From: Virgil on
In article <1161275459.898352.198880(a)f16g2000cwb.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> William Hughes schrieb:
>

> > And here is your problem. Uncountable means unlistable.
>
> Not my problem. Countably infinite means unlistable too.

For a set to be countable requires, by definition, that there be a
surjection from N to the set, which is a listing.

So that "Mueckenh" is claiming that countable means uncountable.

And that is only one of "Mueckenh"'s problems.
From: Virgil on
In article <1161275577.928121.206780(a)f16g2000cwb.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> MoeBlee schrieb:
>

> > You just went right past what I wrote. You just completely ignored what
> > I wrote, which was in direct response to you. We do NOT neglect the
> > exponentiation.
>
> You do. Otherwise there was no reason to explicitly exclude the case 1
> = 0.999. In real numbers we have this equation due to exponentiation.
> With Cantor's diagonal we have not.

In constructing the Cantor diagonal for a given list in decimal
notation, if the construction only uses, say digits, 5 and 6, or any set
of digits excluding 0 and 1 and 9, then the number created will be
different from both forms of any real having two decimal forms.
Since this is the method of construction, no duplication is possible.

Thus "Mueckenh"'s complaint is a straw man fallacy.
From: Virgil on
In article <1161275873.076315.312960(a)b28g2000cwb.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> Virgil schrieb:
>
> > In article <1161200729.156817.65510(a)m7g2000cwm.googlegroups.com>,
> > mueckenh(a)rz.fh-augsburg.de wrote:
> >
> > > Virgil schrieb:
> > >
> > > > In article <1161079685.233073.120000(a)k70g2000cwa.googlegroups.com>,
> > > > mueckenh(a)rz.fh-augsburg.de wrote:
> > > >
> > > > > MoeBlee schrieb:
> > > >
> > > > > > There is no "equal weight" in the proof.
> > > > > >
> > > > >
> > > > > You haven't yet noticed it? Each digit of the infinitely many digits
> > > > > of
> > > > > the diagonal number has the same weight or importance for the proof.
> > > >
> > > > It is only necessary that each digit have non-zero weight in order for
> > > > the difference between the diagonal and one of the list to be non-zero.
> > > >
> > > >
> > > > > In
> > > > > mathematics, the weight of the digits of reals is 10^(-n).
> > > > At least in decimal notation, but as that makes all the weights
> > >
> > > > non-zero, that is sufficient to distinguish the diagonal from each of
> > > > the listed numbers.
> > >
> > > Only for a finite diagonal. In the infinite case we have for example 1
> > > = 0.999...
> >
> > But if the diagonal is not allowed to contain any 1's, 0's or 9's, as in
> > standard constructions of "diagonals", such cases cannot occur.
>
> And why can "such cases" occur if the diagonal is allowed to contain
> these digits? What do you think?

If a real has two decimal representations, a method that merely avoids
one of them is not guaranteed to avoid both, but a method guaranteed to
avoid both will avoid both.
>
>
> > Its digits do not have equal weight, as "Mueckenh" well knows, but each
> > digit has a non-zero weight,
>
> lim {n--> oo} 10^(-n) =/= 0 ???

So which place value, 10^(-n), in that endless sequence of place values
has value zero?
>
> No, I did not yet know,

Then which digit in, say, 0.333... is the first one to have a zero place
value?
From: Virgil on
In article <1161276574.792436.186750(a)i3g2000cwc.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> Dik T. Winter schrieb:
>
> > In article <1161183237.727249.154740(a)b28g2000cwb.googlegroups.com>
> > mueckenh(a)rz.fh-augsburg.de writes:
> > > Dik T. Winter schrieb:
> > >
> > > > In article <1161079802.120515.175530(a)i3g2000cwc.googlegroups.com>
> > > > mueckenh(a)rz.fh-augsburg.de writes:
> > > > ...
> > > > > The inconsistency is that
> > > > > 1) For the balls inserted until noon, you can find the result: It
> > > > > is
> > > > > the set N.
> > > > > 2) For the balls removed until noon, you can find the result: It is
> > > > > the
> > > > > set N.
> > > > > 3) For the balls remaining at noon, the same arguments of
> > > > > continuity
> > > > > which lead to (1) and (2) cannot apply.
> > > >
> > > > There are quite a few obvious reasons.
> > > > (1) 1) is not because of continuity
> >
> > The reason that continuity plays no role is because the function of the
> > number of balls in the vase when written as a function of t is
> > discontinuous
> > at infinitely many positions.
>
> But it is stepwise continuous.
> >
> > > Why then?
> >
> > Depending on the person who writes about the problem. One possibility is
> > defining a set of functions dependent on time t that tells whether at time
> > t (not t now goes from -1 to 0) ball n is in the vase or not. Addition of
> > those functions gives the number of balls present in the vase at time t.
> > But all depends on the exact mathematical formulation of the problem.
> > There
> > is none (in my opinion, but I have already had discussions with others
> > about this point, so I will not repeat them).
> >
>
> > You can *model* the problem in different ways in mathematics, and one of
> > those models leads to 0 balls in the vase at time 0. And there are
> > presumably other models that give different results. So, unless the
> > problem is stated in a mathematically proper way, actually nothing can
> > be said about it.
>
> Translate balls as numbers and vase as set variable. More is not
> required.

Time is required, as every event occurs at a specified time.
>
> If the vase is not empty at noon, than one can never have all natural
> numbers together.

One has them all together twice, but outside the vase, once before the
first insertion and again after all removals.




> If the vase is empty at noon, then can obtain from set theory that the
> limit of a sequence in no way can be determined by the terms of he
> sequence.

The limit of "the sequence" would only be relevant if one required some
sort of continuity, but all the actions described are discontinuous, so
continuity is irrelevant.
From: Virgil on
In article <1161277339.760435.276080(a)m7g2000cwm.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> David Marcus schrieb:
>
>
> > > Therefore, in the limit {2,4,6,...} there are infinitely many finite
> > > natural numbers m > |{2,4,6,...}|
> >
> > Once again you use words and phrases that are not part of standard
> > mathematics, i.e., "in the limit {2,4,6,...} there are infinitely many
> > finite natural numbers m > |{2,4,6,...}|". What does "in the limit
> > {2,4,6,...}" mean?
>
> > You said (repeatedly) that standard mathematics contains a
> > contradiction, so please state the contradiction in standard
> > mathematics. If you use new terms, please define them.
>
> Sorry, I do not know what you state of knowledge is. {2,4,6,...} means
> obviously "all natural numbers". That is the usual notation in
> mathematics.

To me, {2,4,6,...} leaves out as many naturals s it contains.
>
> Binary Tree
> > Unfortunately, it was described in a way that I can't understand it. A
> > wild guess on my part is that you mean to set up a correspondence
> > between edges and sets of paths.
>
> I am sorry, but if you need a wild guess to understand this text, then
> we should better finish discussion. Observe just how the discussion
> runs with all those who understood it, like Han, William, jpale.
> Perhaps you will step by step understand it.
>
> > In standard mathematics, a finite set of natural numbers has a largest
> > element.
>
> Please prove that a set of elements consisting of 100 bits has a
> largest element.

Please prove that it is a set according to ZF or NBG. You will find that
you cannot do it without accepting the existence of N as an infinite set.
>
> > So, what are you actually saying? Are you saying that you don't
> > like standard mathematics?
>
> I am saying that standard mathematics is false.

But you are claiming it without any proof, merely your own set of
unprovable assumptions.

> As I have shown there
> are finite sets in mathematics which have no largest elements.

Claimed but not proven.