of real numbers in [0, 1] is countable. | >> | >> Your proof was neither formal nor valid. | >> | >> Among other things you invalidly assume an infinite the union of | >> finite sets must contain an infinite object, rather than merely | >> containing infinitely many finite objects. | > | > An infinite union of finite sets is an infinite object. | | But that object does *not* have an infinite _member_. `---- So your question "Why should it?" is easily answered: It should not, but as Virgil has pointed out: "you invalidly assume an infinite [...] union of finite sets must contain an infinite object". F. N. -- xyz
From: Virgil on 14 Jan 2007 14:50
In article <1168779322.952374.232460(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Which ones of those specifically "rooted finite paths" does WM claim > > contain infinitely many nodes? > > Which segment {1,2,3,...,n} of N contains infinitely many elements? > Is N the union of all initial segments? If one makes a claim about finite initial segments of N, that claim need not be true for anything that is not a finite initial segment of N. And N is not a finite initial segment of N. > > > > Each of the sets {1}, {1,2}, {1,2,3}, etc., is both a member and a > > subset of N, but N is not. > > And, in addition, it is not green. Yes. Relevance? So that WM claims every property common to all finite initial segments of N must also be a property of N? How about finiteness? > > > > Even the diagonal of the EIT is no longer infinite, if it is > > > inconvenient for set theory? > > > > As properly described, that diagonal is actually infinite however > > inconvenient that may be for WM's version of set theory. > > This diagonal (infinite path) is the union of all digits (nodes) of the > initial segments and the limit of the sequence of the initial segments. The union of all finite initial segments of N is not a finite initial segment of N. > > > > You cannot "unary represent" every real in [0, 1]. You can unary > > > represent every natural number, and you can unary represent real > > > numbers which contain only numerals a_n = const for every n in N > > > (behind the decimal point), like 0.111 or 0.1111111. In addition you > > > can unary represent omega or N = {1, 2, 3, ...}, namely as 0.111... = > > > {0.1, 0.11, 0.111, ...}. Notice that this set does not contain infinite > > > strings of 1. > > > > Then why does WM keep claiming that it does for trees? > > Because it either holds in both cases or in none. WM keeps claiming that the union of all finite intitial segments of N is a finite initial segment of N. Not in ZF or NBG. > > > Then WM's infinite path IS the "infinite" tree, not a member of it. > > Of course. The set of trees is the forest. So WM wants the whole forest to be one just one tree? |