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From: Virgil on 6 Jul 2006 13:33
In article <1152182749.926627.310460(a)a14g2000cwb.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> Dik T. Winter schrieb:
>
>
> > > All sequences, even all single transpositions which I apply can be
> > > enumerated by natural numbers, just like the lines of Cantor's list.
> >
> > Yup, so there are infinitely many.
>
> Are yo sure? How could an enumeration by natural numbers supply
> infinity?

If an enumeration rule exhausts the naturals, the set being enumerated
is Dedekind infinite.

>
> There are only finite natural numbers.

But infinitely many (an endless supply) of those finite natural numbers.
> > In addition to cardinality we have to do here with ordinality. The
> > sequence
> > of transpositions you give has order type w * w.
>
> No. There is a first element and there remains a first element.

The order type w*w indicates a well ordering, so any nonempty subset
has a first element. But this is not the order type of the set being
permuted, but of the set of permutations being applied to it.
>
> > > I have shown that the transpositions can be enumerated by natural
> > > numbers. There is no number oo. There is neither such a number in
> > > Cantor's list. All we do is enumerated. Othewise it would not be
> > > defined at all. Neither in Cantor's list. Therefore it is irrelevant
> > > whether or not something has to be "executed in order". We are in the
> > > countable domain and do not leave it.

But transpostitions do not always commute: As right operators on a list
(a b) (b c) = (a c b) but (b c) (a b) = (a b c)

So changing their order of execution can change their effect.
> >
> > That makes no sense, again. The transpositions have to be execute in the
> > order given.


>
> Of course. But that does not exclude that these transposition can be
> executed and finished (if Cantor's list can be finished).

But it prohibits them from being executed out of their prescribed order.
Since the order of execution induces a well ordering of the set of
transpositions, there would have to be a first transpostion producing
any given effect. What is the first transposition on a well ordering of
the rationals producing an ordering that is dense?


>
> > If you do them in any other order the result can be different.
> > So that you *can* enumerate them does not mean that the result is
> > independent
> > from the order in which you perform them.
>
> That is true but has no relevance.

Perhaps not to you, but considerable to your alleged theorem.
From: Virgil on 6 Jul 2006 13:47
In article <1152183139.346593.62140(a)m73g2000cwd.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> Virgil schrieb:
>
> >
> > Anyone is quite free to reject any axiom set, but no one is free to
> > impose any prohibition of any axiom set on others.
>
> Unless there appears a conradiction.

Not even then. Though if there were any contradiction WITHIN an axiom
system there would be little point in pursuing that system further.

Anyway, "mueckenh" has not shown any contradictions within any set of
axioms, he has only shown conflicts between certain axiom sets and what
"mueckenh" himself believes to be some ultimate "truth".
>
> > > > Pray define an f for which K(f) is not defined, and state *where* the
> > > > definition fails.
> > >
> > > For any f it fails in
> > > f : X --> K, where X is any non-empty set and
> > > K = { x in X | x is not an element of f(x) }
> >
> > This can only fail if the elements f(x) are sets. Suppose a,b, and c are
> > not sets (or classes) and thus cannot have members, then any function
> > from X to X is a function with codomain X = K = {x in X : x not in
> > f(x)}, since x in f(x) is now impossible.
>
> There is only one element K, which, accordng to its definition, is a
> set - it is as impossible a set as the set of all sets.

In my example above, K is an inevitable set, the codomain of the
function.

For f: {a,b,c} -->{a,b,c}: x |--> x, where a, b and c are NOT sets or
classes, then K = {x in {a,b,c}: x not in f(x)} = {a,b,c}

So "mueckenh" muecked up again.
From: Virgil on 6 Jul 2006 13:50
In article <1152183314.431125.225030(a)m79g2000cwm.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> Virgil schrieb:

> >
> > To say n --> oo does not require that there exist any "oo".
> >
> > It is merely an abbreviation for "as n increases unboundedly beyond any
> > given natural".
>
> But always remaining a natural itself, yeah. And the number of naturals
> < n is always a finite number, so that it never does reach infinity
> too.

But it reaches past every finite value!!!
From: Virgil on 6 Jul 2006 13:56
In article <1152183486.731953.63560(a)j8g2000cwa.googlegroups.com>,
mueckenh(a)rz.fh-augsburg.de wrote:

> Virgil schrieb:
>
> > In article <1152100475.267310.35180(a)v61g2000cwv.googlegroups.com>,
> > mueckenh(a)rz.fh-augsburg.de wrote:
> >
> > > Dik T. Winter schrieb:
> > >
> > > > In article <1151865025.167976.75740(a)v61g2000cwv.googlegroups.com>
> > > > mueckenh(a)rz.fh-augsburg.de writes:
> > > > > Dik T. Winter schrieb:
> > > > > > > you need some limit in order to well-order the set of rationals?
> > > > > >
> > > > > > No, but that is unrelated. Transpositions operate on a set. A
> > > > > > sequence
> > > > > > of transposition hence also operates on a set. There is *no*
> > > > > > definition
> > > > > > how an infinite sequence of transformations operates on a set.
> > > > >
> > > > > These transpositions *are* a set. In my example, they do not destroy
> > > > > the bijective mapping between |N and |Q.
> > > >
> > > > This makes no sense. Yes, there is a set of transpositions, but the
> > > > transposition operate in sequence on an ordered set Q1 (which is a
> > > > well-ordered set to start with). After each finite sequence of
> > > > transpositions there is still a bijective mapping between N and the
> > > > resulting ordered set. You have not proven that that is still the case
> > > > after an infinite sequence of transpositions. Much less that it is
> > > > still the case after an infinite sequence of infinite sequences of
> > > > transpositions. You have first to define what that means.
> > >
> > > All sequences, even all single transpositions which I apply can be
> > > enumerated by natural numbers, just like the lines of Cantor's list.
> > > OK? And as long as I can enumerate, the number reached is finite. OK?
> >
> > But the numbers which are reachable include values larger than any given
> > natural. To assume a limit on what is achievable is counterfactual.
> >
> > > Hence I do not need more than a finite set of transpositions, though
> > > its cardinality cannot be given.
> >
> > You, in fact, need more than any finite number can supply.
> >
> >
> > > This case is in general denoted by
> > > "countably infinite" - but the number oo does *not* appear. Please
> > > consider these facts before you demand something to be proved for "oo".
> >
> > WE do not require that it be proved for "oo", which is not a natural,
> > but that it be proved for "for all n in N".
> >
> > You method produces exceptions to "for all n in N", by implying that
> > there is some remote member of N beyond which it is not necessary to
> > proceed in order to establish "for all n in N".
>
> There is not a remote member, but each natural is finite. Therefore,
> how far we may go, there is never an infinite number of transpositions
> accumulated. And the Cantor diagonal has not an infinite number of
> digits.

But it is provably different from each and every number in the list.
If "mueckenh" disputes this, he must find some listed number equal to
the number described by any of Cantor's diagonal rules, or at least show
that some such number must exist.

And after than, "muecken" must show why Cantor's first proof is invalid
and also why Cantor's proof that no set surjects to its power set is
invalid. And then "mueckenh" can take on all those many other proofs of
the uncountability of the reals.

Only when he has finished all of that can "mueckenh"claim that the
Cantor theorem is invalid.
From: mueckenh on 6 Jul 2006 13:58

Dik T. Winter schrieb:


> > > Moreover,
> > > different orderings of the same subset of transpositions will yield
> > > different results. In the formula I gave for the diagonal number,
> > > calculation of one digit does *not* depend on the calculation of
> > > another digit.
> >
> > And what is the consequence of this?
>
> That the calculation of the digits can be done in parallel?

It is done in zero time. It is determined from the beginning. Why
should something "be done"?
>
> > > It is Cauchy's theorem that proves that the limit exists, using the
> > > epsilon argument.
> >
> > But Cantor's arguing is that without epsilon.
>
> I do not know Cantor's argument exactly. I think that he implicitly
> uses that result. In my formulation it was abundantly clear that I did
> use it.

But he needs to consider every digit with equal weight. That is not a
limit process.
> > But recently you raised the problem that the ordinal number of the set
> > of all transpositions is not omega. This is in fact a problem, as your
> > above question shows again, because we cannot count up to or even
> > beyond the first limit-ordinal number. But I think I can remedy this
> > disadvantage in the following way.
> >
> > Start with transpositions of the indices:
> > (1, 2)
> > (2, 3)
> > (1, 2)
> > This orders the set of the first three elements by size. Then order the
> > set of the first 4 elements by size.
> > (1, 2), (3, 4)
> > (2, 3)
> > (1, 2)
> >
> > And so on. After the set of the first n elements has been ordered by
> > size, order the set of the first n+1 element by size. At most n-1
> > transpositions are required for a set of n elements. Now, the complete
> > set of transpositions is of order type omega. And it maintains the
> > well-order by index while it achieves well-order by size.
>
> And again, it maintains the well order as long as you remain in the finite
> domain. So up to every finite n, you have ordered the first n elements
> of the well-ordered list, retaining a finite list. And back again to
> problem 1 with this. How do you define that when n grows without bound?

I remain in omega! So I always stay with finite numbers, they may grow
as far as you like. Wha holds for small natural numbers does equally
well hold for big ones, as far as well-order is concerned.

> How do you define the "limit"? And if you define that, is that "limit"
> also well-ordered? Those are things you have to prove.

I define limit by: *Using all finite natural numbers* just as like as
Cantor does.

Regards, WM

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