An uncountable countable set
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From: Dik T. Winter on 12 Sep 2006 19:30 In article <4506e465(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > Dik T. Winter wrote: > > In article <45004b9b$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > > Dik T. Winter wrote: > > ... > > > > > So, while ...111 may not be considered a standard natural, I see no > > > > > reason why it should not be considered, say, an extended natural. > > > > > > > > Why not use the proper name such numbers already have? 2-adics. > > > > > > No reason, except that 2-adics could possibly have infinite bit > > > positions. > > > > They have not. I have yet to come across basic objects in mathematics that > > have infinite posititions in their representation. > > That's why I invented the T-riffic numbers. It's about time. Perhaps. But you should provide correct definitions and so on before anyone actually can do anything with them. -- 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 12 Sep 2006 19:53 In article <4506d1ae(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > David R Tribble wrote: .... > > Start with a simple proof: Take the set of naturals, N = {0,1,2,3,...} > > and remove one element to get set S = {1,2,3,...}. Now show that > > the "T-size" of N is exactly one less than the T-size of S. In other > > words, find a way to show that every counting of S versus every > > counting of N always leaves one element of N (0) left over. > > Use IFR. N maps to S using f(n)=n+1. The inverse of that function is > g(x)=x-1. So, over the range of 0 to N, |S|=|N|-1. Map N to the Evens E > using f(n)=2n. The inverse function is g(x)=x/2, so over the range of N, > the evens have |N|/2 elements. Isn't that intuitively satisfying? And > gee, it works for finite sets accurately too!! How many elements has the set of primes? Or consider the following mapping: for each n in N write n as prod p_i^e_i where the p_i are prime, now map n = prod p_i^e_i to prod p_(i+1)^e_i. what is the size of the resulting set, and why? (Oh, I should state that 1 maps to 2. And, yes, there is a reverse mapping.) -- 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 12 Sep 2006 19:56 In article <45070bcb(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: .... > What about an unboundedly large but finite number? Let me ask you to, before I can answer such a question. What is your definition of "number"? (I asked the same from Wolfgang Mueckenheim, but his answer was not satisfactory, also not to himself, I think, because he never answered to questions about it.) Once you have answered that question we can probe further into the definition of "unboundedly large but finite number". As far as I understand, for any kind of numbers, they are fixed, so by definition not unbounded. -- 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 12 Sep 2006 20:33 In article <1158102358.446630.158500(a)i3g2000cwc.googlegroups.com> "David R Tribble" <david(a)tribble.com> writes: .... > [http://en.wikipedia.org/wiki/Number] > A number is an abstract entity that represents a count or > measurement. In mathematics, the definition of a number > has extended to include abstractions such as fractions, > negative, irrational, transcendental, and complex numbers. > > I'd also add algebraic numbers, ordinals, cardinals, quaternions, > octonions, matrices, tensors, p-adics, hyperreals, and > surreals to that list. Somthing like that. In my opinion, when you have a set of objects, two operations (+ and *) of which one is distributive over the other, you can talk about numbers. But perhaps some more properties are required? I have no idea. But that is only my opinion. In mathematics, unless the context is clear, the word "numbers" is always qualified. There is *no* mathematical definition of number. > You can see my confusion about Mueckenh's definition of "number". The only definition I have seen is where he stated that a "number" should be larger, equal to or smaller than natural numbers (or something like that). (That is, the trichotomy.) Alas, his definition failed its purpose, to exclude the ordinals from the class of numbers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 13 Sep 2006 03:48
Dik T. Winter schrieb: > You keep confusing the situation where edges do terminate (and so all paths > do terminate) with the situation where edges do not terminate (and so all > paths do not terminate). What do you imagine? Every edge terminates in a node. An example: The digits 1 of the number 0.111... do terminate (they are finite) but the number does not terminate. > > > The paths are not countable because I can > > > not state the n-th path when n is given. The edges within a given path > > > are countable because you can state which is the n-th edge when n is > > > given. > > > > All the edges are countable, because you can state what is the m-th > > edge on the n-th level. > > Say, which of my statements do you contradict here? Impossible to quote a meaningful statement of yours, because you have not yet grasped what a tree, a path and an edge are. This" /" and this "\" are edges. An edge is the connection between two nodes, which represent the digits of a number. A path is an infinite sequence of digits. In my tree every edge is terminating and no path is terminating. > > > > On the other hand, you *claim* that your set of paths is countable, > > > so you should be able to state what the tenth path is. But you refuse > > > to do so. > > > > There are other proofs of 2 + 2 < 10 than executing the addition. > > If I know that 2 + 2 < 5, then I can conclude that 2 + 2 < 10. > > Why should some logic conclusion like that be forbidden in set theory? > > What is the relevance with respect to my question? This example says: Countability can be proven without a bijection. > > > > > What are you talking about? Every edge terminates because it is the > > > > connectio between subsquent nodes of a path. > > > > > > And so all paths terminate, and 1/3 is not in your tree. > > > > But I told you already: The paths do not terminate. The paths are taken > > from the diagonals in Cantor's list. > > Your second sentence makes no sense. Oh yes, it does. My paths are as long as the diagonal of Cantor's list. Every number which can in principle be the diagonal of Cantor's list can be represented by a path of my tree. Why do you accept the axiom of infinity for Cantor's list, but deny it for my tree? > > > > No, I did not know the strange exponentiation used in ordinals. Apparently > > > it can be defined as the set of functions from a set with ordinal number > > > omega can be mapped to a set with ordinal number 2, with the proviso that > > > only finitely elements are mapped to the second element. > > > > Also my paths consist of only finite elements, i.e., edges at finite > > places. Nevertheless the paths are infinite. > > It is well known that the set of *finite* sequences of digits is countable. > You are telling nothing new here. The paths of my tree are not finite paths. If you insist, that no infinite paths do exist, then you must also deny the existence of an infinite diagonal of Cantor's list. Or you should have a firm argument, why the list is infinite but the paths of my tree are not. The axiom of infinity applies in both cases or in no case. Regards, WM |