It is your lack of a proper proof that if each digit of a number can
be indexed that number can be covered. And such a proof does not
exist.

I do not see how I could avoid my conclusion. But if you are so sure
then give us at least one example how you completely index a number
without covering all the preceding numbers.

I have done so many times, and am doing it here again.

You gave an example how a number of the form 0,111...1 with n digits
indexes the n-th digit but does not cover all digits with m =<n ???

No, because that does not exist.

In article <1156689126.335154.133150@h48g2000cwc.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
Dik T. Winter schrieb:
The first part of his first proof shows that a complete ordered field
has cardinality larger than the natural numbers. In his proof he did
not rely an any properties of the reals other than that they form a
complete ordered field (he uses reals to exemplify).
...
Note what I wrote: "in his proof he did not rely on any properties of
the reals other than that they form a complete ordered field". What
other properties of the reals did he use?

He had no field and he used no filed. What is a field without the
axioms of the field? Nothing. Cantor disliked axioms if he did not hate
them. His only concern were numbers, numbers, numbers and their
*reality*. No fields and no axioms.

Perhaps. I thought we were discussing current set theory. And not what
in some time long ago was et theory. He may not have liked them, but in
current terminology "the first proof shows that a complete ordered field
has cardinality larger than the natural numbers".

He wrote that it was a simpler proof for his first theorem, i.e. the
theorem that the reals are uncountable.

And he did *not* write that.

Aus dem in =A7 2 Bewiesenen folgt n=E4mlich ohne weiteres, da=DF

Quoting again only in part. I will quote the translation I gave, with
annotation of the complete paragraph, not only the part you like to quote:
In the article, titled: "Über eine Eigenschaft des Ombegriffs aller
reellen algebraischen Zahlen (Journ. Math. Bd. 77, S. 258) [hier
S. 115], can for the first time be found a proof of the theorem
that there are infinite sets that are not in bijection with the set of
natural numbers 1, 2, 3, ..., v, ..., or, as I say in general, do not
have the same cardinality as the natural numbers 1, 2, 3, ..., v, ... .
So apparently he is thinking that his first article proves the theorem
that there are sets that are not in bijection with the natural numbers.
Which is true.

From what has been proven in section 2 follows that e.g. the
set of real numbers in an arbitrary interval can not be put in
a sequence w_1, w_2, w_3, ..., w_v, ... .
So he is now thinking that that proof was just an example for such a set.
Which it is. The "beispielsweise" is telling.
It is however possible to construct a much simpler proof for
that theorem that is independent from the observation of irrational
numbers.

And the "that theorem" (in German "jenem Satze") can, in my opinion only
refer to the theorem mentioned in the first sentence in this paragraph.

Dik T. Winter schrieb:


Let us remove omega from that set. What is the resulting cardinality?

Not an actually infinite one, because without omega there are only
finite natural numbers.

As there is a bijection between that set and the set including omega, their
cardinalities are the same.

No. Without omega there is no actual infinity.

Your example shows very clear that either both LUBs have to be included
- or none. In my opinion we choose none. But whatever you choose, you
cannot avoid to see the symmetry, can you?

I see the symmetry, I also choose none. And I think I have stated that
already.

Fine. That means there is no actual infinity.

But if there are infinitely many naturals then the infinite sum is
necessary.

Not at all. Why do we need it?

You should know that the n-th natural number is defined by the sum of
the number n-1 plus 1 (Peano). This can be retraced to the sum of n
1's. What is there to be defined? If there are infinitely many naturals
(and if infinity is a number),

then there is at least one infinite sum
of 1's (one from each natural).

OK, I kind of understand. I do not state that infinity is a number
aleph_0.
There exists a host of infinities. And aleph_0 is the infinity that gives
the equivalence class of sets that are in bijection with the set of
natural numbers; the canonical set with cardinality infinity.

If aleph_0 > n for n e N, and if there are even numbers x > aleph_0,

But indeed. If the ordinality of a set is infinite (w or larger), then
so is its cardinality (aleph-0 or larger). On the other hand, if the
cardinality of a set is infinite (aleph-0 or larger), and if it can be
well-ordered, so is its ordinality after well-ordering (w or larger).

As long as the set of natural numbers includes only finite numbers, its
cardinal number is also finite.

Regards, WM

You seem to misunderstand the tree. If the diagonal is in Cantor's
list, then 1/3 is in my tree. Can you give a reason why it should not
(other than that then set theory is inconsistent)?

Either you agree that all the edges of my tree are countable or you
agree that Cantor's diagonal is uncountable.

How then can you believe and assert that the number 0.111... which has
*only finitely indexable positions* could be completely indexed but not
covered.

Dik T. Winter schrieb:

In article <1156689126.335154.133150@h48g2000cwc.googlegroups.com
mueckenh@rz.fh-augsburg.de writes:
Dik T. Winter schrieb:
The first part of his first proof shows that a complete ordered
field
has cardinality larger than the natural numbers. In his proof he
did
not rely an any properties of the reals other than that they form
a
complete ordered field (he uses reals to exemplify).
...
Note what I wrote: "in his proof he did not rely on any properties of
the reals other than that they form a complete ordered field". What
other properties of the reals did he use?

He had no field and he used no filed. What is a field without the
axioms of the field? Nothing. Cantor disliked axioms if he did not hate
them. His only concern were numbers, numbers, numbers and their
*reality*. No fields and no axioms.

Perhaps. I thought we were discussing current set theory. And not what
in some time long ago was et theory. He may not have liked them, but in
current terminology "the first proof shows that a complete ordered field
has cardinality larger than the natural numbers".

In the terminology of the next century it may show even other things,
more general perhaps, with even more insight. I was discussing what
Cantor did. You were accusing me that I had not understood or misquoted
him.

Dik T. Winter schrieb:
....
So you can compute all solutions of a polynomial equation even of
higher than fourth degree in finite time? I doubt that.

I did not state that. I said that the numbers were computable, where
I use the mathematical sense of computable.

And that means you can compute a number even if you need infinite time.

Therefore in the mathematical sense each algebraic number is
computable.

(In reality it is not.)

But even in mathematical sense not
*all* algebraic numbers are computable.

Should you be able to quote a serious scientist who asserted that the
list of all algebraic numbers was computable? (You would prove this
scientist unserious.)

Dik T. Winter schrieb:
In article <virgil-B3927D.12230325082006@news.usenetmonster.com> Virgil <virgil@comcast.net> writes:
....
What I have been able to find was that with indexing of digit WM means
that every digit position can be indexed by a natural number. With
covering WM means that all positions to a certain one can be indexed
by a natural number. What WM is asserting is that when a number can
be totally indexed, it also can be totally covered. Quantifier dislexia
disguised.

Quantifier dyslexia is the assumption that there is a number that
counts all natural numbers, i.e. which is larger than all natural
numbers.

My arguing holds for the unary numbers of my list. If you think not,
then give an example of a finite number n which indexes the digit
position n but does not cover all positions m =< n.

Here is my example: The number 3 = 0.111 in my unary expression indexes
the digit position 3 of any unary number and it covers the digit
positions 1, 2, and 3 of any unary number.

The same can be proven for any finite number n.

In 0.111... there are allegedly only finite digit positions. Hence this
proof is valid for every position of this number. Every position can be
indexed, and every position can be covered. This is dictated by logic:

Position n can be indexed ==> every position m < n can be covered.
Not every position m can be covered ==> not every position n > m can be
indexed.

Dik T. Winter schrieb:

Let us remove omega from that set. What is the resulting cardinality?

Not an actually infinite one, because without omega there are only
finite natural numbers.

As there is a bijection between that set and the set including omega, their
cardinalities are the same.

No. Without omega there is no actual infinity.

Your example shows very clear that either both LUBs have to be included
- or none. In my opinion we choose none. But whatever you choose, you
cannot avoid to see the symmetry, can you?

I see the symmetry, I also choose none. And I think I have stated that
already.

Fine. That means there is no actual infinity.

Wrong. In mathematics the concept of "infinite sum" is not defined,

But if there are infinitely many naturals then the infinite sum is
necessary.

Not at all. Why do we need it?

You should know that the n-th natural number is defined by the sum of
the number n-1 plus 1 (Peano). This can be retraced to the sum of n
1's. What is there to be defined? If there are infinitely many naturals
(and if infinity is a number), then there is at least one infinite sum
of 1's (one from each natural).

OK, I kind of understand. I do not state that infinity is a number
aleph_0. There exists a host of infinities. And aleph_0 is the
infinity that gives the equivalence class of sets that are in
bijection with the set of natural numbers; the canonical set with
cardinality infinity.

If aleph_0 > n for n e N, and if there are even numbers x > aleph_0,
then aleph_0 is a number. Cantor stated that. If you do not do so, then
we are in agreement in that point.

Exactly. And in addition, as long as the numbers are finite, ordinal
and cardinal is the same size. That is just why the cardinal cannot be
infinite as long as the ordinal is finite (as expressed by "the
infinite number of finite numbers").

But indeed. If the ordinality of a set if infinite (w or larger), then
so is its cardinality (aleph-0 or larger). On the other hand, if the
cardinality of a set is infinite (aleph-0 or larger), and if it can be
well-ordered, so is its ordinality after well-ordering (w or larger).

As long as the set of natural numbers includes only finite numbers, its
cardinal number is also finite.

Dik T. Winter schrieb:
In article <1156585631.775313.153510@h48g2000cwc.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
....
I don't understand these questions. 1/3 and 1/5 are infinite paths
consisting of infinitely many edges.

You state that you can number the edges. Each edge terminates at a node,
and if 1/3 and 1/5 are in your tree, there must be edges that terminate at
those nodes.

What do you mean with "those nodes"? 1/3 and 1/5 are not nodes but they
are paths.

In that case you should be able to tell me what the numbers
of those edges are. Or do you now claim that there is *no* edge that
terminates at those nodes? If so, where do the nodes come from? Out of
thin air?

I did already tell you which edges belong to the path 1/3. The nodes
can be enumerated like the edges.

As the set of all edges is countable, you
should be able to state that. The problem with your counting method of
edges (count first level 1, next count level 2, etc.) you will already
have exhausted all natural numbers by counting all edges that are at the
end of finite paths.

There are no finite paths and there are no ends at all in my tree.

During your construction process all paths are finite,

There is no construction process. The tree exists like the real numbers
do exist.

and there is an
edge that terminates at a node terminating the path. I meant the paths
during the intermediate process of construction.

There is no process of construction. If you have difficulties to
comprehend that: it is the same as with Cantor's list. The tree is
defined once and for all. That's it.

you think that 0.1 would be represented by a finite path, what should
then be represented by the path 0.1000...?

The same.

By your arguing you would never be able to count the digits of 1/3
because "you will already have exhausted all natural numbers by counting
all edges that are at the end of finite" sequences.

Wrong. To count digits we need only one number for each level.

A set of countably many countable sets is countable.

Do you believe that the set of edges of 1/3 is uncountable?

That is not the set of all edges in the tree.

No. But would you say that the set of all edges of 1/3 and of 1/5 and
of 1/7 were uncountable?

No. Assuming of course that 1/3, 1/5 and 1/7 are in your tree.

All real numbers of the interval [0, 1] are there, some of them even
twice.

How many paths are required to obtain an
uncountable set of edges, according to your opinion?

Depends quite a bit on a host of questions. In your tree with terminated
edges, also the paths are terminated, so neither becomes uncountable.
But also 1/3 is not in that tree.

You seem to misunderstand the tree. If the diagonal is in Cantor's
list, then 1/3 is in my tree. Can you give a reason why it should not
(other than that then set theory is inconsistent)?

You recently mentioned an interesting aspect: The algebraic numbers,
i.e., the polynoms are countable, because they are finite. And,
therefore, you wanted only to count the finite segments of paths in my
tree. I oppose, unless you agree that the 1's in 0.111... also are
uncountable. You see, the problem is the same: We can count finite
segment but not infinity.

Either you agree that all the edges of my tree are countable or you
agree that Cantor's diagonal is uncountable. Or you state at least how
many infinite path in my tree you would consider to have completely
countable edges.

Any finite number of bit positions produces a finite set of strings.

Any countably infinite set of bit positions produces an uncountable set of
strings.

Which explains why the reals are uncountable (when represented as
infinite binary fractions in [0,1)).

Yes, ala Cantor's second proof of uncountability.

But it does not explain your claim that the naturals are uncountable
when represented as finite bitstrings. It's pretty straightfoward to
show that a countable number of bits produces a countable number
of bitstrings. There is no "in-between" cardinality.

You just contradicted yourself.

You agree that [me] "Any countably infinite set of bit positions
produces an uncountable set of strings", right? That's what [you]
"explains why the reals are uncountable (when represented as infinite
binary fractions in [0,1))"?But, [you] "a countable number of bits
produces a countable number of bitstrings"? Is a countably infinite
number of bits countable or not? If so, then does a countably infinite
number of bit positions produce a countable, or an uncountable, set of
strings?

It also flies in the face of your statements about infinite binary
trees. If each node is numbered with a natural (being the finite
bitstring of the left/right paths traversed from the root to the node),
then the nodes, and thus the naturals, are obviously countable.

On the other hand, you don't understand that the infinite paths in
the tree (the ones that don't have a terminating node) are uncountable
and correspond directly to your infinite bitstrings and to the real
binary fractions in [0,1).

I'll wait for your response on the above, since you're still not ready
for the answer you've already heard for this. Please don't snip it. It's
a good question, and a prime example of standard issues.

Dik T. Winter schrieb:
....
Cantor states (analogously) that *all stairs exist*, that the width of
all ot them is L, but that none of them has hight L. Width [0, 1] and
height [0, 1).

A quote please. But indeed, the width of all of them is L, but there is
none of them with width L. And the heighth of all of them is L, but there
is none of them with heighth L. So width [0,1] and heighth [0,1]. That
is, if you define both as the smallest box containing the completed stair.
If you define both as the largest numbers that can be obtained, you get
width [0,1) and heighth [0,1).

I agree with you. But Cantor said (Werke, p. 409): "So stellt uns
beispielsweise eine veränderliche Größe x, die nacheinander die
verschiedenen endlichen ganzen Zahlwerte 1, 2, 3, ..., v, ...
anzunehmen hat, ein potentiales Unendliches vor, wogegen die durch ein
Gesetz begrifflich durchaus bestimmte Menge (=EF=81=AE) aller ganzen
If "=EF=81=AE" is UTF-8, the Unicode character is U+F06E, which is in the

endlichen Zahlen N das einfachste Beispiel eines aktual-unendlichen
Quantums darbietet.

And I am arguing against this actual understanding of infinity.

Dik T. Winter schrieb:
....
You gave an example how a number of the form 0,111...1 with n digits
indexes the n-th digit but does not cover all digits with m =<n ???

No, because that does not exist.

How then can you believe and assert that the number 0.111... which has
*only finitely indexable positions* could be completely indexed but not
covered.

Dik T. Winter schrieb:
In article <1156689126.335154.133150@h48g2000cwc.googlegroups.com> muecke=
nh@rz.fh-augsburg.de writes:
Dik T. Winter schrieb:
The first part of his first proof shows that a complete ordered=
field
has cardinality larger than the natural numbers. In his proof =
he did
not rely an any properties of the reals other than that they fo=
rm a
complete ordered field (he uses reals to exemplify).
...
Note what I wrote: "in his proof he did not rely on any properties of
the reals other than that they form a complete ordered field". What
other properties of the reals did he use?

He had no field and he used no filed. What is a field without the
axioms of the field? Nothing. Cantor disliked axioms if he did not hate
them. His only concern were numbers, numbers, numbers and their
*reality*. No fields and no axioms.

Perhaps. I thought we were discussing current set theory. And not what
in some time long ago was et theory. He may not have liked them, but in
current terminology "the first proof shows that a complete ordered field
has cardinality larger than the natural numbers".

In the terminology of the next century it may show even other things,
more general perhaps, with even more insight. I was discussing what
Cantor did. You were accusing me that I had not understood or misquoted
him.

Aus dem in =3DA7 2 Bewiesenen folgt n=3DE4mlich ohne weiteres, da=3DDF

Quoting again only in part. I will quote the translation I gave, with
annotation of the complete paragraph, not only the part you like to quote:
In the article, titled: "Über eine Eigenschaft des Ombegriffs aller
reellen algebraischen Zahlen (Journ. Math. Bd. 77, S. 258) [hier
S. 115], can for the first time be found a proof of the theorem
that there are infinite sets that are not in bijection with the set of
natural numbers 1, 2, 3, ..., v, ..., or, as I say in general, do not
have the same cardinality as the natural numbers 1, 2, 3, ..., v, ... .
So apparently he is thinking that his first article proves the theorem
that there are sets that are not in bijection with the natural numbers.
Which is true.

Of course: The set of the reals and the set of the transcendental
numbers are such sets. So he has the right to speak generally about
"sets". Which mathematician would not like to generalize his theorems?

From what has been proven in section 2 follows that e.g. the
set of real numbers in an arbitrary interval can not be put in
a sequence w_1, w_2, w_3, ..., w_v, ... .
So he is now thinking that that proof was just an example for such a set.
Which it is. The "beispielsweise" is telling.
It is however possible to construct a much simpler proof for
that theorem that is independent from the observation of irrational
numbers.

Independent of the *necessary use* of irrational numbers.

And the "that theorem" (in German "jenem Satze") can, in my opinion only
refer to the theorem mentioned in the first sentence in this paragraph.

That theorem the proof of which depends on irrational numbers.

But since no natural requires an infinite bit string, that is irrelevant.

If no natural requires an infinite bit string, even the very largest,
and all bit positions are included in it, then no infinite set of bit
positions is required.

Tony, please drop the "very largest". The rest is ok. If no natural
requires an infinite set then all naturals together do not require an
infinite set. If the contrary is asserted (and if infinity is a number
larger than any finite number), then we may ask which natural is the
first such that all of its predecessors require an infinite number of
bit positions.