Re: Cantor Confusion

In article <1175681058.802334.28720@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

On 2 Apr., 14:43, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1175261756.864645.326...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
mueck...@xxxxxxxxxxxxxxxxx writes:

> On 30 Mrz., 05:14, "*** T. Winter" <***.Win...@xxxxxx> wrote:
...
> > > > > > No, the sum is undefined. If you think it is defined,
> > > > > > *prove* it and
> > > > > > *prove* that the sum is 2. What is the sum of the
> > > > > > "sequence":
> > > > > > "..., 1/4, 1/2, 1"?
> > > > >
> > > > > Well, what is it? If any infinite series ever had a value,
> > > > > then the
> > > > > sum of this sequence is 2.
> > > >
> > > > But it is not a sequence according to mathematical definitions.
> > >
> > > It is a sequence according to mathematical definitions, if you read
> > > it
> > > from the left hand side.
>
> Oh, that should read "from the right hand side"! But you read it as it
> was meant.
> >
> > So there is a first element, being 1, a second element, being 1/2, the
> > only
> > difference is that you apply right to left reading.
> >
> > > Further you can determine a unique limit value by lim{n-->oo}
> > > (1/2^n
> > > + ... + 1/8 + 1/4 + 1/2 + 1}.
> >
> > Yes, because you can revert finite sequences without consequence. You
> > do not even need convergence for that.
>
> And if the series is absolutely converging, then you can exchange all
> terms you like. The result is independent of the order.

Not arbitrarily. The result must also be a sequence. If I start with the
sequence (1, 1/2, 1/4, 1/8, ...) I have a convergent series. If I
apply in sequence the transpositions (1 2), (2 3), (3 4), ... I do not
know what I get because it is not defined in mathematics. On the other
hand, the only reasonable definition for a result I can come up with is
the sequence (1/2, 1/4, 1/8, ...), because there is *no* place for the
element "1". So the sum is changed.

You can mirror the series without change of value. You can decide to
write from right to left.

But you cannot add them unless every term has an index, and WM's
"permutation" leaves at least one term without any index.

> > But in mathematics sequences
> > are defined as having a first element. On the other hand, I wonder
> > how
> > you prove that the series of interchanges on the initial sequence lead
> > to your final "sequence".
>
> It is the same as Cantor's "proof" that he gets ready.

It is not.

It is.

Then Wm does not understand either Cantor or himself.


No. You can *prove* that whatever entry you take the diagonal is
different.

But you cannot prove that for entries which you did not yet take.

Actually, one can, like Cantor did, prove that for any given listing of
binary strings there is a /general rule/ which produces a string not in
that list based on the fact that two such binary strings are different
as a whole if they are different at ay index position.

WM keeps trying vainly to invalidate that argument, but never succeeds.


And, this is independent of the entry you take. So the diagonal differs
from
*all* entries.

Wrong. That holds only fnly for finite segments, not for infinite
segments of the diagonal.

It only requires one bit per listed string, which is easy to accomplish.


You can. If you can prove there is a number that is different from each
member of the list, as is done.

Only for finite segments.

Wm may so limit himself, but he cannot limit others to his own
incapacities.


The same holds for paths. Every finite node of a path means the whole
path because there are no "infinite nodes".

But there are infinite paths in every compete infinite binary tree.


Why? How do you know? If the mapping is (for instance) f: N -> R,
f(n) = sqrt(n), I see no counting involved at all.

That is deplorable. (N is created by counting.)

N is created by axiom. So it is WM who is deplorable.

>
> Every separation takes place at a separation point. No separation
> takes place at any other point. There are only countably many
> separation points. And there is only one initial separate path.

Yes. So what?

Yes. That's it.

A tree with only one path has only one node.

A complete infinite binary tree has as many paths as there are subsets
of N.
.

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