Re: Cantor Confusion

In article <1176302166.249971.221780@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

On 10 Apr., 22:02, Virgil <vir...@xxxxxxxxxxx> wrote:
In article <1176230527.931928.12...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,

All mathematics is equally virtual. Finite trees and their finite paths
are as virtual as infinite trees and their infinite paths, as they exist
only in imaginations.

Wrong. The finite tree
0.
/\
0 1

does exist here, on your screen.

That is only a picture of a tree, and is no more an actual tree than a
picture of a triangle is an actual triangle.

'Actual' trees, like 'actual' triangles are only actual in one's
imagination.

Without pictures or words, there would be no imagination of
mathematics.

There have been blind geometers, who certainly did not rely on pictures.
Without words there would certainly be no communication of mathematics,
but whether that wold prevent imagination is not so obvious.


The finite tree shown above is the best image one can
give, mental or electronical.

Then that one is severely limited.

Cantor's proof does not cover the number behind the last one proved.

Since the proof covers all members of N simultaneously, there is no
"last one"

There is a last position to be recognized so far.

Not when one has done ALL of them simultaneoulsy.

To imagine all
numbers simultaneously can only mean that your are incapable of
imagine anything at all.

That your imagination is so limited, I agree, but your limitation is not
universal.

Same as:
The binary tree does not contain all paths, because otherwise it must
contain a level with uncountably many nodes.

Not in my world:
(1) A tree without all paths is not actually a tree at all.

(2) each partition of WM's set of countably many node levels into an
ordered pair of two sets determines a path in which the first set of the
pair contains those nodes levels from which the path branches left and
the other contains the set of node levels from which the path branches
right. This works equally well for finite or infinite trees.

For a finite tree with n levels, excluding leaves,
this produces 2^n paths.

For an infinite tree with aleph_0 levels, there being no leaves,
this produces 2^aleph_0 paths.

How does the production produce separated paths without separation
points?

WM seems to think that only one path get separated from all others at a
separation point. That only happens at the parent of a leaf node, which
can only happen in finite trees, never in infinite trees, which have no
leaf nodes.

Otherwise each node is the separation point of a set of several paths
from another set of several paths. In finite trees such sets of paths
are necessarily finite, but in infinite trees such sets are necessarily
infinite.

So that even if the set of all paths in an infinite tree were countable,
his demand for "separation" of any one path from the collection of all
others would be impossible.

Thus WM's "separation" arguments are irrelevant to the cardinality of
the set of all paths in a complete infinite binary tree.


That's how it works in mathematics, including ZF and NBG and most other
systems.

Deplorable mathematics.

It is the critic who is deplorable in his misunderstanding and
misrepresenting of mathematics.

If that is not how it works in WM's system, then WM's system is no part
of any standard mathematics.-

No these are unavoidable considerations the logic of which forces upon
us.

As mathematics follows from and is entirely consistent with logic, but
WM's presumptions are not, he speaks only for himself.

For everyone else, avoiding such irrelevant considerateness is quite
easy.
.