Cantor Confusion
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From: Virgil on 1 Dec 2006 15:38 In article <1164984173.975944.13290(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > O.K. we now know that there is no concept of potential infinity > > in ZF (this comes under the heading "the pacific ocean is wet"). > > However, you have not actually given a defintion of potential > > infinity. Do you have one? > > Take the Peano axioms, in particular the axiom of induction. They show > a precise definition of what a potentially infinite set is. Which version? Either the Peano axioms do not mention sets at all, or they require the existence of a set which injects into a proper subset.
From: Virgil on 1 Dec 2006 15:42 In article <45706268.1020005(a)et.uni-magdeburg.de>, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > On 12/1/2006 1:37 AM, Dik T. Winter wrote: > > In article <456EEA86.20001(a)et.uni-magdeburg.de> Eckard Blumschein > > Oh, for once, try to talk mathematics. By the axiom of infinity the > > set of all naturals is neither hypothetical nor fictitious. > > This axiom combines flawless Archimedean reasoning with an at least > questionable replacement of the notion number by the notion set. When EB presents a completed axiom system from which he can generate mathematics, or at least arithmetic, he may join the lists, but until then he is merely a spectator at mathematics, and not competent to be a judge.
From: cbrown on 1 Dec 2006 17:02 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > The binary tree > > > > > > Consider a binary tree which has (no finite paths but only) infinite > > > paths representing the real numbers between 0 and 1 as binary strings. > > > The edges (like a, b, and c below) connect the nodes, i.e., the binary > > > digits 0 or 1. > > > > > > 0. > > > /a \ > > > 0 1 > > > /b \c / \ > > > 0 1 0 1 > > > .......................... > > > > > > The set of edges is countable, because we can enumerate them. Now we > > > set up a relation between paths and edges. > > > > > Relate edge a to all paths > > > which begin with 0.0. Relate edge b to all paths which begin with 0.00 > > > and relate edge c to all paths which begin with 0.01. > > > > So your relation is supposed to be a function mapping edges -> sets of > > paths, correct? > > Yes. But the mapping is not the usual one (one edge --> one path). The > edges are subdivided in shares. Then your eventual claim that you are creating a bijection between edges and paths (or sets of paths) is false; you are instead showing that there is some mapping between "subdivisions of edges" and paths. That's fine; but it says nothing about whether there are equal cardinalities of edges and paths. > > > > > Half of edge a is > > > inherited by all paths which begin with 0.00, the other half of edge a > > > is inherited by all paths which begin with 0.01. > > > > Why is only 1/2 "inherited"? Why not 1, or 1/3, or any other number? > > Because each of the two sets of paths (such paths which begin with 0.00 > and paths which begin with 0.01) shall inherit the same share. This is > the most simple way to execute justice. > > > Isn't "inheriting" a different function than the original relation > > (which was presumably a function from edges to sets of paths)? > > Yes, it is a new idea not yet entertained. I know this because this > idea leads to a contradiction in ZF and hence would have become popular > already. > How does this lead to a contradiction? The "inheriting" function g is not relating the cardinality of edges to that of paths; it is relating the sum of "subdivisions of edges" to paths. > > How is > > it that now we have a function mapping (paths X edges) -> Q? > > I am not sure what you mean here. g is a function that tells us "how much" of edge e is related to path p; e. g., g(p,e) = 1/2, or 1/16, or 0, etc. So g : (paths X edges) -> Q. > We have a function f mapping the > paths onto the real numbers of he interval [0,1]. The domain of f is > the set of all infinite paths. The function is not injective because > some real numbers have two representations. But the function is > surjective. Fine; and with a little tweaking, we can make f a bijection. We also have that there is a bijection h (edges -> N). You then define g so that for every path p, the limit of the sum over all edges e of g(p, e) = 2. None of this is in dispute. What is in dispute is that you can use g, f, and h to show that there is a bijection T from (edges <-> paths), which is required to support your claim that there is a contradiction here (to whit, that there exists a bijection N <-> edges <-> paths <-> R). <snip> > > To clarify, can you indicate which 2 edges are "related" to the path > > which represents the real number 1/3, which are not related to any > > other path? > > > The binary representation of 1/3 is 0.010101... So the edges related to > this set... No, this is where your terminology breaks down. /Edges/ are /not/ "related to this set" in your scheme. Instead, "subdivisions of edges" (e.g., "a/2" below) are "related to this set". > are the limit of the series which has the following partial > sums of edges (compare the sketch below): > > a > a/2 + b > (a/2 + b)/2 + c > ((a/2 + b)/2 + c)/2 + d > ... > > Since every edge counts one unit, the limit is 2. > Hmmm. So instead of g : (paths X edges) -> Q, you have that g : (paths X edges X N) -> R, and that for all paths p, lim n->oo sum (over all edges e) g(p, e, n) = 2. Whatever. It is not disputed that this limit is 2. What is disputed is that /therefore/ there exists a bijection (or surjection) T : (edges -> paths). How do you propose to use g to construct T? Cheers - Chas
From: Dik T. Winter on 1 Dec 2006 20:52 In article <1164974976.838760.48250(a)j44g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1164878462.838895.276420(a)n67g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > The universe of all sets can grow. Define: "The universe of all sets is > > > called the set of all sets", and you see it. > > > > And that is the confusion. If it is a set it cannot grow, but as Fraenkel > > et al. do not define it as a set it is allowed to grow. > > They talk about the set of all sets, but cannot use it for the well > known reasons. No, they do not talk about the set of all sets. That *you* define it as such does not mean that they are talking about that. > > They do not state > > that a set can grow because they do not state that the universe is a set. > > Here is a book on ZF set theory: Karel Hrbacek and Thomas Jech: > "Introduction to Set Theory" What is the relation with what Faenkel et al. did write? > Marcel Dekker Inc., New York, 1984, 2nd edition. they write: The > letters X and Y in these expressions are variables; they stand for > (denote) unspecified, arbitrary sets. Yes, pray reread. If you do not yet understand it reread again. > Therefore we can denote a set by X and we can say that the set X grows. Wrong, again. Consider X a variable that stands for an unspecified integral number. Can you state that the integral number X grows? No. You can state that X grows, but not that the integral numbers grow. In the above X and Y are variables that denote sets in every instance. But X and Y are *not* sets. > That is nothing else than to say that the number of states in the EC > grows. Of course the number 6 has not gown to 25. But it is simply a > matter of definition, how one interprets "to grow" and "number". In that case, please provide a definition. > > That is not "the set of states". You can talk about "the current set of > > states" or about "the set of states in 1957" or whatever. At least > > mathematically. In mathematics, by definition, a set can not grow. > > Wrong. Read my explanation above. Wrong. Read my explanation above. I am talking mathematics. > > You are, of course, entitled to use another definition, but that will > > not clarify the discussion at all (and you are not using standard set > > theory). > > Hrbacek and Jech teach standard set theory including the fact that in > ZF everything is a set. Yes? Do they define sets as allowed to grow? Not in the quote you supply. There they talk about set valued variables that can grow. > > > It is simply a matter of definition. > > > > Yes, with your definition a set can grow, but you put yourself outside > > set theory, and you must at first consider all results from set theory > > unproven theorems in your theorem, and you need to prove them (if > > possible). > > Hrbacek and Jech teach standard set theory including the fact that in > ZF everything is a set. But I see nothing that states that a set can grow. -- 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 1 Dec 2006 20:56
In article <1164975696.538746.93710(a)j44g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <456EEA86.20001(a)et.uni-magdeburg.de> Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> writes: .... > > > So far I recall "the set IN of naturals is countable". > > > I understood set theory means a hypothetical (alias fictitious) set off > > > all natural numbers. > > > > Oh, for once, try to talk mathematics. By the axiom of infinity the > > set of all naturals is neither hypothetical nor fictitious. > > You mean that by stating the axiom this fictitious set gains some > reality? It is assumed to exist. That's all. "Ficititious" is an > adjective describing its state extremely well. By the axiom it is not assumed that it exists, it is stated that it exists. It is similar to the parallel postulate from Euclid that does not assume that there is one line going through a point not on another line and parallel to that other line. It is stated as fact. And that gives us Euclidean geometry. In the same way, the axiom of infinity gives us ZF set theory where the set of naturals does exist as a reality. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |