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An uncountable countable set
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W. Mueckenheim
science forum Guru


Joined: 23 Apr 2005
Posts: 934

PostPosted: Fri Jul 21, 2006 11:42 am    Post subject: Re: An uncountable countable set Reply with quote

Dik T. Winter schrieb:


Quote:
What your
example shows is that an infinite sequence of transpositions can destroy
well-orderedness. What is the difference? The only conclusion is that
Cantor's statement is incorrect or that the reading of Cantor's statement
is incorrect. Take your pick.

Your hinting at any doubt about the clear meaning of Cantor's statement
is outrageous.

Quote:
I have found already a few instances where
Cantor's statements are in conflict with modern set theory.

There are many. See for instance for a very simple one: "On Cantor's
proof of continuity-preserving manifolds" on
http://www.fh-augsburg.de/~mueckenh/

Quote:
And I am not
surprised. He found the building blocks but was not entirely sure about
the way to go. His articles were research in progress, as it happens.

And now it happes that all his way turns out as a deadlock.
Quote:

And now for a debunking of a myth. Nowhere (at least I could not find
any place) has Cantor used the diagonal argument to show that the reals
are not countable. His proof about the reals shows that a complete,
densely ordered set is not countable. That one does not use the diagonal
argument at all.

So far we agreed recently.

Quote:
His diagonal proof shows that the set of infinite
sequences of two symbols is not countable, and as an extension, that the
powerset of a set has cardinality strictly larger than the cardinality
of the original set.

So all your arguments about 0.999... = 1.000... are *not* directed against
Cantor.

Wrong. My arguing is directed against the complete existence of the set
of all naturals, the set of all digits of a real number, the actual
infinity, the first transfinite number. That all is purest Cantor.

Regards, WM
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W. Mueckenheim
science forum Guru


Joined: 23 Apr 2005
Posts: 934

PostPosted: Fri Jul 21, 2006 11:51 am    Post subject: Re: An uncountable countable set Reply with quote

Dik T. Winter schrieb:

Quote:
The "first" successor. Tere is only one "first" successor!

Conveniently snipping your own text again? You wrote in
"news:1153052131.852951.273540@b28g2000cwb.googlegroups.com">:
Hence,
0.111... as the succesor of all naturals must consist of more 1's, than
any natural, if it is to be a number.

Now, if you had written "the first successor" I would not have made any
remarks.

You would have done better.

Quote:
It is not *the* successor.

That depends on the definition of "successor". 3 is the first successor
of 2 if you consider 4 as another successor. It is usual, however, to
define only the first successor at al. Only so we get the completely
defined succession as required by Cantor. He says "völlig bestimmte
Sukzession" and " zu jedem Elemente E' - mit Ausnahme des letzten, wenn
ein letztes vorhanden ist - ein ihm _nächst_ folgendes E'' vorhanden
ist."(italics by Cantor).Therefore omega ist "das Nächstfolgende",
*the* successor.
Quote:

Are you again arguing that the statement
For all p, there is an n such that An[p] = K[p]

That statement is false. (Without K including the index p = aleph_0, k
cannt be different from all natural numbers.)

What has your statement in parentheses to do with the statement I gave?
There is no argument at all about K being different from all natural
numbers. It is just a plain statement:
For all p, there is an n such that An[p] = K[p]
you claim that is false, and again without direct proof. Do you disagree
with the statement:
For all p, Ap[p] = K[p]?
I know that Ap[p] = 1 = K[p]. Now chose in the above n = p and you get
the new statement. Why would this second statement be false?

You said "for all p". If p is a natural number, then the segment K[p]
is the unary representation of a natural number and, hence, is covered
by a natural number. Remember:

For each natural (i.e. finite) number p we know: to index and to cover
are equivalent.
p covers K[p] <==> p indexes p.

Would K have only natural indexes p, then it would be covered by at
least one natural number.

If you deny then please give an example of a *finite* natural number
which indexes the n-th digit but does not cover the digits number 1 to
n.

Regards, WM
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W. Mueckenheim
science forum Guru


Joined: 23 Apr 2005
Posts: 934

PostPosted: Fri Jul 21, 2006 12:02 pm    Post subject: Re: An uncountable countable set Reply with quote

Dik T. Winter schrieb:

Quote:
You cannot exhaust the naturals. Therefore what you write as 0.111...
is not the same as the unary representation which also is denoted by
0.111... but means that N is exhausted and there is at least one 1
which cannot be indexed by a natural number.

Makes no sense to me. I see no exhaustion, see, I did use the symbol
"lim" there, and I had the hope that there would be a proper definition.
Apparently not.

No. You intermingle two different things, all n e N and infinity.
Again: For any *finite* natural n we have a logical equivalence, even
better an inclusion: indexing the n-th digit includes covering at least
all digits from 1 to n.
If each digit of 0.111... is indexed by one natural, then each segment
of 0.111... is covered by at least one natural too.
As long as only natural numbers are concerned there is no outcome.

Quote:

The reason is that

0.1
0.11
0.111
...
gives allegedly the same sum as

0.111... (as representation of aleph_0)
0.1
0.11
0.111
...

Yes, with a proper definition of "...".

But that is not acceptable if aleph_0 is a number.

Quote:
What was the reason again?

If aleph_0 is a number larger than any natural, then it cannot be the
*+ sum of a list containing only naturals excluding itself and
simultaneously the sum of a list including itself.

Quote:
You apparently have another definition in mind. Pray give *your*
definition.

There are two definitions conceivable. Both lead to a contradiction.
1) If "..." means "forall" n, e N then 0.111... is not different from
all n. (Because for naturals indexing includes covering.) But we know
that 0.111... is not ín the list of naturals. Contradiction.
2) If "..." means --> oo including aleph_0 as an index, then the digit
with this index is not defined.


Quote:
But
what you are now meaning is that
lim{n -> oo} SUM{i = 1 .. n} An = 0.111... .

I said above: It is different from n really reaching infinity. But for
1/9 or aleph_0 infinity is reached.

With a proper definition of the limit, the sum will *never* reach 1/9
(or aleph_0). But as you still refrain from giving definitions of the
notations you are using this discussion is less than fruitfull.

Bijection: Every position of a decimal number can be indexed by a

natural number. The set of naturals has cardinality aleph_0. The set of
positions has cardinality aleph_0.

Quote:
I ask questions. Again you refrain from giving answers. Are you unwilling
to give answers?

I ask questions too which you do not answer. The most important one:

How can a finite natural number index digit number n but not cover all
digits from 1 to n?

Quote:
0.1
0.11
0.111
...

As my interpretation of your "..." was apparently wrong above, pray, finally
supply a definition of that notation.

See above. 1) and 2).

Quote:
How many digits has 1/9 in decimal representation?

That is not a natural number.

No, but the set of digits has a cardinality, namely aleph_0.

Quote:
There is no such column. And 0.111... is not a natural.

That is a contradiction. And I told you why:

You said: all digits of 0.111... could be indexed by natural numbers,
but not all could be covered by natural numbers. This is wrong as one
can easily prove: Every natural which indexes a digit covers all digits
up to that one. If all digits of 0.111... can be indexed than all
digits up to every (and that is all digits) can be covered.

What is in *your*
Quote:
opinion the result of *+-ing all natural numbers togther?

As we have seen it is simply impossible to *+ sum all naturals. Every
result yields a contradiction. The set of all naturals does not exist.

Regards, WM
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