An uncountable countable set
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From: Tony Orlow on 19 Aug 2006 22:37 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.
From: Tony Orlow on 19 Aug 2006 22:47 Dik T. Winter wrote: > The original did not arrive her, so I am piggybacking. > > In article <1156009246.842952.233770(a)i42g2000cwa.googlegroups.com> imaginatorium(a)despammed.com writes: > > Actually nothing that I quote. It is more about: > > Tony Orlow wrote: > > > A circle is a regular polygon with an infinite number of infinitesimal > > > sides. The formula for the angle of the vertices of a regular polygon of > > > n sides is 2*pi*(1/2-1/n). The limit of this formula as n->oo is pi. > > Tony, I see you did agree with Wolfgang Mueckenheim in this thread. But > I should warn you. According to WM neither pi nor that limit do exist. > > (Gives me a deja vu about JSH agreeing with EEE. While the latter proved > that FLT was false and the former proved that it was right.) Ha ha. Hi Dik - 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. ;) Indeed, I agree with WM's logic concerning the identity relationship between element count and value in the naturals. He's quite correct in that regard. However, you may notice a couple of questions of mine to him regarding the existence of actual infinities. Wolfgang's a devout anti-Cantorian finitist. I'm a devoted post-Cantorian delver into infinity. While he takes the argument put forth as proving that the set of naturals is finite, he does so with the assumption that infinite naturals cannot exist. For my part, I agree that the set of finite naturals is finite, though unbounded, but that there exists an infinite set of naturals, which includes infinite values. So, don't worry. It's not like either of us is following the other's lead or colluding. We have plenty to disagree on. :) So, what did you think of my answer to Brian? I know that's not what you were responding to, but I'm just curious. Have a nice day! TO
From: Tony Orlow on 19 Aug 2006 23:09 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. There is no point where a countable set of edges produces an uncountable set of paths. Every PAIR of child edges added to an edge adds a single new path. There may not be terminal edges, but there is something terminal in this Virgilogic. > > >> It is unconscionably >> stupid to regard the number of paths as being "uncountable" and >> therefore greater than the "countable" number of edges, when clearly >> there are fewer paths than edges once the first three nodes are added to >> the root node, or binary point. > > 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. 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. > > >> 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. > > Until TO can show that either "interpretation" says anything false about > infinite binary trees or the bijections described, his complaint is mere > sour grapes. > > The issue is not the "interpretations" but the bijections. If the > bijections are as advertized, then the proof holds despite TO's > displeasure at its validity. The last time I went over this with you, you wasted my time pretending I had your proof wrong, when it turned out I was right all along, and then you changed the subject. I'll not waste my time like that again. Bijections without direct consideration of the mapping functions provide no relative measure of sets. If the number of levels to the tree is countably infinite and therefore equal to aleph_0, and there are 2^n nodes on level n (level 0 is the root node), then there are 2^aleph_0-1 nodes, each with 2 edges proceeding from it. That doesn't sound countable to me. It all depends on whether you are treating the set of edges or nodes as countable, or the set of levels of the tree. In the first case paths are countable, and in the second, nodes, edges and paths are all uncountable given the notion that 2^aleph_0=c, which by the way is false outside of Cantorian mathology. Yes, it's a hat trick. You can play it either way you want. The fact remains. There are half as many paths as edges, asymptotically, in the balanced binary tree. TO
From: Virgil on 19 Aug 2006 23:49 In article <ec8gln$oc9$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Dik T. Winter wrote: > > Let L be the list of all natural numbers > > Let K be the sequence of digits such that for each p in L, K[p] = 1. > > What is *wrong* with that definition? > > Hi Dik - > > Sorry to interject, but could you tell me how long a sequence K is? If > it's finite, then you only have a finite list L, and if it's countably > infinite, then you have an uncountable list of naturals? How does having a countable list of naturals require an uncountable list of naturals?
From: Virgil on 19 Aug 2006 23:53
In article <ec8h1q$prm$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > It would seem to me that any good theory > of infinite sets could be applied to infinite sets of points, such as > the reals in (0,1] or those in (0,2], and be able to draw conclusions > such as that there is twice as many points and twice as much space in > the second as in the first That would require each point to occupy enough space that only a finite number of them could fit into any finite interval. |