Re: Cantor Confusion

On 26 Mrz., 17:01, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1174671186.397336.238...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueck...@xxxxxxxxxxxxxxxxx writes:

> On 23 Mrz., 16:10, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> > In article <1174655745.265882.25...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueck...@xxxxxxxxxxxxxxxxx writes:
>
> > > The union of all *finite* paths gives the same infinite set of nodes
> > > as the union of all finite trees. But there is no infinite path
> > > involved / required.
> >
> > Not in the uniting, no. But the union *does* contain infinite paths. Just
> > like the union of all finite initial segments of the natural numbers does
> > have as a subset an infinite initial segment of the natural numbers.
>
> Just like the set of all negative unit fractions does cover zero?

What does *this* mean? The set of all negative unit fractions does contain
infinite subsets. *That* is what we are talking about.

But these infinite subsets do not cover zero.

> > > The union of all finite paths is an infinite union of finite paths,
> > > but not an infinite path.
> >
> > It is not even a path. It is the infinite tree.
>
> The union of all paths with only zeros 0., 0.0, 0.00, ... does not
> contain and not possess an infinite path. And it does not require an
> infinite path to pass every finite level.

Again that word "contain". The union of all those paths *is* an infinite
path.

The union of all finite paths is an infinite union of finite paths.
Why should it be an infinite path?
The union of all finite natural numbers is an infinite uion (with
cardinality aleph_0) but it is not an infinite natural number.

Regards, WM

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