Re: Cantor Confusion

In article <1168551871.231366.88600@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Virgil schrieb:

In article <1168511743.391940.58070@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Franziska Neugebauer schrieb:

Since there is no largest element in "potentially" infinite sets (in
"actual/complete/finished", too) this sentence makes no sence at all.

A potentially infinite quantity (set or not) is always finite.

There is no time in maths.

So you write these letters and develop new ideas in zero time? There is
no existence outside of time.

Therefore in a linearly ordered set here is a last element. Contrary
to the claim of set theorists, a set is not fixed in reality.

When your arguments (by the way: What exactly are your arguments?)

Here you can read it:

Theorem. The set of real numbers in [0, 1] is countable.

Lemma.
Each digit a_n of a real number r of the real interval [0, 1] in binary
representation has a finite index n.
r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}.

This is only true if N is actually infinite, but is quite false
otherwise.

It is always true. If N is not actually infinite, then there are no
actually infinite strings, i.e., in particular there are no irrational
numbers.

The rational numbers with the usual addition and multiplications form a
field, which must be closed under addition and multiplication.
This cannot happen with a finite N.
So WM would scuttle standard arithmetic to maintain his delusions.



I corrected that already: 0 is not a natural number.

Looks natural enough to me. There is nothing in mathematics which
requires the naturals to start at 1 rather than 0, and there are many
reasons in math, computer science, engineering and science why starting
with 0 is more natural.

The union of binary trees is defined as the union of levels.

Not in general.

But here.

Give a precise definition then. For example, how does one form the union
of a trinary tree with a decimal tree?

The union of two or finitely many different finite binary trees simply
is the largest on.

Not for disjoint trees. In that case the union is not a tree at all.

But here.

Not without a more precise definition of the WM-union of trees thatn has
been so far presented.

For example, it may be in finite binary trees that neither of two such
trees is tree-isomorphic to any subtree of the other, What is the union
then?


The set of /finite/ paths in the union is countable.
But when one takes the union of sets of finite paths one only gets
finite paths in that union. There are no infinite paths in that union.

The union is the set of infinite paths.

No nodes? No edges? Just paths that were not in any of the original
trees?


The same thing happens with ordinals. When one takes the union of all
finite ordinals (like unary trees), there is no infinite ordinal IN that
union

The union is the infinite ordinal.

Which is not one of the objects being unioned, so there is no reason in
that for the union of finite trees to be a tree itself.
.

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