Re: Cantor Confusion


*** T. Winter schrieb:

> The diagonal digits a_nn of Cantor's argument have to be
> multiplierd by 10^-n in order to yield the diagonal number SUM a_nn *
> 10^-n. But there is no problem with yielding zero for n --> oo?
> --- Remarkable.

Apparently you do not understand the working of limits.

Apparently your limits work only for real numbers in lists but no for
real numbers in trees.

It has been
proven that *every* Cauchy sequence yields a real number. And it is
easy to see that *every* decimal expansion is in fact a Cauchy sequence.
10^-n is *never* zero. You need limits.

Yes. And they say: As long as 10^-k is not zero, the sum {n=1 to k} a_n
+ 10^-n is not an irrational number.


> There is no edge in the *finite* domain which is "passed in full". In
> the infinite series there is an edge passed in full.

Which edge is passed in full to the path that goes at each step to
the left?

I don't claim the existence of this representation of 0.

> Otherwise the path
> does not exist at all.

That is what you are trying to prove. But you can not just state it. You
hav to *prove* it.

I need not prove that two entities which have no different property,
cannot be distinguished.

> The infinite sum 1 + 1/2 + 1/4 + ... yields 2, although there is no
> smallest term 2^-n.

Indeed.

> It can be applied to calculate the edges per path. You should know that
> absolutely converging series can be reordered.

You should know that you can *not* reverse absolutely converging series.

Proof?

And that is what you are doing. Because that would result in a series
without start.

No reordering of an absolutely converging series changes its sum.

> ===============================
> > I now understand what you were trying to do.
>
> It lasted rather a while.

Consider it a complaint about your clarity and lack of definitions.

Here are some readers who understood it quite easily.

> > But in the complete tree
> >the parts of edges that are assigned to an infinite path are not 1,
> >1/2, 1/4 etc. There is *no* edge that is completely assigned to a
> > path,
>
> Then the tree is incomplete. You fail to understand infiniteness
> already on this low level?

Circular reasoning abounds here. Care to give a *proof* that the tree
is not complete in that case? This statement is completely similar to
the statement that the completed set of natural numbers should have a
last element.

There is no largest element, but the infinite path is nothing than the
union of all finite paths. Therefore we can calculate the union of all
finite sequences 1 + 1/2 + 1/4 + ...+1/2^n.

Please state when you intend to no longer reason within ZF (especially
the axiom of infinity). As long as you are reasoningin in ZF, there
is *no* largest element of the (completed) set of natural numbers.

> > Note that in each finite tree, the complete edge assigned to a path is
> > the last edge on the path. As there is no last edge in an infinite
> > path, this does not hold in the infinite tree.
>
> So there are different paths without different edges?

No. I never did state that.

So at least one path has its own edge?

Regards, WM

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