Cantor Confusion
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From: Virgil on 25 Dec 2006 19:16 In article <1167082643.909029.231380(a)f1g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > 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: > > > > 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. > > > What is an infinite path? You have been going on about them for so long and now you don't even know what they are? A path is infinite if there is an injection from the set of naturals (or any other set which has injections into proper subsets of itself) to the set of its edges (or its nodes). > > Do you think it has some fairy tale character? Some very special > properties, different from any thing we can observe in the universe? Yes! Quite different from anything we can "observe in the universe"! > > 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. As every "personalization" separates a set of infinitely many paths which contain that edge from a set of infinitely many which do not, such an edge is not "personal" to a single edge, but only to a tree which is , except for a finite initial sub-path, tree-isomorphic the the entire tree. > > 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. ZFC or NBG set theory might corrupt the peculiar faiths of those who believe as WM does, but is perfectly consistent with standard logic.
From: Virgil on 25 Dec 2006 19:35 In article <1167083104.107146.181640(a)h40g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > 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? > > Everybody knows it - or could know it. There is no n-ary representation > of an irrational with more than 10^100 bits. One can give an exact continued fraction of sqrt(2). Who says that the only way to "represent" a number is in terms of some base like decimal? > > >> You like to snip in such a way that a wrong meaning of my words becomes > >> manifest? > > > Since your words are full of wrong meanings, anything that reveals this > > truth is to be supported. > > Yeah. THAT's it. Heil Virgil! > > Would you consider to change also your wrong opinion on: > > > If WM were the least bit amenable to reason, he would see > > his logic requires that every infinite path in a infinite binary > > tree must have a terminal (leaf) node at its "end". > > Because then you could see that every separated path (or bunch of > paths) must have an initial node where it starts as a separated path > (or bunch of paths). There is no such thing as a single "separated path" if it is to be "separated" by a single node or edge, there are only "bunches" of such paths starting for some node with each bunch being equinumerous with the whole tree. There is an easy bijection between the set of paths through any edge (or node) and the set of paths of the entire tree. There is no surjection from the set of edges to the set of paths, nor from the set of nodes to the set of paths. These things are all easily seen if one notes that any infinite path is an infinite sequence of left/right branchings. For any natural n, cutting off the first n branchings of any (infinite) path leaves an (infinite) path, and for fixed n, the set of all such truncated paths is identical to the original set of all paths.
From: Newberry on 25 Dec 2006 19:49 David Marcus wrote: > Newberry wrote: > > Virgil wrote: > > > In article <1166895046.650593.195620(a)a3g2000cwd.googlegroups.com>, > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > Virgil wrote: > > > > > In article <1166854303.474151.267360(a)h40g2000cwb.googlegroups.com>, > > > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > > Is it true that the ratio of edges over paths converges to two as we > > > > > > approach infinity? > > > > > > > > > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > > > > > > > > > It is true that the ratio of terminal nodes to paths converges to 1 as > > > > > the path lengths increase towards infinity. > > > > > > > > What about the ratio of all the edges to all paths? Does it converge to > > > > 2? > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > > > > > It does not matter. > > > > Why does it not matter? > > The cardinality of the inexes in the limit is aleph0, and the > > cardinality of the nodes in any infinite path is aleph0. It means that > > in calulating the limit > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > we transversed all the infinite paths. > > What does "traversed" mean? And, how is it relevant to determining the > cardinality of the set of paths? It means that we have taken into account the entire tree and we determined that the number of edges in said entire tree is twice as higher as the number of paths. > > -- > David Marcus
From: Newberry on 25 Dec 2006 19:55 David Marcus wrote: > Newberry wrote: > > paths edges > > level 1: 2 = 2^1 2 > > level 2: 4 = 2^2 6 > > level 3: 8 = 2^3 14 > > > > level n: 2^n (not sure what the formula is) > > > > Does the ratio edges/paths converge to 2 for n --> infinity? > > Yes, as WM is fond of repeating ad nauseum. > > > It certainly makes it highly couterintutive that there are more paths > > then edges although I do not know if it generates a flat contradiction. > > Yes, it is counterintuitive (depending on your intuition). No, there is > no contradiction. Just because a system avoids a contradiction of the type P & ~P does not mean that it is justified. For example an omega-inconsistent system may not produce any P & ~P and would be still unacceptable. Similarly a system in which we can prove that #edges = 2 * #paths and at the same time that the cardinality of paths is greater than the number of edges is unacceptable. You can't take limits without justification. Here is a > nice fallacy of WM's from page 5 of > http://www.arxiv.org/pdf/math.GM/0403238: Group the natural numbers > greater than 1 as follows. > > (2) (3 4) (5 6 7 8) ... > > If we number the groups starting with zero, then there are 2^n numbers > in group n. So, for finite n, we must have > > |N| >= |group n| = 2^n. > > Now, let n go to infinity to get > > aleph_0 >= 2^{aleph_0}. > > -- > David Marcus
From: Dik T. Winter on 25 Dec 2006 19:52
In article <1167082643.909029.231380(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > 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: > > > > 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. > > 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? > > A path is an infinite path if you can follow it without ever reaching > an end - and that's all. Yes. So it is essentially different from a finite path. The same with your trees, Each finite tree has a natural number as maximum level and has no infinite paths. The completed infinite tree has no maximum level and has infinite paths. So any conclusion you can make from the finite trees does not necessarily hold for the infinite tree. > 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. For each two paths you can identify an edge where they separate. But there is *no* edge that separates (for instance) the path leading to 1/3 from all other paths. And through each and every edge run infinitely many paths. > 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. You are extending conclusions valid for finite trees to the infinite tree without proof, it is just your intuition. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |