Re: An uncountable countable set

In message <vmhjr2-1B2D19.14091601072006@xxxxxxxxxxxxxxxxxxxxxx>, Virgil <vmhjr2@xxxxxxxxxxx> writes
In article <1151758813.109076.7170@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

*** T. Winter schrieb:


> I wrote: P(k) = lim{k -> oo} (k,k+1)(k,k+2)...(k,k+n) where I explicitly
> wrote that I used some intuitive form of limit here. This is the only
> way I can make sense of "infinitely many transpositions".

A transposition is an element of the set of transpositions, i.e. a set
of pairs of numbers. The set af rationals is also a set of pairs of
numbers. Do you raise any questions concerning this infinite set? Do
you need some limit in order to well-order the set of rationals?

If one stars with the normally ( and denslely) ordered rationals, no
sequence of transpositions will make it any less densely ordered.

If one stars with the rationals well ordered, no sequence of
transpositions will make it any less well ordered.

Thus you alleged conversion form one form to the other does not and
cannot occur.

I'm not sure about this. Suppose we use the following sequence of permutations on the naturals with the usual ordering.

(1 2), (1 3 2), (1 4 3 2), ...

After the nth permutation, the ordering becomes

n+1, n, n-1, ..., 2, 1, n+2, n+3, ...

Given any i, j with i < j, for any n>j, i comes after j in this ordering and so, using the obvious definition, i comes after j in the limit ordering. Thus the limit ordering is the reverse of the usual one. A sequence of permutations can destroy a well-ordering.

Each permutation can be broken down into a chain of transpositions so, with a little care, we get a sequence of transpositions which destroys the well-ordering. (We cannot replace each (1 n n-1 ... 2) by (1 2), (1 3), ..., (1 n) as the resulting sequence of orderings doesn't have a limit. But if we break it down as (n 1), (n 2), ..., (n n-1) then the limit exists.)

Looking now at the rationals with usual ordering. Apply the sequence of transpositions (0 2), (1 2), (0 3), (1 3), (0 4), (1, 4), ...
The limit ordering of Q has 0 > 1 > q for ever q /= 0,1 and so is not densely ordered.

Now suppose q_1, q_2, ... is an enumeration of Q and suppose q_1 < q_2 < ... < q_(n-1) in the usual sense. If q_(n-1) < q_n we do nothing, otherwise apply the sequence of transpositions

(q_i q_n), (q_(i+1) q_n), .. , (q_(n-1) q_n)

where i is the least integer such that q_i > q_n. The new ordering has q_1 < q_2 < .. < q_n.

If we do this for each n from 2 upwards, we get a sequence of transpositions which gives a sequence of well-orderings whose limit is the usual ordering.

The set of rationals is taken to be well-ordered. The following
transpositions operate on that set simultaneously (given are the
indices):
(1,2), (3,4), (5,6), ...
where Elements q 2n-1 and q 2n are interchanged, if they deviate from
order by size.
In the next step the pairs
(2,3), (4,5), (6,7), ...
are ordered by size, in the next step the pairs
(1,2), (3,4), (5,6), ...
are ordered by size, and so on. There are exactly as many steps as are
required to define the diagonal of a Cantor list. And there is the same
definition of "infinitely" as in Cantor's diagonal proof.

Then there must be a ->first<- transposition after which there is no
longer any first element in the current ordering of the rationals, if
the final result is to be the usual ordering of the rationals.

But that is clearly impossible.

I'm afraid this argument is on a par with "each set {1,2,..,n} has a maximum, therefore N has a maximum". I don't think WM's construction works, but even if it does, it proves nothing because sequences of transpositions needn't preserve well-ordering.

> So you define a series of mappings. But the series is infinite. How do
> you define such an infinite series?

Ask Cantor.

The difference being that Cantor's sequence of non-diagoal digits need
not be created seqeuentially, but can be done independently and
simultaneously, whereas yours cannot.

> Again you are starting with a conclusion about the finite and extending it
> to the infinite. You have to prove that you can do that.

Ask Cantor, how he can be sure that his diagonal is different from
every number of the list in case of infinitely many numbers.

One does not have to ask Cantor to explain what is transparently clear.
All Cantor needs do is specify a rule which can be applied
simultaneously to ever number in a given list to produce something that
cannot be a member of the list.

I am not an advocate of infinity.

You are not an advocate of sanity either.

Regards, WM

--
David Hartley
.

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