Re: Cantor Confusion

In article <1175681058.802334.28720@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
On 2 Apr., 14:43, "*** T. Winter" <***.Win...@xxxxxx> wrote:
....
> 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.

Yes. In that case the mirroring is just virtual and the first number is
on the right. But that means that the '1' applies to the *first* node
in the path. I do not think you want that...

> > 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.

If so, *prove* it. How do you *prove* that your series of interchanges on
the initial sequence leads to your final "sequence"?

> > Yes, you use limits to show something which you can not determine.
> > The strange thing is that the first node contributes nothing, as
> > does the second nodes, as do all nodes in a finite distance from
> > the root. And as all nodes in an infinite path are a finite
> > distance from the root. Nevertheless you maintain that all nodes
> > together contribute 2 because there is no node that contributes one,
> > there is no node that contributes 1/2, etc. So there is a sequence
> > of no nodes that contribute 2. And in some mysterious way you
> > conclude that that sequence of no nodes is the same as the sequence
> > of nodes.
>
> Take the geometric series. It contains exactly the same terms as my
> reverted series.

What is that in relevance to my remark?

To show tha its sum is 2.

What is the relevance to my remark? I will quote again:
"... And as all nodes in an infinite path are a finite distance from the
root. Nevertheless you maintain that all nodes together contribute 2
because there is no node that contributes one, there is no node that
contributes 1/2, etc. So there is a sequence of no nodes that contribute
2. And in some mysterious way you conclude that that sequence of no
nodes is the same as the sequence of nodes."
Now what?

> Yes, if you want to conclude that it is different from any other list
> entry, then it must be finished. Otherwise you only know that it
> differs from some entries.

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. And
that is the majority and remains so.

You take them all at once, by taking an "arbitrary" entry. As the entry is
arbitrary the proof goes for each and every entry, so in effect you have
proven it for all entries. This kind of proof is pretty abundantly available
in mathematics.

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.

There is only one infinite segment of the diagonal, and that is the complete
diagonal. So what are you talking about?

> You cannot say: "there is a real number that is not in the complete
> list" unless you have searched the complete list.

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.

What do you mean? I am talking about numbers, not about segments.

> > "Every finite part" means
> > just that: "every finite part", it does not apply to "infinite parts",
> > which is "all". On the other hand with Cantor we have "every finite
> > element" and that means "all elements", because there are no
> > "infinite elements".

You have no remark to this, which *shows* you are wrong?

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

Yes, so what? Every finite node of a path means all the nodes of the path.
Every finite part of the path does *not* mean all parts of the path.

> > > How would you know which element the n-th element is.
> >
> > By putting n in the mapping given.
>
> The mapping given includes and requires the counting.

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.)

Oh. So you require counting to use sqrt(500)? Strange.

> > > Therefore we know that only countably many are there.
> >
> > A proof, please, for once.
>
> 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.

No that is not a proof.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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