Cantor Confusion
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From: Virgil on 23 Dec 2006 20:25 In article <1166908718.682782.155790(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > In short: If any natural number n does exist, then there exists a line > > > in which this and any smaller number does exist. > > > This holds for every number n. > > > > Every natural n. > > > > > You know that the logical symbol for every is forall. > > > > I don't see how this could support your "famous" quantifier reversal. > > Hence that does not prove the existence of a _single_ line (*). > > Show me a number which is not in a single line. Number n+1 is not in line n.
From: Virgil on 23 Dec 2006 20:28 In article <1166908897.512565.192500(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Gc schrieb: > > > mueckenh(a)rz.fh-augsburg.de kirjoitti: > > > > > Gc schrieb: > > > > > > > > > Every two > > > > paths separete on some finite level edge, but only when you got the > > > > whole countably infinite path (the union of it`s all finite subpaths > > > > starting from the beginning) all infinite long pathes separate from > > > > each other. > > > > > > Of course. And therefore all the paths can be counted by the number of > > > split positions. > > > > No. There is no point (except union of all the points) where a path > > separates from all the other pathes and becomes a unique path. > > This union is covered by the limit n --> oo, i.e., by the infinite > tree. So is WM able to point to a point at which any one path has finally separated from all others? In finite trees these would be terminal nodes. but infinite trees don't have any.
From: Dik T. Winter on 23 Dec 2006 22:13 In article <1166907306.745536.315650(a)h40g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Virgil schrieb: .... > > > Dik T. Winter schrieb: > > > > > > > Indeed. No axiom is necessarily true. > > > > > > How then should the truth of AC be proved? > > > > One does not prove axioms, one either assumes them or does not assume > > them. > > You like to snip in such a way that a wrong meaning of my words becomes > manifest? You think it is necessary? I wonder what the correct meaning of your words are. > It was Dik who insisted that a well ordering of the reals can be done > (not only be proved), if AC is true. Here is the discssion: Indeed, it can be done *when AC is true*. But perhaps we have a different view on the meaning of the sentence "can be done"? I can go shopping *when the shops are open*. Does that mean that I need to prove that the shops are open? To recap: > > > > But I explicitly stated how you could construct > > > > a well-ordering when AC is true. You can not do it in general, and > > > > nobody can, because AC is not necessarily true. > > > > Are there axioms which are necessarily true? Not according to your > > > point of view. > > > Indeed. No axiom is necessarily true. > > How then should the truth of AC be proved? Why do I need to prove that the shops are open when I state that I can do shopping when the shops are open? > >> The sum 1 + 1/2 + 1/4 + ... has no last term. > > >It is not a sum, but is often used to represent an infinite sequence of > >partial sums. Such a sequence may have a defined limit value, but that > >limit is not "the sum" except by abuse of language. > > Whatever you may call it. The value < 1 required by set theory can only > be attached to it by total abuse of human brain But there is a total abuse of the human brain when you think that the sum of the part edges edges assigned to a path is equal to that sum. What edge has a value of 1 in that sum when assigned to the path that goes only to the left? Strange enough, with your reasoning the sum has no first term. Starting at level 1 and assuming finite trees, we find: 1 1/2 + 1 1/4 + 1/2 + 1 and going on to: ... 1/8 + 1/4 + 1/2 + 1 so there is no starting point. So again my question. To how many paths is the first left branch from the rood assigned? To how many paths is the second left branch assigned? -- 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 23 Dec 2006 22:16 In article <1166921237.502878.48560(a)h40g2000cwb.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > Dik T. Winter wrote: > > In article <1166845904.426550.122020(a)48g2000cwx.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: .... > > > No, I don't. What I am showing here is that each path will accumulate > > > the weight of two edges as it approaches infinity. > > > > No, you do not. In the finite case you do, approximately. In the > > infinite case you do not because *all* parts you pass to a path are > > reduced to size 0. Consider what we pass on to the paths that go > > left only. Look at the following: > > level 1: 1 edge at level 1 > > level 2: 1/2 edge at level 1 + 1 edge at level 2 > > level 3: 1/4 edge at level 1+ 1/2 edge at level 2 + 1 edge at level 3 > > etc. for each finite n, but what about the infinite case? > > level 'oo': 1/oo edge at level 1 + 1/oo edge at level 2 + ... > > What is level oo? The levels are indexed by natural numbers. The completed infinite tree? As long as you are talking about finite trees, you are welcome. But in that case 1/3 is not a path in the tree. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 23 Dec 2006 23:18
In article <1166909252.274423.224430(a)i12g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Are you saying that x^2 = 2 does not have a solution? > > Yes. But that is another topic. (In short: There are less than 10^100 > bits in the universe. Therefore, no irrational number can be > approximated better than to about 10^-100. Who says one has to be satisfied with an approximation? |