From: WM on
On 17 Dez., 22:02, YBM <ybm...(a)nooos.fr.invalid> wrote:
> WM a écrit :
>
>
>
>
>
> > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote:
> >> In article
> >> <2cd3c08a-a072-4ef5-bc0b-f0aaa9126...(a)a21g2000yqc.googlegroups.com>,
>
> >>  WM <mueck...(a)rz.fh-augsburg.de> wrote:
> >>>> 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. And an infinite union of that
> >>> kind may contain the path 0.111...
> >> So according to WM a union of sets may contain an object not contained
> >> in any of the sets being unioned.
>
> > For paths and initial segments to contain and to be is the same. In
> > fact:
> > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...}
> > and also
> > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}.
>
> As usual you confuse to be an element and to be a party of...
>
I am concerned with numbers. Your absurd distinctions are irrelevant.
Either N exists or not. If yes, then it is the union of all natural
numbers as well as the union of all initial segments of the ordered
set of all natural numbers as well as the union of all these and
itself, because

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

If you see that this equation is true, then you acknowledge my
argument. If you cannot see that, then either you are a matheologian
or you have too little intelligence to understand it or both.

Regards, WM

From: Quaestor on
In article
<9dde10e2-2ab6-4714-9dd1-d3ebaea51757(a)u7g2000yqm.googlegroups.com>,
WM <mueckenh(a)rz.fh-augsburg.de> wrote:

> On 17 Dez., 22:07, Quaestor <quaes...(a)qqq.com> wrote:
> > In article
> > <a737fea3-6dd2-4a1d-9506-2df7cb6d4...(a)v30g2000yqm.googlegroups.com>,
> >
> >
> >
> >
> >
> > �WM <mueck...(a)rz.fh-augsburg.de> wrote:
> > > > > It is. I constructed a finite path from the root node to each other
> > > > > node.
> >
> > > > Yup, you constructed a finite path, and that does not represent 1/3.
> >
> > > It 1/3 can be represented by a bit sequence, then it is represented in
> > > my tree by the union of the node sets representing
> > > 0.0, 0.01, 0.010, ...
> >
> > > > > Then I appended an infinite tail.
> >
> > > > Whatever that may be, it is *not* a path according to your explicit
> > > > statement
> > > > that the tree did not contain infinite paths.
> >
> > > You have misread my construction. I use finite paths and then I append
> > > an infinite tail.
> >
> > We have not misread the consequences of your construction.
> >
> > Attaching only one infinite tail , or even a countably infinite number
> > of infinite tails, to each finite subpath does not produce all possible
> > paths.
>
> Please prove your assertion by showing which path I did not produce.
> But prove it by knowing only all nodes of all paths.

Until you give some explicit way of telling which paths you do produce,
ther is no way of telling which ones you miss.

> Information exceeding the infinite sequence of nodes of each path is
> not acceptable in mathematics.

I do not accept your fiats on what is or is not acceptable in
mathematics, having seen you to be so often wrong in such claims.
>
> >
> > Consider the following model for a complete infinite binary tree:
> >
> > � �The set of nodes is N = {1,2,3,...}, each such natural being a node.
> >
> > � �For each node n in N,
> > � � � its �left child is defined to be 2*n + 0 �in N and
> > � � � its right child is defined to be 2*n + 1 in N.
> >
> > � �A path is, by definition, a set of nodes which contains 1, and,
> > � �for each node in it, contains exactly one child of that node.
>
> Which path do you see in the tree that has not that property?

In my tree, none. But in there are sets of nodes satisfying my condition
of pathhood which do not occur in your tree.
> >
> > Until WM can produce some cogent reason why the above definition fails
> > to be a binary tree,
>
> It does not fail. Every node has two child-nodes.

Then WM concedes his errors.
From: Quaestor on
In article
<25016582-3715-4b47-a02f-5fe79b0a594b(a)m3g2000yqf.googlegroups.com>,
WM <mueckenh(a)rz.fh-augsburg.de> wrote:

> On 17 Dez., 22:02, YBM <ybm...(a)nooos.fr.invalid> wrote:
> > WM a �crit :
> >
> >
> >
> >
> >
> > > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote:
> > >> In article
> > >> <2cd3c08a-a072-4ef5-bc0b-f0aaa9126...(a)a21g2000yqc.googlegroups.com>,
> >
> > >> �WM <mueck...(a)rz.fh-augsburg.de> wrote:
> > >>>> 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. And an infinite union of that
> > >>> kind may contain the path 0.111...
> > >> So according to WM a union of sets may contain an object not contained
> > >> in any of the sets being unioned.
> >
> > > For paths and initial segments to contain and to be is the same. In
> > > fact:
> > > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...}
> > > and also
> > > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}.
> >
> > As usual you confuse to be an element and to be a party of...
> >
> I am concerned with numbers. Your absurd distinctions are irrelevant.

They are neither absurd nor irrelevant to mathematics.

But then, comprehensible mathematics and WM seem to be incompatible.


> Either N exists or not. If yes, then it is the union of all natural
> numbers as well as the union of all initial segments of the ordered
> set of all natural numbers as well as the union of all these and
> itself, because
>
> {1, 2, 3, ...} U {1, 2, 3, ...} = {1, 2, 3, ...}
>
> If you see that this equation is true, then you acknowledge my
> argument.

On the contrary, that equation is true AND your argument is FALSE.

> If you cannot see that, then either you are a matheologian
> or you have too little intelligence to understand it or both.

Actually, what we have is too much intelligence to "understand" that
which is incapable of being understood, i.e., is nonsense..
From: K_h on

"Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message
news:Kusoo8.xz(a)cwi.nl...
> In article <GLidnXhuo5t347TWnZ2dnUVZ_sWdnZ2d(a)giganews.com>
> "K_h" <KHolmes(a)SX729.com> writes:
> > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message
> > news:KuqwG3.qv(a)cwi.nl...
> ...
> > > > > The definition you provided for a sequence of sets
> > > > > A_n
> > > > > depends on whether
> > > > > each A_n is or is not a set containing a single
> > > > > set as
> > > > > an
> > > > > element.
> > > > >
> > > > > Your definition leads to some strange
> > > > > consequences. I
> > > > > can
> > > > > state the
> > > > > following theorem:
> > > > >
> > > > > Let A_n and B_n be two sequences of sets. Let A_s
> > > > > =
> > > > > lim
> > > > > sup A_n and
> > > > > A_i = lim inf A_n, similar for B_s and B_i. Let
> > > > > C_n
> > > > > be
> > > > > the sequence
> > > > > defined as:
> > > > > C_2n = A_n
> > > > > C_(2n+1) = B_n
> > > > > Theorem:
> > > > > lim sup C_n = union (A_s, B_s)
> > > > > lim inf C_n = intersect (A_i, B_i)
> > > > > Proof:
> > > > > easy.
> ...
> > > Well, the above theorem is still not valid with your
> > > definition.
> >
> > What case did you have in mind? I found cases where it
> > works fine.
>
> Let's have some arbitrary object 'a' and the natural
> numbers. Create
> the sequence A_n where A_n = {a} and the sequence B_n
> where B_n = {n}.
> According to your definition:
> lim sup A_n = {a}
> and
> lim sup B_n = {N}.
> Now create the sequence C_n: C_2n = A_n, C_2n+1 = B_n.
> Again according
> to your definition:
> lim sup C_n = {a}
> which is not equal to union (lim sup A_n, lim sup B_n).

This is a good example, thanks. Your theorem only applies
in special cases for the definition I have offered (although
my definition satisfies some different but interesting
theorems).

k


From: K_h on

"Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message
news:Kusoty.176(a)cwi.nl...
> In article <yrydnX_FwfczULTWnZ2dnUVZ_vmdnZ2d(a)giganews.com>
> "K_h" <KHolmes(a)SX729.com> writes:
> ...
> > lim(n ->oo) n = N is true for the standard definition of
> > natural numbers using just the wikipedia definitions for
> > the
> > limit of a sequence of sets.
>
> But not with Zermelo's definition of the natural numbers.
> Nor when we take
> the natural numbers as embedded in the rational numbers.

Yes, and that was never denied. I just point out that any
limit definition for a sequence of sets will give answers
that violate the intuitive notion of a limit for certain
constructions. There are two ways one can deal with that.
First, for a given class of constructions, have another
definition that does not violate the limit notion. The
second way is to look at the meaning that the wikipedia
definitions have for the constructions. I now think that
the latter approach is superior to the first since the first
caused so much misunderstanding. With the second option we
can restrict ourselves just to the wikipedia definitions and
define the sets n by:

n = 0 = {}
n = 1 = {{}}
n = 2 = {{{}}}
....

Wikipedia gives liminf(n-->oo) n = 0. What this means is
that nothing is `accumulated' in a limit set since each set
does not persist past its introduction. So the notion of n
growing bigger, as one tends to the limit, is not embodied
here since |n| is 0,1,1,1,1... as one proceeds and is 0 in
the limiting case. Under the standard construction each
natural is `accumulated' in the limit set, since each set
persists beyond its introduction, and this preserves the
notion of n growing bigger as one proceeds: |n| is
0,1,2,3,... and is N in the limiting case.

k