Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: David Marcus on 12 Nov 2006 01:48 Virgil wrote: > In article <1163282892.731472.33380(a)i42g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > William Hughes wrote: > > > > > > > a: Any infinite set of numbers must contain an infinite number > > > > > > > > b: It is possible to have an infinite set of numbers that does > > > > not contain an infinite number. > > > > > > WM, please tell us if you agree or disagree with the statements a and b > > > above. > > > > You will not understand it. But I'll try. > > > > The potentially infinite set N does not contain an infinite number. In > > this case (b) is correct. > > Sets cannot be merely potential. If not actual, then not sets. > > Whatever it is that WM is talking about when he misspeaks of "potential > sets", it is not sets. Probably nothing that WM talks about is a set (where the word "set" has its standard mathematical meaning), but I wonder whether WM's notion of "potential set" isn't closer to the mathematical concept than is his notion of "actual set". After all, in mathematics (b) is true, and WM says (b) is true if we change "set" to "potential set". Not sure if WM says (b) is true if we change "set" to "actual set". > > An actually infinite set is a set which has a cardinal number like > > omega. In set theory every set is actually infinite. So N is actually > > infinite. If its cardinal number omega is claimed to count the members > > of N, then there is a contradiction, because omega counts all n in N > > but does not belong to N while the initial segments of N are counted by > > natural numbers. > > There is nothing in mathematics that says that the "number" of elements > in a set must be a member of that set. > WM wants to have the number of members of a natural to be a member of > itself, which does not happen in ZF or NBG. Kind of an odd thing to want, since 5 isn't in {0,1,2,3,4}. -- David Marcus
From: mueckenh on 12 Nov 2006 11:58 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > [...] > >> Are there really three vertices in WM's "triangle"? > > > > If finished infinities [...] > > Verbiage. Yes. But, sorry to see, it is the fundament of modern mathematics. Regards, WM
From: mueckenh on 12 Nov 2006 12:08 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > > > > > Lines with finitely many indexes cannot exhaust the column with its > > > > infinitely many > > > > > > An infinite number can. > > > > A God can do everything he wants. Infinite numbers are of the same > > power? > > > > No. God can do anything. Infinite numbers cannot. However, > Infinite numbers can do some things. > In particular an infinite number of finite lines can > "exhaust": an infinite number of indexes. > > > Can you apply your reasoning to the following matrix? > > > > 1 > > 21 > > 321 > > 4321 > > ... > > > > Reversing the order of the lines does not change thing. It should show you that without an infinite line there is no infinite column. > There > are an infinite number of lines. Each line is finite. There is an infinite number of initial segments of columns. Each one is finite. And nothing more is there. > There are > are omega lines and omega columns. There is no last line. > There is no omega th line. There are no omega lines. > There is no inifinite line. > > > What ordinal numbers have the following sequences? > > > > None. These are not ordinals. > An ordinal number must consist of an > initial segment of ordinals. However, in > a slightly sloppy way they can be said to > represent ordinals (note that this > representation is no longer uniqe). Note that you can > represent any countable ordinal (not just omega) by > using just finite integers, you do not need (athough > you can use if you want) infinite ordinals. > > > 3,4,5,...,1,2 > > A infinite set consisting of finite integers. > This represents omega+1 No, the sequence above represents omega + 2. > (For n>=3 you have put n in the spot one > would usually have n-2, 1 is in the spot where > you would usually have omega, and 2 is in > the spot where you would usually have omega+1) > > > > > 3,4,5,...,omega,1,2 > > > > An infinite set consisting of the finite integers and > the non-natural number omega. This represents > omega+2 No, the sequence above represents omega + 3. But you see the problem. > (For n>=3 you have put n in the spot one > would usually have n-2, 1 is in the spot where > you would usually have omega+1, and 2 is in > the spot where you would usually have omega+2) ? > > What matters is not which ordinals you use > to occupy places, but where you put the "..." > For example. The both sequences > > 5,4,3,2,1,6,7,8 ... > omega, omega*omega, omega+5*omega^3 + 7, 1,2.3, ... > > represent the ordinal omega. > The first is omega. The second is not omega. You see it if you replace omega by 1,2,3,... Perhaps you wanted to say that they both represent the *cardinal* omega. That is correct. Regards, WM
From: mueckenh on 12 Nov 2006 12:10 David Marcus schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > > David Marcus schrieb: > > > > > If you think that your "mathematical proof" is valid and you prove > > > something that is absurd, this only shows that your mathematical > > > reasoning leads to absurdities. Everyone who uses mathematical reasoning > > > consistent with ZFC cannot prove your absurdity. So, if you learn ZFC, > > > you will stop proving absurdities. > > > > Only then the absurdities would begin. But I would not notice. > > > > > > > We need no experts in trees. This tree does nothing else but represent > > > > the real numbers of an interval. Here it is in a form which can be > > > > understod by ever mathamatician. > > > > > > > > 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. 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. > > > > > > Half of the paths include edge a. One quarter of the paths include edge > > > b. > > > > > > Suppose we have a finite tree of height n. Then there are 2^n paths. > > > Consider a path (e1,e2,...,en). Edge e^j is contained in 2^-j of the > > > paths. We can define a real-value function on paths by g(e1,e2,...,en) = > > > sum_{j=1}^n 2^-j = 1 - 2^-n. Not sure what to do next... > > > > > > > Continuing in this > > > > manner in infinity, we see by the infinite recursion > > > > > > > > f(n+1) = 1 + f(n)/2 > > > > > > What is f(n)? > > > > f(n) is the number of edges related to the initial segment of one path > > which has passed through the first n edges. > > Are you saying that f(1) is the number of edges "related" to the paths > that contain edge a? What is the value of f(1)? f(1) = 1 is the number of edges related to initial segment (of the path) that contains edge a. We start with the first edge a of a path. Then we see the path splits into two paths. So half of edge a is related to each one and so on. Of course we never get ready, but there is no edge which remains unconsidered. f(n) is a recursion formula like many others. Do you know how to calculate the square root of 2 by x_n+1 = x_n/2 + 1/x_n? You start with x_1, calculate x_2 and so on. Of course we never get ready, but there is no digit o sqrt(2) which remains uncalculated. It is not impossible to achieve more with infinite strings belonging to representations of irrational numbers. Regards, WM
From: mueckenh on 12 Nov 2006 12:13
David Marcus schrieb: > > > What is wrong with my reasoning? > > > > Nothing. Wrong is only the assumption that "goes on forever" could be > > considered a finished infinity, i.e., could be denoted by a fixed > > cardinal number being larger than any natural number. > > I'm not sure what you mean. In particular, I'm not sure what > "considered", "finished" (or "finished infinity"), "denoted", and > "fixed" mean here. I see. But recently you used the word "completed infinity". Be sure that my finished infinity means the same. > > > If you look at the diagonal, you always see that it is only there where > > lines are. Therefore you will always see that it is of finite length. > > "It goes on forever" does not mean actually infinite length. The > > lengths of the lines also increase from line to line forever. > > Nevertheless, an infinite length will never be reached. > > To me, "infinite length" just means "goes on forever". To you, the two > phrases have different meanings, it seems. So, what do you mean by the > phrase "infinite length"? "Goes on forever" is a property of the set of natural numbers, as well of the elements n as of initial the segments 1,2,3,...,n. It does not make you believe that there are infinite natural numbers n. Why does it make you believe that there is an infinite initial segment 1,2,3... ? Regards, WM |