Re: An uncountable countable set

In article <1151865025.167976.75740@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
*** 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.

Well, in the first place, this is false as written. The transpositions
are conditional. But indeed, given the well-ordered set of rationals
Q1, this gives a different ordering of the rationals, say the well-ordered
set Q2. We can call that an infinite permutation.

But all transpositions to be applied to a given set are completely
defined by the few lines of my definition and they are determined from
the scratch. There is not a single one undetermined.

No, but the "infinite permutations" have to be executed in order. Or can
you prove that they need not? Moreover, you have to show that an infinite
sequence of such "infinite permutations" leave well-orderedness. You have
not shown that.

You keep assuming that the definition of the anti-diagonal of a list of
reals is an ongoing steps. That is false. There is only a single step,
the definition.

The same is true for my example. There is only one single definition.
If you do not like "steps" then drop that word.

You *explicitly* say: apply aleph-0 times. Consider the Cantor diagonal
on a list l_n of reals in the range [0, 1):
I define two subfunctions (only to shorten the definition line):
remainder(x, a): x - entier(x / a) * a
and
digit(k): if k = 4 than 5 else 4
and now:
D = sum{k = 1...oo} digit(remainder(entier(l_n * 10^k), 10)) / 10^k.
there are no steps involved.

After that we can determine each and every digit of
the diagonal number without the need to know other digits. Moreover,
by Cauchy, such a sequence of digits defines a real number.

Leave Cauchy out of the play. There is no justification to reach all
last digits of the diagonal number by his epsilons. His specification
would only show that all columns of the list up to nn have been
covered.

Apparently you do not understand what Cauchy involves.

I do not know of any definition of a sequence of overlapping "infinite
permutations" that allows an inifinite sequence. Pray provide one.

There are no "overlapping" permutations.

There are. The first sequence of transpositions defines the first
"infinite permutation". Indeed, the transpositions do not overlap.
The second sequence of transpositions defines the second "infinite
permutation", but that one overlaps the first "infinite permutation".

Because there are not infinitely many comparisons, there is a single
definition.

This single definition leads to infinitely comparisons, unless the
numbers in the list are all known.

By representing a list it is assumed that all members are known. If they
are not all known, it is not a list.

> Ask Cantor.

No, I ask you.

I do it just like Cantor does, by one single definition.

No. Give a single definition as I have presented above for the anti-diagonal.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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