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An uncountable countable set
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W. Mueckenheim
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Posts: 934

PostPosted: Tue Jul 18, 2006 10:54 am    Post subject: Re: An uncountable countable set Reply with quote

Franziska Neugebauer schrieb:

Quote:
mueckenh@rz.fh-augsburg.de wrote:

Dik T. Winter schrieb:
[...]
Yes, it is larger than all naturals, but I would not call it *the*
successor, but *a* successor, or, if you wish, *the smallest*
successor.

However, not all of its 1's in unary representation can be indexed by
natural numbers because they are smaller.

My understanding of unary represenations is:

strange!
Quote:

n e omega: unary(n) def= (a_i)
a_i = 1 if 0 <= i < n

a_i =/= 1 if n = 0. Hence (what means the same as therefore etc.) you
should write "a_i = 1 if 0 < i <= n".

Quote:
a_i = 0 if n <= i < omega

a_i = 1 if i = n. Example: 0.11 is the unary represenation of 2, not of
3.
Quote:

omega: unary(omgea) def= (a_i) a_i = 1 A i e omega

This is the deep dilemma of set theory: There is no actually infinite
set of finite numbers.

Non sequitur. Ever considered _your_ representation theory broken?

If you insist that 0.111 represents 4, then something with your
representation theory must be broken. How many letters x do you see
here: x. Is zero the correct answer? But even this approach of yours
does not prevent the dilemma, because even there the *+ sum of all
naturals is omega, the *+ sum of all naturals and omega is omega too.
This is a contradiction, and it is solved only by the observation that
the *+ sum of all naturals is not omega, because a *+ sum of finite
numbers alone cannot result in an infinite number.

Regards, WM
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 11:01 am    Post subject: Re: An uncountable countable set Reply with quote

Virgil schrieb:


Quote:
The successor of all naturals is not a natural and, therefore, must be
larger (because it is not less).

There is no such thing as the successor of all naturals any more that
there is a single successor common to both 3 a and 6.

This is an expression coined by Cantor: "a number following after all
natural numbers".

But he does not call it a 'natural' number any more than he calls it an
even number or a prime number.

I said above: The successor of all naturals is not a natural.

Quote:
There is no number following all naturals and, hence, there is no
number omega at all.

There is no such 'natural' but there is such a cardinal or ordinal. Both
are generically 'numbers' not all cardinals or ordinals are naturals.

Why do you emphasize this? Of course it cannot be a natural if it
follows after all naturals.
Quote:

One has the set of all naturals, and that set can have a successor under
the definition that the successor of any set, x, is (x union {x}).

That is nonsense too. One cannot have a set of all naturals, because
the step x U {x} defined by the axiom does never result in this set.

No one said that the set of naturals was a successor, stupid, we only say
that it has one.

Some say that omega is the set of naturals. And Cantor said that omega
was the smallest successor of all naturals.
Quote:

Misrepresentations like that are the tools of trolls.

Those trolls which troll set theory.
Quote:

You recently asserted that such stepwise processes can never exhaust a
set. Now you changed your mind? Just in time?

One can have a set of all of an inexhaustable supply of objects even
though one cannot exhaust that supply by serial operations.

such as a well-ordering undoubtedly is, because one number follows the
other in a sequence or a chain or in steps.

Regards, WM
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 11:40 am    Post subject: Re: An uncountable countable set Reply with quote

Virgil schrieb:

Quote:
WM's version of logic is quite different from the sort used in ZF of NBG
or any other part of mathematics.

No the logic used in mathematics is the same as mine. It is only in
modern set theory that such silly things as the rejection of the binary
tree occur.
Quote:


Consider the columns spanned by the digit positions of 0.111... Either
there is a column with only zeros, or there is at least one 1 in each
column, or ?

If you mean to list 0.0, 0.1, 0.11,0,111,... so that the 0's line up
vertically, then every "column" has a least one 1. And each column to
the right of the '.' column has the same "number" of 1's as all the
others.

Here the existing number 0.111... with its aleph_0 digits can be
realized by its not existing natural predecessors in the list?

Regards, WM
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 11:44 am    Post subject: Re: An uncountable countable set Reply with quote

Virgil schrieb:


Quote:
The difference is that our "traditions and folklore", which we chose to
call axioms and definitions, are logically consistent,

Only if you decide *arbitrarily* which infinite set can be exhausted
and which cannot.
Regards, WM
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 11:51 am    Post subject: Re: An uncountable countable set Reply with quote

Virgil schrieb:

Quote:
In article <1153148551.942037.97110@s13g2000cwa.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

This is the deep dilemma of
set theory: There is no actually infinite set of finite numbers.

But the existence of this "dilemma" can only be established by assuming
it.

So for those who do not chose to assume it, it does not exist.

Those who choose to close their eyes enjoy always the mercy of not
being forced to see the sheer misery of mathematics.

Regards, WM
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 12:04 pm    Post subject: Re: An uncountable countable set Reply with quote

Franziska Neugebauer schrieb:


Quote:
aleph_0 def= | omega |
0.111... =def (a_i) having a_i = 1 A i e omega
a_ij means the well known matrix of figures

IF aleph_0 does exist, THEN 0.111... covers aleph_0 columns.

This is as meaningful as
If i exists then sqrt(-1) is i.

IF 0.111... covers aleph_0 columns, THEN aleph_0 columns do exist.

This is as meaningful as
If sqrt(-1) is i then i exists.

IF aleph_0 columns do exist THEN we can consider their contents.

This is as meaningful as
If i exists then we can consider its value.

IF we can consider the contents of each column, THEN we can ask how
many 1's are therein.

Lotta questions.

Unknown word.
Quote:

IF we can ask how many 1's are in each one, THEN the answer can be
"zero 1's" or "not zero 1's".

We can.

IF the answer is in each case is "not zero 1's", THEN in each column
at least one 1 must be present.

This is the case, since every a_jj = 1 j e N by definition.

However, there is no natural numbers with this property,

Could you precicely _define_ which /property/ you are talking about?

To contribute a 1 to each column.
Quote:

For every column j e N a_mj has the 1 in position m(j) = j, since a_jj =
1 A j e N. Where exactly lies your problem?

I told you recently: In a list like mine a *+ sum is defined. This *
sum is equal to the largest number of the list. And if a larger number
is included, then the sum grows. You pretend that this is not valid for
0.111... because without 0.111... the list has the *+ sum 0.111... and
when you include it in the list (as diagonal number or as the first
line or anywhere else) the *+ sum does not grow.

Regards, WM
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 12:10 pm    Post subject: Re: An uncountable countable set Reply with quote

Virgil schrieb:

Quote:
In article <1153168957.805313.57460@p79g2000cwp.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Dik T. Winter schrieb:


Now you use an entirely new term: "can be exhausted". I think you mean
that you can take out elements one by one and doing this at some stage
the infinite set becomes empty. Howver, I think, that if that can be
done, that there is a last element you can take out. And, according
to the axiom of infinity, that is not possible, so infinite sets can
not be exhausted in this sense.

But in another sense?

In the sense of having a set of all of them, as per the axiom of
infinity, an axiom does it.

The axiom is not responsible for the well-order of *all* elements of
the set, because well-order is a stepwise process, you see? You take
one element and then define its successor and then you define the
successor of this one and so on. Of course you can give a compact
expression (as I did with my transpositions, for instance) but that
doesn't change the principle character of the process.

Regards, WM
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 12:18 pm    Post subject: Re: An uncountable countable set Reply with quote

Dik T. Winter schrieb:

Quote:
In article <1153145345.281803.129520@35g2000cwc.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
Franziska Neugebauer schrieb:
mueckenh@rz.fh-augsburg.de wrote:
...
Indeed. I recall. Therefore the linearity of the list numbers enforces
a column with zeros.

Nice try, but non sequitur. There is no such column. Otherwise show one
or prove its existence.

The proof requires logic. Therefore I am afraid you will not accept it.
It reads: Either there is a column with only zeros, or there is at
least one 1 in each column spanned by the digit positions of 0.111... ,
isn't it?

Yes, right. And now the remainder of the proof, please?

If the *+ sum of the list is a unary representation of some 1's, then,
by the construction of the unary numbers of the list, this sum must be
a unary numer of the list. This is so whether you pretend it would not
be so or not.

A mathematic which does not reflect this fact does nt deserve its name.

Regards, WM
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Franziska Neugebauer
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PostPosted: Tue Jul 18, 2006 12:26 pm    Post subject: Re: An uncountable countable set Reply with quote

mueckenh@rz.fh-augsburg.de wrote:

Quote:
Franziska Neugebauer schrieb:
mueckenh@rz.fh-augsburg.de wrote:

Dik T. Winter schrieb:
[...]
Yes, it is larger than all naturals, but I would not call it *the*
successor, but *a* successor, or, if you wish, *the smallest*
successor.

However, not all of its 1's in unary representation can be indexed
by natural numbers because they are smaller.

My understanding of unary represenations is:

strange!

n e omega: unary(n) def= (a_i)
a_i = 1 if 0 <= i < n

a_i =/= 1 if n = 0.

Of course. unary(0) = 000...
The unary representation of 0 (the empty set) is the empty string
(stiputalting only 1's are written).

Quote:
Hence (what means the same as therefore etc.) you should write "a_i =
1 if 0 < i <= n".

If you are confused by the range you may drop "0 <=" entirely, since
A i (i e omega -> 0 <= i).

+--------------------------------------------+
| a_i = 1 if i < n. (sic!) |
+--------------------------------------------+

Quote:
a_i = 0 if n <= i < omega
a_i = 1 if i = n. Example: 0.11 is the unary represenation of 2, not
of 3.

I presume omega = { 0, 1, 2, ... } and I presume i e omega.
Therefor unary(2) = 11. The positions 0 and 1 are 1.

If you are confused by the range you may drop "i < omega" entirely,
since A i (i e omega -> i < omega).

+--------------------------------------------+
| a_i = 0 if n <= i (sic!) |
+--------------------------------------------+

summing up:

a_i = 1 if i < n (1a)
a_i = 0 if n <= i (1b)

A tabulation may help you to understand what I mean:

unary(n) (only 1's
n 1-allocated i's 0-allocated i's written)
--------------------------------------------------------------
0 (none) 0 1 2 ... "" (empty string)
1 0 1 2 3 ... 1
2 0 1 2 3 4 ... 11
3 0 1 2 3 4 5 ... 111
... .... ... ....

If you want to drop index and number 0, you can simply use
omega \ { 0 } instead of omega.

Quote:
omega: unary(omgea) def= (a_i) a_i = 1 A i e omega

With (1a) and (1b) it is even not necessary to define unary(omega)
specially.

Quote:
This is the deep dilemma of set theory: There is no actually
infinite set of finite numbers.

Non sequitur. Ever considered _your_ representation theory broken?

If you insist that 0.111 represents 4,

I never did and will not.

Quote:
then something with your representation theory must be broken.

Dead horse.

Quote:
How many letters x do you see here: x. Is zero the correct answer?

saucy.

Quote:
But even this approach of yours does not prevent the dilemma, because
even there the *+ sum of all naturals is omega, the *+ sum of all
naturals and omega is omega too.

,----[ http://en.wikipedia.org/wiki/Dilemma ]
| A dilemma is a *problem* offering two solutions, neither of which is
| acceptable
`----

Please explain in detail what the *problem* is. I can't see any problem.

Quote:
This is a contradiction,

Once again: I can't see a "problem" or a "contradiction". Would you like
to explain, what exactly _is_ the problem?

Quote:
and it is solved only by the observation that the *+ sum of all
naturals is not omega,

You are trying to "solve" a yet non-existing "problem". BTW: This can
not be "ovserved" since every "observation" of the present *-sum of all
unaries of naturals _is_ obviously representing omega. You confuse
observation by hallucination.

Quote:
because a *+ sum of finite numbers alone cannot result in an infinite
number.

A *-sum of _infinitely_ _many_ representation of finite numbers does in
the present case result in a representation of an infinite number. This
is a fact.

F. N.
--
xyz
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Franziska Neugebauer
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Posts: 52

PostPosted: Tue Jul 18, 2006 12:49 pm    Post subject: Re: An uncountable countable set Reply with quote

mueckenh@rz.fh-augsburg.de wrote:

Quote:
Franziska Neugebauer schrieb:

[...]

Quote:
IF we can ask how many 1's are in each one, THEN the answer can be
"zero 1's" or "not zero 1's".

We can.

IF the answer is in each case is "not zero 1's", THEN in each
column at least one 1 must be present.

This is the case, since every a_jj = 1 j e N by definition.

However, there is no natural numbers with this property,

Could you precicely _define_ which /property/ you are talking about?

To contribute a 1 to each column.

Should be a property of which object?

Quote:
For every column j e N a_mj has the 1 in position m(j) = j, since
a_jj = 1 A j e N. Where exactly lies your problem?

I told you recently: In a list like mine a *+ sum is defined. This *
sum is equal to the largest number of the list.

The notion of "largest number" is misleading. There is no largest
number. Neither in the list nor in omega. If it makes sense to talk
about "a largest number", then please

1. name it (show us its name or position _in_ the list or _in_ omega),
and/or
2. present a _proof_ that it exists.

Until then we cannot (meaningfully) talk about "a largest number in the
list" or "a largest number in omega". Wittgenstein applies here.

Quote:
And if a larger number is included, then the sum grows. You pretend
that this is not valid for 0.111... because without 0.111... the list
has the *+ sum 0.111... and when you include it in the list (as
diagonal number or as the first line or anywhere else) the *+ sum does
not grow.

Postponed until existential status of "largest number" is clarified by
WM.

F. N.
--
xyz
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Dik T. Winter
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PostPosted: Tue Jul 18, 2006 2:15 pm    Post subject: Re: An uncountable countable set Reply with quote

In article <1153225087.356970.296750@h48g2000cwc.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
Quote:
Dik T. Winter schrieb:
In article <1153145345.281803.129520@35g2000cwc.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
Franziska Neugebauer schrieb:
mueckenh@rz.fh-augsburg.de wrote:
...
Indeed. I recall. Therefore the linearity of the list numbers enforces
a column with zeros.

Nice try, but non sequitur. There is no such column. Otherwise show one
or prove its existence.

The proof requires logic. Therefore I am afraid you will not accept it.
It reads: Either there is a column with only zeros, or there is at
least one 1 in each column spanned by the digit positions of 0.111... ,
isn't it?

Yes, right. And now the remainder of the proof, please?

If the *+ sum of the list is a unary representation of some 1's, then,
by the construction of the unary numbers of the list, this sum must be
a unary numer of the list.

As long as your list is finite. Suppose the list is infinite, and the
*+ sum of all members of the list is a member of the list. That would
mean that the *+ sum of the list is the last element of the list,
contradicting that the list is finite. Hence the assumption is false.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 5:01 pm    Post subject: Re: An uncountable countable set Reply with quote

Dik T. Winter schrieb:

Quote:
Sorry, we are now talking about 0.999..., please remain with the argument.
You said that the definition is silly. Why is the definition above silly?
And that there are undefined digits in irrational numbers does not bother
me in the least. I have no idea why you have a problem with that.

In this case not the digits are undefined but the indices are
undefined. See the discussion of the *+ sum of my list of unary
numbers.
Quote:

But to go on:

And therefore we have *always* undefined digits in any
irrational number and in the diagonal of Cantor's list. There are
*always* most of its digits unknown, i.e., always there are more digits
unknown than are known. No matter how small an epsilon you select.

For Cantor's diagonal no epsilon is needed, neither is their for irrational
numbers. The only thing we need to know is that the digits are all in the
range [0, 9],

The only thing we need to know for the binary tree is that no path can
split without two additional edges. Why don't you use this kind of
abstract mathematics in that case too?

Quote:

Hence you cannot prove that the digits of the diagonal are all
different from those of the line numbers.

Why not? For the proof no exhaustion is needed. Just like (in another
thread) the proposition

For the binary tree no exhaustion is requird either. It suffices to see
that

|
o
/\

is the ever repeating element which cannot be outwitted.

Quote:
For all p there is an n such that An[p] = K[p]
does not need exhaustion. In the case of the list and the diagonal the
similar statement is
For all p there is an n such that An[p] != D[p]
proving that D is different from all An. BTW, in the first case also the
proposition
For all p there is an n such that An[p] != K[p]
proving that K is also different from all An.
In the first and second case take n = p, in the third case take n = p + 1.
This is *not* a proof by exhaustion. It is simply stating a fact, with
an easy proof.

You can prove it for each one
but you cannot prove it for all.

If something is true for each one, it is true for all.

It is true for each natural, that it cannot index all 1's of 0.111... .

Quote:
Let's reason
with the excluded middle (because that would complicate matters). Assume
it is not true for all, then there must be some for which it is not
true (say z). On the other hand, it is true for each one, so also
true for z. A contradiction, hence the assumption is false. And in
the cases above you can proof it for each one in one sweep, because
you prove it for an arbitrary element.

It is true for each natural, that it cannot index all 1's of 0.111... .
Quote:

That is the same problem as with the
*+ sum of my list. You can prove the sum is 1 for each column but you
cannot prove it for all columns.

But you can.

Because stepwise exhaustion of
infinite sets is impossible. Otherwise my (and Cantor's) reordering
could be completed.

If stepwise exhaustion is impossible, your reordering could be completed.
But you can with your *+ operation give the proof in one sweep, but you
failed to answer to my question: "what happens at infinity?".

1) There is o infinity. Ever Transposition has a natural number.
2) The same happens as with Cantor's diagonal. There is absolutely no
difference.
Quote:

But, that was precisely what I intended when I asked you what you meant
with "*+ ..." in 0.1 *+ 0.11 *+ 0.111, because you gave no definition.
As you gave no meaning to that infinite "sum", only to finitely many
of them (and in that case going to infinity is impossible), there was
no way with your definitions.

But the assertion
that the digits of these limits could be used to construct a diagonal
number is simply nonsense.

Only assertion and meaning. No content.

Proven by epsilon. Set theory lives by contradiction. Some exhaustions
of nfinite sets are accepted other are not.

You can exhaust an infinite set if you take out all elements at once.

You may think that would be possible, but it is not. You always need
the belief expressen by the three "..."

Quote:
That is why the function f(n) = n + 1 is a bijective mapping from
{1, 2, ...} to {2, 3, ...},

sic!


Quote:
every definition works at once, you need
not to count to 100 to know what f(100) is. And the same is the case
with Cantor's diagonal number. You need not look at the first 100
elements to find the 101-st element to know that the 101-st digit
of the diagonal is. That is what the definition of a list is. A
mapping from N to the members of the list.

You need not look at the transposition number 100 to see what it does.
Quote:

Your set of transpositions are different. They do not work at once,

They do work at once. Once they are defined the result cannot be
changed.

Quote:
you
need sequencing (transpositions are in general non-commutative). In the
case of a sequence you need limiting procedures to show what would be
the "final" result (like lim{n -> oo} 1/n = 0). But the problem with
limiting procedures is that the "final" result has not necessarily the
same properties as each and every finite result.

Airy-fairy. My transpositions converge, i.e., a partial order by size
can never become worse but can always only become better. I think there
is no proof equired to see that.
Quote:

Let me give an example with transpositions. Let's say that (a, b)
means that we interchange the a-th and b-th element in an ordering.
Let us start with the ordered set of natural numbers {1, 2, 3, ...}
(yes, not Bourbaki this time). Let's define a sequence of transpositions:
(1, 2)(2, 3)(3, 4)(4, 5)...
with a proper measure and a proper definition of limit, I think that we
can show that that leads to the ordered set {2, 3, ..., 1}. Now interchange
the first two elements of the transpositions, in this case we get
{1, 3, ..., 2}. So sequencing plays a crucial role.

But that is completely uninteresting, because my transpositions do
never lead off an ordering by size. What is achieved will and can never
again be destroyed. Your example fails.

Regards, WM
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W. Mueckenheim
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PostPosted: Tue Jul 18, 2006 5:07 pm    Post subject: Re: An uncountable countable set Reply with quote

Dik T. Winter schrieb:

Quote:
In article <1153148551.942037.97110@s13g2000cwa.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:

Dik T. Winter schrieb:

In article <1153052131.852951.273540@b28g2000cwb.googlegroups.com> muecke> > nh@rz.fh-augsburg.de writes:
Dik T. Winter schrieb:
...
(1): 0.111... is not *the* successor of anything, we may sloppily say that
it is *a* successor of all naturals, just like 10 is *a* successor of
2

It was Cantor who coined this expression saying " daß omega die erste
ganze Zahl sein soll, welche auf alle Zahlen nu folgt." (Works, p.
195) So it must be larger anyhow.

Yes. Not *the* successor, but *a* successor.

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

The successor of all naturals is not a natural and, therefore, must be
larger (because it is not less).

Yes, it is larger than all naturals, but I would not call it *the*
successor, but *a* successor, or, if you wish, *the smallest* successor.

However, not all of its 1's in unary representation can be indexed by
natural numbers because they are smaller.

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.)

Quote:
is false, or are you arguing that the statement
For all p, there is an n such that An[p] != K[p]
is false?

That statement in the form An != K is correct. The seqence An[p] does
not exist for all p which are in K.
Quote:

There is no actually infinite set of finite numbers.

Axiom of infinity.

In contradiction with mathematics, obviously.

Regards, WM
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PostPosted: Tue Jul 18, 2006 5:20 pm    Post subject: Re: An uncountable countable set Reply with quote

Dik T. Winter schrieb:

Quote:
In article <1153169497.380974.32190@i42g2000cwa.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
Dik T. Winter schrieb:


I don't like definitions which define nonsense like the corner of a
circle.

Check the manhattan-measure and you will find square circles.

Then I'll better do without.

Quote:
But whatever,
if you do not like definitions that define nonsense in your opinion, you
should have a hard time with mathematics.

Yes, it was sloppy terminology. What happens when n grows without bound?

Nothing happens with the *+ sum.

What is lim{n -> oo} SUM{i = 1 .. n} A_i ? How do you define that?
Without such a definition I have no idea what the result is when I *+
all An.

If there is at least one 1 in a column then the *+sum is 1. You need
not investigate how many 1's are there to follow if you want to
calculate the *+ sum.

That is not an answer to my question. But I will take it in good faith,
so
lim{n -> oo} SUM{i = 1 .. n} A_i = 0.111...
tell me where I have gone wrong.

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.

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
....

Quote:

Define: If any case includes at least one 1 the the *+ sum is 1.

Again, in the finite case. You have not defined what you mean with the
infinite sum.

Hell and devil! Can't you read? The definition for *+ sum = 1 is: at
least one 1 must be encountered. That is enough in any case, finite or
not.

Damnation leaving aside, you defined what SUM{i = 1 .. n} An is. You did
*not* define what happens when n grows without bound. Now you state that
in each column, whenever there is a 1, the final result should be 1.


Here is another example for the *+ sum:

0.1
0.11
0.111

The sum is 0.321, the *+ sum is 0.111.

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.

Quote:

So the statement
For all n, An[n] = K[n]
is true? As is the statement
For all p there is an n such that An[p] = K[p]
also true?

No. Then 0.111... would not differ from every n.

No to which question? Is the first statement false? And if so, why?
Is the second question false, and if so why? And how do you come at
your conclusion?

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
....

This could not happen if 0.111... was a number with more 1's than any
natural number (in unary representation).
Quote:

Why than do you write that it is false?

Because it is not correct. According to the axiom of infinity [...]
infinite sets can not be exhausted in this sense. (Dik T. Winter)
Therefore, a unary representation of a natural number can never reach
the line next to the unary representation of aleph_0.

You are arguing on two different lines at the same time, confusing one
with the other. When we consider the An as unary representations of
natural numbers the result is aleph_0, not a natural number. When we
consider them as being decimal numbers, the result is 1/9.

How many digits has 1/9 in decimal representation?

Quote:
In both
cases the resulting number is not in the list. What is the problem?

Therefore, a unary representation of a natural number can never reach
the line next to the unary representation of aleph_0. The latter does
exist according to set theory.

Can you prove that? If the line next to the unary representation of
aleph_0 does exist (and I think you mean preceding line) you have to
show that aleph_0 does have a predecessor. But it has not. Unless
you show how it can be proven through set theory.

The natural next to it does not. Hence,
the sum of

0.1
0.11
0.111
...
is *not* 0.111...

With your definition (finally extracted) above (if in any column there is
a 1, there is a 1 in the final result), it is. If you think that is false,
please show me a column where there are only 0's.

That column which cannot be covered by a natural number.
(f all couldcovered then 0.111... was a natural.)

Regards, WM
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Virgil
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Joined: 24 Mar 2005
Posts: 5536

PostPosted: Tue Jul 18, 2006 6:38 pm    Post subject: Re: An uncountable countable set Reply with quote

In article <1153220508.661035.130520@s13g2000cwa.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Quote:
Virgil schrieb:


The successor of all naturals is not a natural and, therefore, must be
larger (because it is not less).

There is no such thing as the successor of all naturals any more that
there is a single successor common to both 3 a and 6.

This is an expression coined by Cantor: "a number following after all
natural numbers".

But he does not call it a 'natural' number any more than he calls it an
even number or a prime number.

I said above: The successor of all naturals is not a natural.

No one else even says that such a thing exists. There is a set of all
naturals but if is not the successor of any of them.
Quote:

There is no number following all naturals and, hence, there is no
number omega at all.

Perhaps not in your philosophy, but you are not God, to command what is
true or false.
Quote:

There is no such 'natural' but there is such a cardinal or ordinal. Both
are generically 'numbers' not all cardinals or ordinals are naturals.

Why do you emphasize this? Of course it cannot be a natural if it
follows after all naturals.

I emphasize it because WM needs to learn it.
Quote:

One has the set of all naturals, and that set can have a successor under
the definition that the successor of any set, x, is (x union {x}).

That is nonsense too. One cannot have a set of all naturals, because
the step x U {x} defined by the axiom does never result in this set.

Does WM claim that the only sets that can exist must be of form
(x union {x})? Then WM is wrong.
Quote:

No one said that the set of naturals was a successor, stupid, we only say
that it has one.

Some say that omega is the set of naturals.




Not quite, omega is the cardinality of the set of naturals. Do try to
get things right, WM. It makes you look so silly when you make such
silly mistakes.

Quote:
And Cantor said that omega
was the smallest successor of all naturals.

Not quite. Cantor said that omega was the smallest cardinality greater
than that of any natural. WM is being silly again.
Quote:

Misrepresentations like that are the tools of trolls.

Those trolls which troll set theory.

Which is what WM is doing.
Quote:

You recently asserted that such stepwise processes can never exhaust a
set. Now you changed your mind? Just in time?

One can have a set of all of an inexhaustable supply of objects even
though one cannot exhaust that supply by serial operations.

such as a well-ordering undoubtedly is, because one number follows the
other in a sequence or a chain or in steps.

Well ordering is defining an order in which each non-empty subset has a
first member. There is nothing that requires the defining process to be
sequencial. One can, for example, well order the rationals by injecting
them into an already well ordered set like the set of naturals, which
will induce a well ordering on the rationals
Quote:

Regards, WM
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