An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 20 Aug 2006 00:06 In article <ec8hlf$qq8$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > If the formula applies to an infinite number of finites, then does the > sum of the aleph_0 finite naturals equal (aleph_0^2+aleph_0)/2? Only if (aleph_0^2+aleph_0)/2 = aleph_0. But even then, one must have some prior definition of addition of infinite cardinalities, and some prior definition multiplication of infinite cardinalities (or at least squarings of them), and some prior definition of division infinite cardinalities by finite cardinalities (or at least halving of infinite cardinalities), none of which TO has provided, so his "formula" is all nonsense. > I think that what WM is pointing out here is that aleph_0 does not > behave like a normal number Only a nutcase would suppose that an infinite cardinal need behave entirely like a finite cardinal anyway. > Given an injection f: N -> P(N), why does the set > M(f) = {n in N | n !in f(n)} > not exist? It does. But for a non-surjective f: N -> P(N), which all functions from N to P(N) must be, there will not be any x in N such that f(x) = M(f)
From: Virgil on 20 Aug 2006 00:30 In article <ec8hsd$r5t$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Dik T. Winter wrote: > > In article <ec7gek$frg$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > > > Inductively provable: Any set of consecutive naturals starting at 1 > > > always has its largest element equal to size of the set. > > > > What is the size of the empty set? What is the largest element of the > > empty set? > > The empty set is not a set of consecutive naturals starting at 1, > because it does not include 1 as an element, and therefore it does not > stand as a counterexample. However, even in your von Neumann ordinals, > the empty set is 0, and contains zero elements, and could be considered > the 0th in this sequence of sets. You fail to address the inductive > proof that the non-empty sets of this series, of which the naturals is > the limit, always have this property. TO fails to note that induction does not apply. The form of induction is: Let F(x) be a predicate whose argument, x, is a member of a minimal inductive set S.* If F(x) is true for the first member of S and if whenever F(x) is true then F(successor(x)) is also true. then the principle of induction says that F(x) is true for each member of S. It does not say anything about what is true for the set S itself, only for its members. In particular it does not say that F(S) is true. * Note a minimal inductive set is a non-empty set, S, together with a "successor" function s:S -> S such that (1) there is a unique member of S which is not in the image of s. (2) s is injective. (3) If T is a subset of S such that (a) T contians the unique member of S not in the image of s and (b) for all x in S, if x \in T then f(x) \in T then T = S Note that the set of von Neumann naturals, N, with s(x) = x \/{x} and the non-image element {}, are a minimal inductive set under the above definition. There is no equivalent model in which the ordinal of each element is equal to the element itself as that would require sets to be members of themselves, which is impossible for ordinals.
From: Virgil on 20 Aug 2006 00:32 In article <ec8if0$rtd$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Oh, how I love being compared to JSH! Shall I begin ranting about the > conspiracy against me, and how you will all suffer for your mathematical > sins against my superior intellectual achievements? No, I'll save that > for when I'm actually schizophrenic, thanks. ;) Tomorrow? > > Indeed, I agree with WM's logic concerning the identity relationship > between element count and value in the naturals. Except that ordinal numbers cannot work that way, as it would require each ordinal to be a member of itself, which is not possible for ordinals.
From: Virgil on 20 Aug 2006 00:55 In article <ec8jpl$t2j$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil wrote: > > In article <ec7pmg$rj2$1(a)ruby.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > >> Virgil wrote: > > > >>> The sequence of edges leading up to any edge is finite. So a finite > >>> sequence of 0's and 1's, 0 for left child, 1 for right child, uniquely > >>> identifies each edge, but it takes an endless sequence of 0's and 1's to > >>> identify any endless path. > >>> > >>> Thus in infinite binary trees, the set of edges is countable but the set > >>> of endless paths is not. > >> Ah yes, we've been over this. Since one extra path is created for every > >> two edges added to the tree, there are clearly half as many paths as > >> edges (floored, for a balanced tree, of course). > > > > That only holds for finite trees and for terminal edges in such trees. > > But in infinite trees there are no terminal edges, so TO's analysis > > again fails. > > > > > > No, it is inductively provable that for every edge added to your > countable set, only a countable set of paths may be produced at that > point. For every edge in an infinite binary tree, there are uncountably many non-termnating paths passing through (or including) that edge. The number of terminating paths is irrelevant. > There may not be terminal edges, but there is > something terminal in this Virgilogic. Is TO so simply unable to comprehend such a simple proof, or is his ego too fragile to accept that he is wrong again. > > > > Does To then choose to ignore that there is an explicit bijection > > between the set of paths and an uncountable set and an explicit > > bijection between the set of edges and a countable set? > > > > I haven't ignored it at all. I pointed out at the time that it had > everything to do with using two different interpretations of the tree. Since each "interpretation" correctly "counts" what it purports to count, interpretation is irrelevant and only the counting counts.. > If you view the problem using one or the other, and not changing horses > midstream, then your proof falls flat. It's a hat trick. That TO thinks of all of mathematics as being a collection of "hat tricks" is well known. That TO is wrong is equally well known. > >> The proof you have offered in the past > >> is dishonest in the sense that it bijects the edges with the naturals > >> using one interpretation of the tree, and then bijects the paths with > >> the reals using another incompatible interpretation. The issue is whether it actually does biject the edges with the members of naturals, which it does, and actually bijects the paths with the subsets of the naturals (not with the reals as TO would have it). If both bijections are bijections, THAT is all the proof needs to pay attention to. WE already have a satisfactory proof that there are injections but no surjections from any set to its power set, so that the cardinality of any set is strictly less that the cardinality of its power set. TO chooses to reject all of standard mathematics and all standard axiom systems in which that mathematics is valid without having anything but his intuition as foundation. When TO can produce an axiom system in which any of his claims can be proven, only then will he have anything but unproven claims.
From: Dik T. Winter on 20 Aug 2006 08:26
In article <ec8gln$oc9$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Dik T. Winter wrote: .... > > What is the wrongness of the definition? Let me refrase: > > Let L be the list of all natural numbers > > Let K be the sequence of digits such that for each p in L, K[p] = 1. > > What is *wrong* with that definition? > > Sorry to interject, but could you tell me how long a sequence K is? Can you not determine it? I would think it follows immediately from the definition. > If > it's finite, then you only have a finite list L, and if it's countably > infinite, then you have an uncountable list of naturals? What is your problem? From the definition I would say that the size of the sequence of digits in K is equal to the size of the sequence of natural numbers in L. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |