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Topic:
Matheology § 190
Replies:
15
Last Post:
Jan 14, 2013 3:35 PM



Virgil
Posts:
8,833
Registered:
1/6/11


Re: Matheology � 190
Posted:
Jan 12, 2013 4:00 PM


In article <c061586061904c10918578ed2f6a28ea@x10g2000yqx.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> Matheology § 190 > > > The Binary Tree can be constructed by aleph_0 finite paths. > > 0 > 1, 2 > 3, 4, 5, 6 > 7, ...
Finite trees can be built having finitely many finite paths. A Complete Infinite Binary Tree cannot be built with only finite paths, as none of its paths can be finite. > > But wait! The Binary Tree has aleph_0 levels. At each level the number > of nodes doubles. We start with the (empty) finite path at level 0 and > get 2^(n+1)  1 finite paths within the first n levels. The number of > all levels of the Binary Tree is called aleph_0. That results in > 2^(aleph_0 + 1)  1 = 2^aleph_0 finite paths.
Wrong! At any finite level one has a finite number of finite paths but in the Complete Infinite Binary Tree one has no finite paths at all but does have 2^aleph_0 INfinite paths.
> > The bijection of paths that end at the same node proves 2^aleph_0 = > aleph_0.
No two paths end in the same node in any binary tree, and in the Complete Infinite Binary Tree no path "ends" at all. > > This is the same procedure with the terminating binary representations > of the rational numbers of the unit interval. Each terminating binary > representation q = 0,abc...z is an element out of 2^(aleph_0 + 1)  1 > = 2^aleph_0.
But so are all the nonterminating ones.
And the terminating ones only make up aleph_0 of that 2^aleph_0 total. > > Or remember the proof of divergence of the harmonic series by Nicole > d'Oresme. He constructed aleph_0 sums (1/2) + (1/3 + 1/4) + (1/5 + ... > + 1/8) + ... requiring 2^(aleph_0 +1)  1 = 2^aleph_0 natural numbers.
While I can see a need for aleph_0 natural numbers in that proof, and in others needing all the natural numbers.
Neither I nor anyone else outside of WMytheology can see any need for more naturals than exist in any proof.
> If there were less than 2^aleph_0 natural numbers (or if 2^aleph_0 was > larger than aleph_0) the harmonic series could not diverge and > mathematics would deliver wrong results.
WRONG! > > Beware of the settheoretic interpretation which tries to contradict > these simple facts by erroneously asserting aleph_0 =/= 2^aleph_0. Beware of those so ignorant of mathematics, like WM, that they claim to be able to surject any set onto its power set.
There is an easy bijection between the set of paths of a complete infinite binary tree and the set of all subsets of N: For each path, the set of levels at which it branches left will be a subset of N and each such subset will correspond to a unique path.
So that if WM wishes to establish his claims, he must PROVE that aleph_0 = 2^aleph_0, or show a bijection between N and 2^N, which he often claimed but has never proved (shown a bijection for). 



