Cantor Confusion
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From: David Marcus on 25 Jan 2007 12:27 G. Frege wrote: > On 25 Jan 2007 01:47:48 -0800, mueckenh(a)rz.fh-augsburg.de wrote: > > > I am interested whether there are irrational numbers. > > Well, fact is, no one has ever seen one of them so far. > > (On the other hand, the same is true for natural numbers, integers, > and rational numbers, too. :-) Responding to WM? Isn't that pointless? :) -- David Marcus
From: G. Frege on 25 Jan 2007 12:45 On Thu, 25 Jan 2007 12:27:49 -0500, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: >>> >>> I am interested whether there are irrational numbers. [WM] >>> >> Well, fact is, no one has ever seen one of them so far. >> >> (On the other hand, the same is true for natural numbers, integers, >> and rational numbers, too. :-) >> > Responding to WM? Isn't that pointless? :) > Sure it is! :-) F. -- E-mail: info<at>simple-line<dot>de
From: Franziska Neugebauer on 25 Jan 2007 12:57 mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 15:34, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > On 24 Jan., 21:05, Franziska Neugebauer >> > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: >> [...] >> >> Using Virgil's trees we have the result that the induced std-tree >> >> (M(P), E(P)) may be the same if different sets of paths P are >> >> given. >> >> > Then it is not what I am talking about.So what _are_ you then >> > talking about? >> >> > If T(oo) = T concerning the set of nodes and edges,Axiom of >> > Extensionality: Two sets a and b are equal if >> >> A c (c e a <-> c e b) >> >> There is no room in equality for "concerning" this or that. > > Tell it Virgil. This reply goes to your post, so I tell you. If you adopt an argument you have to defend it regardless of its origin. > But you are in error. The set {1, 0, 1} is identical > to the set {0, 1, 1}, concerning the elements. There is no notion of "equality concerning this or that". There is solely equality. > Concerning the order of elements, both sets differ. 1. ZFC sets have no order at all. (Please refresh your basic knowledge of set theory). Hence it is *meaningless* to speak of "equality concerning the order of elements" on sets. 2. The notation S = { a, b, ... } does not imply that there is an order "a before b". It is just a necessity of that particular notation to choose an order in which the elements are written. 3. If you are talking about ordered sets you are talking about "a set together with its ordering" (S, Order(S)). This is an ordered pair and hence a set of cardinality 2: {{S}, {S, Order(S)}}. 4. General remark according to Korzybski: "The notation is not the set". >> > then they are identical with respect to paths too. There is no >> > further parameter to fix. If you do not accept this identity, then >> > further discussion is not meaningful. > >>Have you scrapped your plan to show a contradiction in contemporary >>set >> theory? > > Why? We read nothing new from you. >> >> [...] >> >> >> >> Since the union of V* is the std-union we know that that U V* >> >> >> contains only "finite" paths. >> >> >> > Correct.This is correct _only_ for Virgil's trees. For std-trees >> >> > there the >> >> std-union is not a tree and for David's tree-union the tree-union >> >> is not the std-union and the tree-union of all finite trees is the >> >> binary tree G.[indentation correction:] >> >> >> > Hence 0.[10] is _not_ in UV*. >> >> Using Virgil's trees this is correct. >> >> Using David's tree-union this is not correct. >> >> > I do not intend to talk about this or that tree which may veil the >> > facts.If you still want to show some contradiction in contemporary >> > set theory >> you have to commit to an appropriate formalization of what you >> consider a tree. > > So I did. Perhaps someone at google is hiding all your definitions and commitments from the public? I cannot recall an appropriate formalization of what you consider a tree. >> > In my proof there are two sorts of trees, solely defined by >> > nodes and edges. Let the other trees grow, exist, contain, or >> > whatever they like, when and where they can, if they can. > >> >> > It is valid for all existing natural numbers. Counting over to >> >> > the infinite is nonsense.[indentation corrected] Counting occurs >> >> > in the finite. >> > > What you do _is_ "counting over to the infinite" >> >> > By no means! Every n in N is finite. > >>The set of finite trees V* = { T(n) | n e N } is not a finite set. >> Induction over its members does not prove a property of V*. > > Then it does not consist of only finite levels. Levels are a "derived" property. You did not put them in. So please show what you mean by your statement. If you mean that all trees of V* have only a finite depth than you are right. Nonetheless V* ist not a tree. Hence: So what? What do you want to posit? >> This is quite the same as in the case of the infinite set >> N = { n | n e N }. Induction over n for some property p(n) does not >> prove p(N). Maybe you forgot that. > > No. All I require is that a property holds (or not) for all the > natural numbers or all the levels or nodes or edges of trees which can > be enumerated by natural numbers. Which is not sufficient to show what UV* means considering all you have committed to so far. >> >> when you claim that >> >> induction proves existence of U V* where in fact is does only >> >> prove existence of UV(n) A n e N. > > I am fully satisfied with the latter. I call it U T(n) for A n e N = > T(oo). This is already *formal* nonsens. Are you in the Design Department of your Augsburg University? > Nothings remains to be desired. Can you spot a difference to > your "UV*"? Perhaps not a node but a path? Let it be. I am not > interested in fairy tales. You have a monopoly on that. >> > Yes. >> >> >> V* is not a finite set, i.e. a set of the >> >> form V(n). Hence induction does not "reach" V*. >> >> > Induction reaches and includes every tree which exists. > >> What *exists* in ZFC is defined by the axioms and the rules of >> inference. If you still want to show a contradiction _in_ ZFC you >> must adhere to that axioms and rules. > > Only if you find a difference between T(oo) and T. Define/Commit/Prove than we may see. >> In ZFC/graph theory the tree T(oo) = G (the infinite binary tree) >> exists but is only "reached" when using the non-std-union of David. >> When using the std-union of Virgil, the _induced_ tree (M(P), E(P)) >> is "reached" but the Virgil-tree (i.e. the union of all finite >> _paths_) is not identical to the paths induces by the infinite binary >> tree T(oo)/G. > > I think this gives a delicious impression of the present state of > mathematics. If you have any questions I will surely explain it to you in more detail. Simply ask! >> I must admit this is probably not simple enough at applied university >> level but nonetheless valid. > > Better fitting to scientology or matheology? Well fitted in mathematics. >> > As I said above: If you conclude from one and the same tree that >> > there are different paths, > >> There exist different sets of paths (Virgil's trees) which *induce* >> the same standard-tree (M(P), E(P)). > > There exist different systems of paths which may induce the same tree. Indeed there exist different *sets* of paths which induce the same *standard* tree (M, E). E.g. the standard union P* of the sets of paths of all finite trees T(n) induces the standard tree G (the full infinite binary tree). As one easily sees is the set P^omega of paths induced by G does not equal P*. You may notice the parallel between Sigma* and Sigma^omega. > A a simple finite example: > 0.1 and 0.11 and 0.111 are paths which unioned result in the path > 0.111. Oho. Now you want to use a union of paths. How is your path defined and how is your union? Please check which does apply: [ ] A path is a sequence of nodes. [ ] A path is a sequence of edges. [ ] A path is a set of edges. [ ] Other. Please define:_______________________________________ > Of course you can also get to this same result when starting > from the paths 0.1 and 0.111 or from the paths 0.11 and 0.111. That is > clear. But once the union tree T(oo) is constructed, the set of path > contained in it is uniquely defined. You are using more and more undifined notions. > You always express very carefully that Virgls approach "leads to the > same tree". Please state explicitly wether you are willing to assert > that this same tree, T(oo), after construction, does contain different > sets of paths, induced by the mental constitution of the observer? It is up to you to disambiguate your use of "tree" and "union". F. N. -- xyz
From: G. Frege on 25 Jan 2007 12:53 On 25 Jan 2007 03:13:19 -0800, "Tez" <terence.hoosen(a)gmail.com> wrote: >> >> Consider a bit sequence b_n where b_0 is the least significant bit >> and b_n = 1 (n e N). >> >> and the number defined by m = b_0 + 2 * b_1 + ... 2^n * b_n ... >> ~~~~~~~~~~ Here's the first error. There _is no_ number "defined" by the expression b_0 + 2 * b_1 + ... + 2^n * b_n + ... (*) >> >> Is m a natural number? >> The question is pointless, since "m" is not defined. > > [...] I'd say the your sum, doesn't converge. > Indeed. Hence (*) can't be used to define "m". > > "m = " part on the left hand side of your expression is misleading if > the sum diverges (doesn't converge), which is the case here. > Right. > > Usually, the way these infinite sums (infinite series, technically) are > thought about is to consider the infinite sequence whose nth term is > the sum of the first n terms of the series, ie, consider > c_n = sigma (2^n b_n) for k = 0 to n > and determine whether c_n converges. > So the OP would first have to _prove_ that the expression at the RHS actually denotes exactly one (real) number. Then he might formulate the definition: m = <expression> . F. -- E-mail: info<at>simple-line<dot>de
From: Franziska Neugebauer on 25 Jan 2007 13:26
mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 15:57, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > WM ] >> | The union of two natural numbers is defined to be the larger one. >> | This is a set theoretic union. >> `---- >> >> So the union of two natural numbers is _not_ defined to be the larger >> one but the "larger" of two distint naturals (sets) a and b is >> defined to be a iff >> >> b c a >> >> or b iff >> >> b c a. >> >> So the order _is_ relevant to successfully prove you wrong. > > The union of two different natural numbers is defined to be the larger > one. Repeating a falsified claim. "Union" in contemporary set theory is generally defined (not restricted to natural numbers). This definition does by no means use a notion of "larger". Using e. g. von Neumann ordinals you can define the less than relation by the subset-property. Not the other way round! F. N. -- xyz |