Re: abundance of irrationals!)

In article <MPG.1cead42041fe5c92989c03@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:

> Virgil said:
> > In article <MPG.1ceaad4d31013157989bf7@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> >
> > > Randy Poe said:
> > > >
> > > > mueck...@xxxxxxxxxxxxxxxxx wrote:
> > > > > Randy Poe wrote:
> > > > > > A number x being in your list means that there is m
> > > > > > such that the m-th number in your list is x.
> > > > >
> > > > > No, these numbers exist in the same way as Cantor's antidiaonal.
> > > >
> > > > Then it's not a list of real numbers.
> > > >
> > > > First, Cantor's antidiagonal exists in the sense of NOT
> > > > being on the list. So you have constructed a list by
> > > > which all real numbers have the property of not being
> > > > on the list?
> > > >
> > > > There is one and only one meaning of "a list of real
> > > > numbers". That is that you have an association between
> > > > every natural number m and every real number a_m on your
> > > > list.
> > > >
> > > > If you respond to this remark with "no", as you did:
> > > > > > A number x being in your list means that there is m
> > > > > > such that the m-th number in your list is x.
> > > >
> > > > then x is not on your list. Because that is the meaning
> > > > of "on a list". There is no other.
> > > >
> > > > If you want to come up with some other construction
> > > > (some sort of infinitely deep binary tree, according
> > > > to comments from others), that's fine. But it's not
> > > > a list, and it doesn't advance your argument that
> > > > a list can contain all real numbers in (0,1).
> > > >
> > > > - Randy
> > > >
> > > >
> > > Excuse me, but the nodes of an infinitely deep binary tree can
> > > obviously be enumerated in a linear manner, corresponding to binary
> > > integers and thus to the naturals.
> >
> > But how does one enumerate the paths (from root node downward forever)
> > in such a tree? It is not at all clear that this can be done linearly
> > with an ordering to match that of the binary integers, as each path
> > would be infinitely long, and not representable by a binary integer.
> >
> > And it is the paths that turn out not to be countable, not the nodes.


> Absolutely wrong. We start at the top, element 1. The two child nodes are 10
> and 11. The two nodes off 10 are 100 and 101, and the two off 11 are 110 and
> 111. For each node, we add a digit to the right for each cild node, a 0 for
> the
> left and a 1 for the right. At each level n (starting with 1) we have 2^(n-1)
> nodes, numbered from 2^(n-1) to (2^n)-1, which are all the binary numbers
> with
> n significant digits. By normal tree traversal, an infinite binary tree
> cannot
> be traversed one branch at a time, but like the rationals in a 2D array, they
> can be traversed "diagonally".

Given any string, an infinite sequence of zeros and ones, by what rule
do you determine which will be the NEXT string in your "diagonal"
traverse of all of them?

For rationals a variety of specific rules can be found, but for such
infinite strings, there will be no such rule possible.

There is no difference in enumerability

There is unless you can provide a specific and concrete "next string"
rule which applies to every possible such string.

Hand waving, not acceptable.
> >
> >
> > > If you want to complain that it would require infinite digits for
> > > most values, that's irrelevant. it's still an enumeration of the reals.
> >
> > WRONG!
> That's a pretty weak argument. perhaps you'd like to expand on that thought.

Give us the "next string" rule or give up.
> >
> > > Enumeration is irrelevant to the level of infinity, except insofar as we
> > > must
> > > rearrange the quantities in order to enumerate them linearly. Don't look
> > > in
> > > your bag of axioms for proof of this. It's been conveniently excluded.
> >
> > Only by concentrating on counting the wrong things.


> The "wrong" things? define what is right and wrong, oh purveyor of
> mathematical
> morality.
> >
> > Each path of your infinite binary tree corresponds to a real number, so
> > until you count paths, not merely nodes, you are not done.
> >
> I counted the nodes in a way that traverses the paths all together. Check it
> out.

I did, it doesn't. You need a rule to proceed from each string of
infinitely many binary digits to a unique "next" string.

Give us the rule or give it up.
.

Relevant Pages

  • Re: abundance of irrationals!)
    ... but the nodes of an infinitely deep binary tree can ... But how does one enumerate the paths ... it's still an enumeration of the reals. ... Each path of your infinite binary tree corresponds to a real number, ...
    (sci.math)
  • Re: A definition of an infinity, perhaps!?
    ... a string is infinite iff it does not have a largest index). ... If you mean binary trees and unions of node sets, it is possible, at ... The result will be the node set of a complete n-leveled binary tree ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... an infinite binary tree cannot ... Cantor's diagonal proof which cimple proves there are more reals than naturals. ... >> rearrange the quantities in order to enumerate them linearly. ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... You are bringing digits into this, ... >>> it prove if a complement of an infinite binary string exists? ... >>string of digits. ... an infinite tree with finite branchings, such as a binary tree. ...
    (sci.math)
  • Re: The Binary Tree analysed by nodes and levels
    ... The Binary Tree is complete. ... node alone is a null string. ... The "infinite set of finite paths" has ... But there the infinite sequence of finite sequences ...
    (sci.logic)