Re: abundance of irrationals!)

In article <MPG.1ceac1efbfdb7808989bfb@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:

> Russell said:
> > aeo6 Tony Orlow wrote:
> >
> > [snip]
> >
> > > 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.
> >
> > Yes.
> >
> > If you want to complain that it would require infinite digits for
> > > most values,
> >
> > Huh? What does that even mean? What values are you
> > talking about? And digits of what?
> Duh. The digits of the binary number that represent each node on the tree,
> where each bit represents a right or left branch as a 0 or 1. Infinite
> strings
> of such digits represent elements infinitely far out on the branches.

Wrong! Every node is represented by a finite string of binary digits,
since there are no "nodes at infinity". An infinite such string can only
represent a never ending pathway through infinitely many nodes. A string
cannot stop on at any node without coming to an end and being finite.


> What did you think, when I referred to correspondence between the
> nodes of a binary tree and binary integers?

When you mentioned infinite strings representing nodes, we knew you were
not referring to anything but your own delusions.

> >
> > > that's irrelevant. it's still an enumeration of the reals.
> >
> > No. The obvious linear enumeration of *nodes* is not an
> > enumeration of the reals. For that, you need to enumerate
> > the different paths. That's not merely a rearrangement of
> > the enumeration of nodes; if you think it is, then please
> > tell us how you associate each node with one and only one
> > path. Can't be done.
> Each bit enumerates a step in the choice of path. Think about it.

But the infinite strings do not stop, so cannot repredsent any
particular node.
> >
> > > 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.
> >
> > Yes, it would be rather inconvenient to make that an axiom,
> > since it's inconsistent with the other axioms. I think you've
> > gotten stuck thinking about rearrangements that turn out not
> > to exist in fact.
> >
> >
> Think what you like.

We would like you to think, but so far there has been no evidence that
you can.
.

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