Re: An uncountable countable set
- From: mueckenh@xxxxxxxxxxxxxxxxx
- Date: 15 Aug 2006 11:02:10 -0700
*** T. Winter schrieb:
In article <1155485742.529974.141050@xxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> *** T. Winter schrieb:
> > > Further: Indexing the digit number n is equivalent to covering the
> > > string up to digit number n. A finite string can never cover an
> > > infinite string.
> >
> > But *that* is irrelevant.
>
> Wrong.
Why is it wrong
Because it is very relevant. You state that an infinite string could be
indexed by finite strings. Complete indexing includes covering.
> > Consider K = 0.111... . For each n we can
> > index digit number n of K and so cover K up to digit number n.
>
> Of course you can index each n. But your "each n" stems from the true
> list. And we know that 0.111... is not in the true list, because it is
> distinguished from any element of the true list.
Pray read what I state. Again, what you state is irrelevant.
> Therefore your assertion is correct but void of power to prove that any
> digit of 0.111... can be indexed.
Eh? This tells me exactly nothing. My assertion "each digit of K can be
indexed, but K can not be covered" is void of power to prove that any
(why any here?) digit of 0.111... can be indexed? If each digit of K
can be indexed, any digit of K can be indexed.
Then any digit of K can be covered by a member of the true list. K is
nothing else but all its digits. Then K is in the true list.
> > > Don't forget: The infinite set of natural numbers contans only natural
> > > numbers, i.e., finite strings of 1's in unary representation.
> >
> > Yes, but there is no bound on the index positions, and hence the number
> > of index positions is infinite.
>
> The number of digit positions is irrelevant. The digit positions are
> all finite - in the true list. All numbers with merely finite digit
> positions are in the true list. 0.111... is not there.
Yes, I never said otherwise. Why do you always come back with this
statement that I do not contradict?
Because your assertion that K could be completely indexed and covered
is equivalent to K being in the true list.
> > > Why can't you learn that the
> > > asserted infinity of the number of numbers is completely irrelevant.
> >
> > That is just opinion.
>
> That is not opinon. In oder to index or to cover, only the digit
> postions of the numbers are relevant. It is completely irrelevant how
> many othe numbers are there. Because we always consider only one
> special number.
Again, makes no sense to me. *If* there are infinitely many natural
numbers, there are also infinitely many digit positions.
An infinite many of digit positions would result in an infinite number.
That is what set theory and the definition of natural number exclude.
To cover a
number by a natural it is required that the number to be covered has
only finitely many digit positions. So a number with infinitely many
digit positions can not be covered by a natural number.
Likewise a number with infinitely many digit positions can not be
indexed by a natural number.
> My basic assumption is the existence of this axiom. I *derive* a
> contradiction.
You do not. In all your derivations somewhere you assume that that axiom
is false.
The axiom of infinity states that there are infinitely many numbers,
hence infinitely many differences of 1. It is obvious that an infinite
number of differences of 1 leads to an infinite number. I need not do
more than show this implication.
> > > > But if you claim that the set if finite numbers is finite, there is
> > > > a largest number.
> > >
> > > There is no largest number. The set of of natural numbers is infinite,
> >
> > Eh? Above you wrote that there is *not* an infinite set. Contradicting
> > yourself?
>
> Important: There is no largest number. The axiom says only that the set
> is infinite.
I said, there is no largest number. The axiom says the set does exist. In
and of itself the axiom does *not* state that the set is infinite. It is
a theorem that the set is not finite, and so must be infinite.
Why should a set which is not finite be actually infinite (i.e. have a
cardinal number)?
> > > There is no infinite natural number. Because there is no infinite
> > > finite number - and every natural number is finite. Even the axiom of
> > > infinity does not state the contrary. (But if the axiom of infinity had
> > > to be satisfied, then a finite infinite number would be required.
> > > Compare the staircase.)
> >
> > Again an assertion, without proof.
>
> The staircase is a proof.
No.
You may deny this, nevertheless it is true. If there is an actual
number of steps, then the width is equal to the lenghts is aleph_0.
What I do not understand is how you can believe that someone (except,
perhaps, Virgil) could share your opinon that only in one dimension the
infinity was actually realized, but not in the other.
Regards, WM
.
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