Re: An uncountable countable set

In article <1155664930.866986.157410@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
*** 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.

Nope. You claim the last, but it is false. Each finite segment can be
covered, and each finite index can be covered. But the total is not
finite, but contains only finite digit positions.

> > 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.

Again, you keep asserting such things without *ever* providing a proof.

> > 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.

My assertion is that K can be completely indexed but not completely
covered. Again you misquote my assertion.

> > 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.

But it is not a natural number.

That is what set theory and the definition of natural number exclude.

So? It is not a natural number. Those two do indeed tell that it is
not a natural number. So what?

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.

Why not? Each and every digit position is finite, so can be indexed,
and, by *your* definition the number can be indexed. But there are
infinitely many digit positions, as there are infinitely many
natural numbers.

> 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.

Yes. When you can add all those numbers. But the result is not a natural
number.

> 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)?

A set that is not finite is infinite, that is the definition of the word
infinite. So yuor question is actually, why are there cardinal number?
Well, they come in handy when comparing infinite sets. And (by definition)
a cardinal number is the equivalence class of sets that can be put in
bijection with each other. A set being in bijection with another set is
easily shown to be an equivalence relation, and so we can split the sets
in equivalence classes. That is all elementary mathematics. And those
equivalence classes are called cardinal numbers. Just as the equivalence
classes of Cauchy sequences are called real numbers and the equivalence
classes of ordered pairs of integers are called rational numbers (with
some particular equivalence relation in mind). All extremely elementary.

> > 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.

You keep asserting that.

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.

Just because there is no last line. If there were a last line, it would
have width aleph-0, but there is no last line.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.