Re: Cantor Confusion

On 9 Mai, 04:42, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1178463202.150771.47...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> WM <mueck...@xxxxxxxxxxxxxxxxx> writes:
> On 6 Mai, 04:33, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> > > > Your conclusion is invalid. Indeed, each node is passed by a
> > > > mutlitude of paths. But the path "0.10101010..." is nevertheless
> > > > separated from all other paths.
> > >
> > > No. Only from those few you can ask for.
> >
> > And those are (in your opinion) only finitely many. Or can I ask for
> > infinitely many paths?
>
> No. You may ask for a set of infinitely many paths, but not for
> infinitely many paths.

So in your opinion there are only finitely many paths in the infinite tree.

I did not say that.

Great. You reject the axiom of infinity, as I wrote already many times
before. (Note: reject, not refute.)

I did not reject anything. I ask you whether you can put infinitely
many questions. Please answer. If the answer is no, then you should
recognize that you cannot ask whether all paths which accompany
0.101010... will get separated from it.

> > > > So what? For each two different paths it is possible to state a
> > > > node where they do separate. For each path there is no node where
> > > > it separates from all other paths. There is no contradiction.
> > >
> > > No. Only from those few you can ask for.
> >
> > Opinion again.
>
> It is proven by Cantor that there are uncountably many real numbers.
> It is also clear that there are only countably many questions, no?

I would think there are only finitely many questions. But all this is not
mathematics but philosphy.

No. That is mathematics, because you imply to be able to ask
infinitely many question.

> > There is a subtle difference. With the diagonal proof we always remain in
> > the finite.
>
> That's why it fails for numbers with infinitely many digits.

No.

Assertion, no argument.

> > With the paths we can also always remain in the finite because
> > any two different paths have a node where they differ *in the finite*. So
> > for each node there is a path p' that has it in common with p. But also
> > for each p' != p there is a node in p that is not in p'.
>
> No. Only for those p' you can ask for. There must be others, because
> there is no node which belongs only to p alone.

I do not ask for each individual path. I show it for all. Or do you reject
the formula sum{i = 1..n} = (n + 1) * n / 2? You can not ask the validity
for all n, but only for finitely many n, according to your thinking. So for
each and every n you have to prove it again.

This formula is obviously only correct for finite sets f nubers,
becasue for the whole set the result aleph_0. Do you think it would be
possible to sum an infinite set by (n + 1) * n / 2?

> > > "completed" means nothing remains.
> >
> > Perhaps.
>
> Sure.

Oh. A mathematical proof, please.

The set of natural numbers is completed such that there is no natural
number outside of that set. Do you disagree?

> > > In the present state of affairs we know that at every node of path p
> > > which we look at, path p is not unique. And we know, that we can test
> > > and test for different paths p', but we cannot carry out more than
> > > countably many tests (our list is limted).
> >
> > You can do countably many tests? I thought you could only do finitely
> > many tests.
>
> Please read carefully. We cannot do more than countably many. And in
> finite time, we cannot do more than finitely many.

Perhaps. You are getting more and more philosophical each turn.

You refuse to talk and think about what mathematicians can do?

> > But again, in mathematics it is *not* necessary to test
> > each and every individual. Otherwise you could not even prove that
> > sum{i = 1..n} i = (n + 1) * n / 2
> > for all n, but only for finitely many n.
>
> Of course this is only true for finitely many n. For infinitely many
> natural numbers, we get aleph_0.

You are wrong. We do not get that. If so, pray show a proof.

See, for instance, Hrbacek and Jech, p. 188, if you do not believe me.
(There are more books which treat that topic, but this in my
possession:

"It is not very difficult to evaluate infinite sums. For example,
consider
1 + 2 + 3 + ... + n + ... + ... (n in N).
It is easy to see that this sum is equal to aleph_0."

What else should this sum be? It cannot be finite and it cannot be
uncountable.

What is the relevance? But try to find the first occurrence where pi(x) and
Li(x) switch place. Nevertheless, it *is* a natural number

Are you sure? Look at

What evidence is there that 2^65536 is a natural number?

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ndjfl/1093634481&abstract=

Regards, WM

.

Quantcast