Date: Dec 18, 2009 9:19 AM Author: Dik T. Winter Subject: Re: Another AC anomaly? In article <2cd3c08a-a072-4ef5-bc0b-f0aaa9126ea9@a21g2000yqc.googlegroups.com> WM <mueckenh@rz.fh-augsburg.de> writes:

> On 17 Dez., 15:08, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:

> > In article <6a57309a-a136-430c-a718-e38518c65...@q16g2000vbc.googlegroups=

> .com> WM <mueck...@rz.fh-augsburg.de> writes:

> > > On 16 Dez., 03:44, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:

> > > > > ({1} U {1, 2} U {1, 2, 3} U ... = ({1} U {1, 2} U {1, 2, 3} U

> > > > > U {1, 2, 3, ...}

> > >

> > > That is a matter of taste.

> >

> > No, it is a matter of convention. In mathematics

> > a, b, c, ..., z

> > means a, b, c, continue this way until you reach z. But starting

> > {1}, {1, 2}, {1, 2, 3}

> > and going on you never reach

> > {1, 2, 3, ...}

>

> That is true. Therefore it does not exist.

That you can not get there step by step does not mean that it does not

exist.

> However, see Cantor,

> collected works, p 445:

> 0, 1, 2, 3, ... w_0, w_0 + 1, ..., gamma, ...,

> He seems to reach far more.

Right, he uses a convention that is no longer used.

> > > > A sequence of paths is not a path.

> > >

> > > But a union of paths is.

> >

> > No. Suppose we have the paths 0.000 and 0.100, what is their union? And

> > is it a path?

>

> No. But the union contains two paths.

Wrong. If we look at the paths as sets, they are sets of nodes. Their

union is a set of nodes, not a set of paths. And as a set of nodes we

can form from them seven different paths.

> And an infinite union of that

> kind may contain the path 0.111...

The union will contain a set of nodes. With the nodes we can form paths

(but they are not elements of the union). But by your statements an

infinite sequence is not a path, and so 0.111... is not a path.

> > > > Here, again, you err. You can not construct something in aleph_0

> > > > steps; you will never complete your construction. You *cannot*

> > > > get at aleph_0

> > > > step by step.

> > >

> > > But you can make a bijection with all elements of omega?

> >

> > Yes, but not with a step by step method that will ever be complete.

>

> The construction of the tree can be done within one step. Define: Let

> there be every finite path. And every finite path is. The construction

> of a path does not depend on a preceding step.

It was in the construction *you* made. But what is the definition of path

in this definition of the tree?

> > > > > This hold for every limit of every sequence of finite paths.

> > > >

> > > > A limit is not a step by step process.

> > >

> > > Then assume it is a mapping from omega.

> >

> > Which mapping?

>

> Let every finite path of every infinite path be mapped on the elements

> of omega. That was simple.

By your statements infinite paths do not exist. But pray give such a

mapping. Until now you have only asserted that such a mapping exists

without showing that.

--

dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131

home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/