Cantor Confusion
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From: Andy Smith on 24 Jan 2007 06:29 In message <gccer2dvlhm5q04g7deea6ro0t0c5b31p6(a)4ax.com>, G. Frege <nomail(a)invalid.?.invalid> writes >On Wed, 24 Jan 2007 08:43:43 +0100, G. Frege <nomail(a)invalid> wrote: > >>> >>> If that's what he means, then I'll agree he could be close. It isn't a >>> proof, but as a heuristic it is OK. >>> >> Though by using (almost) the same heuristic he might conclude that >> rational numbers aren't countable too. Well... >> >Or even better: > >Consider a countable subset of the set of real numbers containing only >irrational numbers. > >Then the argument (the heuristic) of the OP might lead him to the >conclusion that this subset is not countable. > > >F. > Well we were talking about the uncountability of the reals, not necessarily proposing a computer science perspective as a replacement methodology for your number theory ... But if your subset is countable, then its definition implies that the subset can be packed into a smaller address space (though whether it would be easy to determine that from the definition is TBD) -- Andy Smith
From: mueckenh on 24 Jan 2007 06:47 On 23 Jan., 13:35, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > ** The std-union of the set of sets of paths of every finite binary > std-tree is not identical to the set of paths of the infinite > binary std-tree G. ** Summary 1) Every complete infinite binary tree T (containing all nodes and edges) contains all paths. 2) The union tree T(oo) of all finite trees is well defined (as I have shown elsewhere) and yields the complete infinite binary tree containing all nodes and edges: T = T(oo). 3) The union of all finite trees includes the union of all nodes and, with it, the union of all such subsets which are paths (because every path is a well defined subset of the set of nodes if the structure of the tree is well defined). 4) The set of paths in T(oo) is a subset of the countable set of finite sets of all paths in the finite trees. 5) A countable union of countable sets is a countable set (according to ZF with AC). ==> The set of all path is countable. (==> The real numbers are countable.) Going on, we can say: 6) T(oo) = T contains only finite paths. 7) T(oo) = T contains all paths including all infinite paths. ==> There are no infinite paths. (There are no irrational numbers.) Nothing further remains to say. Regards, WM
From: Andy Smith on 24 Jan 2007 06:51 In message <q2cer25laem96n2mig6mvlc189jbjk6rnh(a)4ax.com>, G. Frege <nomail(a)invalid.?.invalid> writes >On Wed, 24 Jan 2007 09:27:59 GMT, Andy Smith ><Andy(a)phoenixsystems.co.uk> wrote: > >>> >>> So I think he's reached the (correct) conclusion that you >>> can't denumerate the reals (in [0,1]) using naturals, >>> albeit in a somewhat clumsy way of saying it. >>> >> Yes! Thank you. >> >*sigh* > >Yes, you reached a correct "conclusion"; but by a faulty reasoning. >That's certainly not something to be proud of (at least not in >mathematics). > If you want to index 2^n numbers, you require 2^n indices. You require as many bits to define your indices as log(no of numbers,base 2). With the reals defined as an infinite series of bits, you require an infinite series of bits to define your indices. But if the indices are to be natural numbers they cannot have an infinite number of bits; all natural numbers are finite. If you prefer, there is a 1:1 correspondence between the countably infinite set of bits used to define the reals and their indices (established by systematically counting the reals bit0, bit 0 + bit 1, bit 0 + bit1 + bit2, etc). But that implies that the indices are infinite. Is that faulty reasoning (I am asking, not being snotty) ? -- Andy Smith
From: mueckenh on 24 Jan 2007 06:58 On 23 Jan., 13:42, "William Hughes" <wpihug...(a)hotmail.com> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > There is no path which ends at a node which you can specify. Every path > > > > of the union tree has maximum length. 0.11 is not a path in the tree of > > > > three or more levels. > > > > Since T1 is not a tree, you cannot use a statement about the > > > properties of of a tree to say something about the properties > > > of T1. 0.11 is a path in T1. > > > Whatever T1 may be called. 0.11 is, by definition, not a path in T1.It does not matter what we call T1. It does matter what properties T1 > has. > Since T1, the union of all finite trees, is the union of all finite > paths, and 0.11 is a finite path, > T1 has the property that 0.11 is in T1. No. The union of {1} and {1,2} does not contain the maximum sequence {1}. Paths in trees without leaves are maximum sequences. T1 is a tree without leaf-nodes. > > I guess you have a different definition for the union of all finite > trees. > > Let R be a set of finite trees with the property that: > there is a (fixed) tree t_D in R, such that: > if s is in R, then > s is a subtree of t_D > > Definition i: > > The union of all finite trees is the tree which has all nodes and edges which are in at least one finite tee. > > If you have a different definition for T1 > would you care to share it? See above. To simplify notation we should call the complete tree T and the union of all finite trees T(oo) in order to be able to express the identity: T = T(oo) > > The problem is that you want a definiton > under which > > -the set of paths in the union of > all finite trees is the union of all finite paths One path is the set of subsets of all nodes of T which belong to it (defined by the guiding edges). > > -the set of paths in the union of all > finite trees is the same as the set > of paths in the infinite tree T2 > > However, you cannot have both. I can and I have: Summary 1) Every complete infinite binary tree T (containing all nodes and edges) contains all paths. 2) The union tree T(oo) of all finite trees is well defined (as I have shown elsewhere) and yields the complete infinite binary tree containing all nodes and edges: T = T(oo). 3) The union of all finite trees includes the union of all nodes and, with it, the union of all such subsets which are paths (because every path is a well defined subset of the set of nodes if the structure of the tree is well defined). 4) The set of paths in T(oo) is a subset of the countable set of finite sets of all paths in the finite trees. 5) A countable union of countable sets is a countable set (according to ZF with AC). ==> The set of all path is countable. (==> The real numbers are countable.) Going on, we can say: 6) T(oo) = T contains only finite paths. 7) T(oo) = T contains all paths including all infinite paths. ==> There are no infinite paths. (There are no irrational numbers.) Nothing further remains to say. Regards, WM
From: imaginatorium on 24 Jan 2007 07:04
Andy Smith wrote: > In message <ep6a6p$aef$1(a)news.msu.edu>, stephen(a)nomail.com writes > >Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > >>> > >>>Since the integers are finite, you cannot represent a real requiring an > >>>actually infinite number of bits, is what I meant. Maybe that is too > >>>simplistic? > >>> > >>> > >> To be more explicit, to represent (= address) all the reals in say > >> [0,1] you would need as many bits for your integers as the reals occupy > >> . But that would require integers with an actually infinite bit length > >> e.g. say, the reflection of the reals about the decimal point to give > >> "numbers" like ...1101 > > > >Why would it require that? Can you explain your reasoning? > >To represent a real number between 0 and 1, you need 1 bit for > >each positive integer. > > > > x = b1*(1/2)^1 + b2*(1/2)^2 + b3*(1/2)^3 + b4*(1/2)^4 .... > > > >Why do you think this requires an "actually infinite" integer? > > > Because you need binf to complete the sum; inf is not a natural number > and you need an actual infinity of bits to describe it. Really? binf? What is the value of binf? Tell us more about the difference between x = b1*(1/2)^1 + b2*(1/2)^2 + b3*(1/2)^3 + b4*(1/2)^4 .... (A) and x = b1*(1/2)^1 + b2*(1/2)^2 + b3*(1/2)^3 + b4*(1/2)^4 .... + binf*(1/2)^[?!@?] (B) Do you understand that (A) is the 'sum' of an unending series (defined properly in terms of limits, and with not an 'inf' in sight)? It goes on without end - there is no 'inf' digit when you get to the end, because you can't get to the end, because there isn't one. (Gosh, I've typed this before, I suspect). So if you insist on adding something called 'binf', it is quite separate from the unending series, and its function is actually [!] rather obscure. Now go back and rephrase your statement above, eliminating the word "actual". Thinking clearly about whether something has an end or not should help. Brian Chandler http://imaginatorium.org |