Re: abundance of irrationals!)

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.



> 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!

> 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.

Each path of your infinite binary tree corresponds to a real number, so
until you count paths, not merely nodes, you are not done.
.

Relevant Pages

  • Re: Cantor and the binary tree
    ... > Cantor and the binary tree. ... > to assume that one of these numbers becomes uncountably infinite while ... , and are thus as uncountable as the reals. ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... By normal tree traversal, an infinite binary tree ... Given any string, an infinite sequence of zeros and ones, by what rule ... >>> rearrange the quantities in order to enumerate them linearly. ...
    (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: how to list all of the real numbers
    ... the reals between 0 and 1, ... The only thing the binary tree contains in the usual sense of the ... The set of primitive components is countably infinite. ... The subset of that power set containing precisely the paths from ...
    (sci.math)
  • Re: An uncountable countable set
    ... infinite-length paths as H-riffic numbers. ... the H-riffics in "Well Ordering the Reals" as a proposed well ordering. ... infinite paths of your binary tree as well as the finite-length paths ... can linearly order the reals in this way, but eliminating any "countably infinite descending sequences" appears to be a matter of infinite regression, and I don't see how to prove that it's a *countably* infinite regression. ...
    (sci.math)
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