Cantor Confusion
in [Math]
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From: mueckenh on 17 Mar 2007 09:57 On 16 Mrz., 21:18, Virgil <vir...(a)comcast.net> wrote: > In article <1174054064.244699.153...(a)n59g2000hsh.googlegroups.com>, > > > The function of all cross sections, f: n |--> 2^n, is "continuous" in > > the sense that never a jump by more than a factor 2 can occur because > > the nodes of the tree are connected by an untearable network. > > Following WM's argument, g:n |--> n is even more continuous in that it > can never "jump" by a difference of more than 1, so can never become > infinite at all. It does never "become" infinite. This function "is" infinite, i.e., n is always finite but not bounded. > > > The > > domain is the same as the range, namely N. > > The range of f: n |--> 2^n can never be the same as the domain, unless > both are empty. > > > That is fact, not by claim > > but by construction of the tree. That's why I constructed it. > > A construction which requires N for both the domain and range of > f: n |--> 2^n is fatally flawed. You need only consider the number of pairs of parentheses and of unit fractions per pair of parentheses in the proof by Oresme. Regards, WM
From: mueckenh on 17 Mar 2007 10:02 On 16 Mrz., 21:31, Virgil <vir...(a)comcast.net> wrote: > In article <1174055000.972069.261...(a)y80g2000hsf.googlegroups.com>, > > > > > The function of all cross sections, f: n |--> 2^n, is "continuous" in > > > > the sense that never a jump by more than a factor 2 can occur because > > > > the nodes of the tree are connected by an untearable network. The > > > > domain is the same as the range, namely N. That is fact, not by claim > > > > but by construction of the tree. That's why I constructed it. > > > > You constructed the tree to show that 2^{n+1} <= 2*2^n ? Well, that > > > really must have been fun. Ok, I agree on this. Now we know a property > > > of the function > > > > f : N -> N > > > n |-> 2^n. > > > > This does not tell us anything about 2^aleph_0. > > > aleph_0 is not a natural number. > > Are you just discovering that? Every index of a digit of a real number or of a node, however, is a natural number. Therefore it is of no interest at all to speculate about what happens "at aleph_0". That should you try to discover. Regards, WM
From: Virgil on 17 Mar 2007 17:39 In article <1174137052.814421.258150(a)l75g2000hse.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 16 Mrz., 15:54, Carsten Schultz <cars...(a)codimi.de> wrote: > > mueck...(a)rz.fh-augsburg.de schrieb: > > > > > > Don't mistake the infinite number of finite paths with infinite paths. > > > > I don't. > > I am not sure. A set of finite paths is bounded by aleph_0. Aleph_0 cannot be a bound on an ordered set of paths, or even on a partially ordered set of pahts, unless it is itself a path. Some sets of finite paths are not bounded at all. > > > > > In the union of all finite trees every path has a finite length, given > > > by a natural number of nodes. Presently we are considering the union > > > of all such finite paths. (The union of all finite natural numbers is > > > an infinite union - nevertheless this union cotains only finite > > > numberrs.) > > > > The set of all finite subsets of N is countable, the set of all subsets > > of N is not, > > Alas, these subsets, as represented by indexes of paths, cannot be > distinguished. They are not represented by "indexes of paths" but are, for each path, the sets of levels at which it has left branchings. As such as sets for different paths must be different, and different sets produce different paths, we have the set of paths bijecting with P(N). The set of all finite paths is identical with the set > of all finite and infinite paths. How can that be when one set contains members the other lacks? > > As long as the paths exist as sets of nodes within the tree, they are > countable. Since the set of all subsets of the set of all nodes is uncountable, and paths 'are' such subsets, WM has much yet to prove and no talent for proving. > After having left the tree and lost any contact to nodes, > the paths do no longer represent real numbers. And this will not > change, regardless of the number of opposing believers. As it is pure nonsense, no one but WM believes it. We, who do not accept WM's unproven claims as gospel, do not need to have any path separated from its tree or its nodes in order to find WM's errors.
From: Virgil on 17 Mar 2007 17:42 In article <1174139500.430721.317350(a)o5g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 16 Mrz., 21:06, Virgil <vir...(a)comcast.net> wrote: > > In article <1174042350.684943.146...(a)l75g2000hse.googlegroups.com>, > > If n can jump from finite to aleph_0, > > then 2^n can jump from finite to 2^aleph_0. > > Not if the function is continuous as is the number of paths-bundles in > the tree. Between uncountable and finite, there is mor than a factor > of two. Between finite and countably infinite there is more that a summand of 1. WM's path bundles and WMs are irrelevant in determining the number of paths in the CIBT.
From: Virgil on 17 Mar 2007 17:50
In article <1174139848.400476.4970(a)o5g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 16 Mrz., 21:18, Virgil <vir...(a)comcast.net> wrote: > > In article <1174054064.244699.153...(a)n59g2000hsh.googlegroups.com>, > > > > > > The function of all cross sections, f: n |--> 2^n, is "continuous" in > > > the sense that never a jump by more than a factor 2 can occur because > > > the nodes of the tree are connected by an untearable network. > > > > Following WM's argument, g:n |--> n is even more continuous in that it > > can never "jump" by a difference of more than 1, so can never become > > infinite at all. > > It does never "become" infinite. Then lim n {n --> oo} = aleph_0 is false. This function "is" infinite, i.e., n > > > > The range of f: n |--> 2^n can never be the same as the domain, unless > > both are empty. > > > > > That is fact, not by claim > > > but by construction of the tree. That's why I constructed it. > > > > A construction which requires N for both the domain and range of > > f: n |--> 2^n is fatally flawed. > > You need only consider the number of pairs of parentheses and of unit > fractions per pair of parentheses in the proof by Oresme. There has been a long discussion of how to define functions. According to EVERY definition considered in that discussion, the alleged function described by WM as having domain and range equal to N and having f: n |--> n^2, is NOT a function at all. WM's ignorance of even the most primitive of mathematical ideas leads him to make a fool of himself too often. |