Cantor Confusion
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From: Virgil on 26 Dec 2006 23:43 In article <1167188553.482447.243720(a)a3g2000cwd.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Virgil wrote: > > > The number 2 has been claculated for the entire > > > infinite tree. > > > > Your claim is based on the assumption that > > > > lim n -> oo (f(n)/g(n) = (lim n -> oo f(n))/(lim n -> oo g(n)) > > It is not. Then your claim says nothing at all about the number of edges or paths in the limiting case.
From: Virgil on 26 Dec 2006 23:48 In article <1167188902.725881.21050(a)79g2000cws.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Virgil wrote: > > In article <1167167417.484441.153260(a)i12g2000cwa.googlegroups.com>, > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > Virgil wrote: > > > > > > Newberry cannot deny that the logic of as many leaf nodes as paths is > > > > the same as the logic of more nodes than paths, so if one holds they > > > > both must, and if one fails, both fail. > > > > > > Let me get this straight. How many terminal nodes are there in an > > > infinite binary tree? > > > > Depends on whose rules one is going by. > > > > By my rules, none. > > > > By your rules, one per path. > > I have explicitly stated that there are no terminal nodes in an > infinite tree. The only thing that prevents us from reaching consensus > on this one your insistence that I have said the opposite of what I > have clearly written above (that there are no terminal nodes in an > infinite tree.) But your limit arguments require them. You can't have your same limit claim prove two contradictory results. > > Now let us assume that there are no terminal nodes in an infinite tree. > Correct me if I am wrong but it seems to me then that the set of the > terminal nodes in an infinite tree is empty and its cardinality is > zero. > > Having said that, what does it have to do with the cardinality of the > paths? You have just falsified your limit argument for the number of nodes being greater than the number of paths. If either limit argument is valid they both are, and if either is invalid they both are.
From: Virgil on 26 Dec 2006 23:53 In article <1167189286.045893.92280(a)h40g2000cwb.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > David Marcus wrote: > > Newberry wrote: > > > David Marcus wrote: > > > > Newberry wrote: > > > > > stephen(a)nomail.com wrote: > > > > > > In what system can you prove that #edges = 2 * #paths? > > > > > > > > > > In ZFC. Theorem: In an infinite binary tree there are twice as many > > > > > edges as there are paths. > > > > > Proof: At the level n the ratio of eges over paths is expressed by > > > > > 2*2^n - 2)/2^n > > > > > For the entire tree we calculated the limit for {n-->oo} > > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2. QED > > > > > > > > That's fine up to where you write "QED". Why does the limit you > > > > calculated tell us anything about the cardinality of the edges and paths > > > > in the infinite tree? > > > > > > I was only proving that the ratio of edges to paths in an infinite > > > binary tree is 2. The number 2 has been claculated for the entire > > > infinite tree. The number of nodes in each infinite path is aleph0 same > > > as the number summands in our infinite sequence. There is one-t--ne > > > mapping between N and the levels in the tree. The limit uses all > > > members of N. If you disagree please tell me which node in which path > > > has been left out. > > > > No idea what you mean by "left out". However, you didn't answer my > > question: Why does the limit you calculated tell us anything about the > > cardinality of the edges and paths in the infinite tree? > > Because if #edges/#paths = 2 then #edges > #paths. Then equally if #leaf-notes/# paths = 1 then #leaf-nodes = # paths. > Everybody on this board, including the Cantorists, insist that |edges| > = aleph0. And you are claiming that #leaf_nodes =/= # paths in the limit. > > > > > I really don't see how you can think that you've proved that "the ratio > > of edges to paths in an infinite binary tree is 2". The number of edges > > and paths in the infinite tree are both infinite, so why is the ratio > > even defined? > > That is what limits are for. Where do you find that in any standard mathematics??? > > > > Are you trolling?
From: David Marcus on 27 Dec 2006 01:51 Newberry wrote: > David Marcus wrote: > > No idea what you mean by "left out". However, you didn't answer my > > question: Why does the limit you calculated tell us anything about the > > cardinality of the edges and paths in the infinite tree? > > Because if #edges/#paths = 2 then #edges > #paths. Sure. But, you've only proved it for finite trees. > Everybody on this board, including the Cantorists, insist that |edges| > = aleph0. Sometimes the cranks get something right. > > I really don't see how you can think that you've proved that "the ratio > > of edges to paths in an infinite binary tree is 2". The number of edges > > and paths in the infinite tree are both infinite, so why is the ratio > > even defined? > > That is what limits are for. Do you mean that you want to define "ratio of number of edges to number of paths in the infinite tree" to be the limit as n -> oo of the ratio of number of edges to number of paths in the tree with n levels? Well, I suppose you are welcome to do so, but then you must prove that this limit tells you something about the cardinalities of the sets of edges and paths in the infinite tree. Hint: it doesn't. -- David Marcus
From: David Marcus on 27 Dec 2006 02:00
mueckenh(a)rz.fh-augsburg.de wrote: > What is an infinite path? > > Do you think it has some fairy tale character? Some very special > properties, different from any thing we can observe in the universe? You mean like "personalization"? > A path is an infinite path if you can follow it without ever reaching > an end - and that's all. > > Therefore: If you follow some path you will see that whenever it > separates itself from another one, this happens by two edges .- one > edge for the path, and the other edge for the other path. This process > repeats and repeats without end. Nothing else happens. From the > unavoidable and "inseparable" connection of separation and appearance > of another edge we can conclude that every path which can be > distinguished from another path runs through an edge which does not > belong to the other path; call it "personalization of an edge. > Therefore distinguishability of paths and personalization of edges are > unavoidably connected. > > To assert that there are more paths separated by edges than are edges > existing at all, is an error which cannot be explained other than by > the desastrous and corrupting influence of the "logic" of set teory. Which edge personalizes the path that always goes left? -- David Marcus |