Re: abundance of irrationals!)

aeo6 Tony Orlow 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,

Ok; if that's what you mean, then each node on the tree
is represented by a *finite* sequence of digits.

> 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.

Infinite strings represent *paths*. Not nodes. You talk
about elements (nodes, I assume) that are "infinitely far
out", as if they exist. But they don't! Every node in the
tree is a finite distance from the root.

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

Well, if you had said "nodes" instead of "values", I could
simply have told you that you were wrong, as I am doing now.
I was giving you the benefit of the doubt.

> >
> > > 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.

Yes, I did. Now *you* think about it. If the path leads
to a node and ends there, that path identifies the node.
But if the path has no end, what node does it identify?

There are *more* paths than there are nodes.

[snip]

.

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