Cantor Confusion
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From: David Marcus on 25 Dec 2006 14:28 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. 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: David Marcus on 25 Dec 2006 14:41 cbrown(a)cbrownsystems.com wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > cbrown(a)cbrownsystems.com schrieb: > > > > > What "authorities" do you claim I am quoting when I state "the function > > > f(x) = 2*x is a bijection between the set of all naturals, and the set > > > of all even naturals"? > > > > > > I am simply applying specific definitions of the terms "function", > > > "bijection", "the set of all naturals" and "the set of all even > > > naturals", using the usual logic. > > > > This bijection between sets (initial segments {1,2,3,...n} and > > {2,4,6,...2n}) is only valid for finite sets. > > Suppose I instead counter-claim that this bijection is obviously "not > valid" for finite sets; only for infinite sets. > > Is that what you call a "correct mathematical argument"? Simply > claiming something is valid or not valid? How is it different from your > own argument? Here is WM's argument from his amusing manuscript "The Meaing of Infinity": For finite positive n, the set {2,4,6,...2n} contains at least one number that is greater than |{2,4,6,...2n}|. Now, let n go to infinity to conclude that the set of even numbers contains a natural number greater than aleph_0 (let's call it W). The map f(x) = 2x on N doesn't have W in its image, so f is not a bijection from N to the evens. Apparently, WM calls the preceding a correct mathematical argument. Or, perhaps he only thinks that mathematicians think it is a correct mathematical argument. Or, perhaps he only thinks it is correct in ZFC. -- David Marcus
From: David Marcus on 25 Dec 2006 14:44 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? -- David Marcus
From: mueckenh on 25 Dec 2006 16:37 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. 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. Regards, WM
From: mueckenh on 25 Dec 2006 16:42
Dik T. Winter schrieb: > > 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"? "can be done" is not "does exist". We had cleared this point long ago. > > I can go shopping *when the shops are open*. Does that mean that I need > to prove that the shops are open? Do not state that you can go shopping (that had already been claimed by Zermelo) but demonstrate it by going shopping - or confirm that never a shop is open. > > 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? 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. 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. Regards, WM |