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Message |
| Guest |
 Posted: Thu Jun 22, 2006 12:19 am |
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Jens Kruse Andersen schrieb:
[quote:c8651d601b]By definition, an infinite set M is uncountable if and only if:
There is no bijective mapping f from N to M.
[/quote:c8651d601b]
Correct. What do you think, why this is so? My point is only to show
that the impossibility of a mapping involving Hessenberg's trick does
not prove anything about cardinalities of sets involved. It shows that
the sets |N and P(|N) do not exist.
[quote:c8651d601b]
Note that M is given first, and *then* it is said that no bijective f exists
for that M, with no other condition on f than it has to be bijective.
[/quote:c8651d601b]
Oh, I see. Let's apply this new insight:
2) 0.12324389
3) 0.23123123
4) 0.85348714
5) 0.11133333
6) 0.31415161
...
1) 0.24446...
So this is the proof, that the proof that a surjective mapping f: |N
--> [0,1] does not exist, does not exist, isn't it?
[quote:c8651d601b]
In the well-known proof that P(N) is uncountable, P(N) is a fixed given set
before any function is even mentioned.
[/quote:c8651d601b]
That is the question.
[quote:c8651d601b]The proof then shows that no matter
which function f is chosen, f will not be a bijection from N to P(N).
[/quote:c8651d601b]
And my proof shows that no matter what function f is chosen, there will
not be a bijection from N to M. You will have obtained from my sketch
of Cantor's diagonal proof above that it is *not* allowed to switch the
mapping after the set has been fixed.
Regards, WM |
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| Guest |
| Posted: Thu Jun 22, 2006 12:23 am |
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Dik T. Winter schrieb:
[quote:227948c9bf]
A *surjective* mapping does not exist. The same does Hessenberg.
A *surjective* mapping does exist, but it is not f. There exist a
surjective mapping g -> M(f).
[/quote:227948c9bf]
Like this one?:
2) 0.12324389
3) 0.23123123
4) 0.85348714
5) 0.11133333
6) 0.31415161
...
1) 0.24446...
This is the proof, that the proof that a surjective mapping f: |N -->
[0,1] does not exist, does not exist, isn't it?
Regards, WM |
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| Virgil |
| Posted: Thu Jun 22, 2006 1:09 am |
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Guest
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In article <1150957437.491685.293410@g10g2000cwb.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:9cef1db1f1]Dik T. Winter schrieb:
A *surjective* mapping does not exist. The same does Hessenberg.
A *surjective* mapping does exist, but it is not f. There exist a
surjective mapping g -> M(f).
Like this one?:
2) 0.12324389
3) 0.23123123
4) 0.85348714
5) 0.11133333
6) 0.31415161
..
1) 0.24446...
This is the proof, that the proof that a surjective mapping f: |N --
[0,1] does not exist, does not exist, isn't it?
[/quote:9cef1db1f1]
No. |
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| Dik T. Winter |
| Posted: Thu Jun 22, 2006 3:43 am |
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Guest
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In article <1150957437.491685.293410@g10g2000cwb.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
[quote:99e69935ba]
Dik T. Winter schrieb:
A *surjective* mapping does not exist. The same does Hessenberg.
A *surjective* mapping does exist, but it is not f. There exist a
surjective mapping g -> M(f).
Like this one?:
2) 0.12324389
3) 0.23123123
4) 0.85348714
5) 0.11133333
6) 0.31415161
..
1) 0.24446...
This is the proof, that the proof that a surjective mapping f: |N --
[0,1] does not exist, does not exist, isn't it?
[/quote:99e69935ba]
This is plain nonsense.
--
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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| Guest |
| Posted: Thu Jun 22, 2006 5:22 am |
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Virgil schrieb:
[quote:08e8245982]So the resolution is that either f, K_f and M_f are self-contradictory
and do not exist at all
[/quote:08e8245982]
if f is defined to be a surjection. That is right. And why do you
believe that a self contradictory approach should prove anything (for
instance about the surjectivity of certain mappings)?
Regards, WM |
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| Guest |
| Posted: Thu Jun 22, 2006 5:26 am |
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Dik T. Winter schrieb:
[quote:5d6d5c1c79]In article <1150957437.491685.293410@g10g2000cwb.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
Dik T. Winter schrieb:
A *surjective* mapping does not exist. The same does Hessenberg.
A *surjective* mapping does exist, but it is not f. There exist a
surjective mapping g -> M(f).
Like this one?:
2) 0.12324389
3) 0.23123123
4) 0.85348714
5) 0.11133333
6) 0.31415161
..
1) 0.24446...
This is the proof, that the proof that a surjective mapping f: |N --
[0,1] does not exist, does not exist, isn't it?
This is plain nonsense.
[/quote:5d6d5c1c79]
Maybe. It is, however, the same approach as yours. If it is impossible
to find a direct surjective mapping, then first define the set and then
try to find a surjection. I can proudly declare in your words:
A *surjective* mapping does exist, but it is not f. There exists a
surjective mapping g -> M(f).
Regards, WM |
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| Guest |
| Posted: Thu Jun 22, 2006 5:36 am |
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Dik T. Winter schrieb:
[quote:f91363e5d5]1. Given a mapping f: N -> P(N), the set K(f) constructed according to
Hessenberg *does* exist.
but it is not possible to determine that K(f).
Why not?
[/quote:f91363e5d5]
Because, in case f should be surjective, a number k need be contained
in a set in which it must not be contained.
[quote:f91363e5d5]
The proof above.
It does not disprove a surjection but the completeness of |N and P(|N).
You are completely wrong. How does it prove that?
[/quote:f91363e5d5]
Because it is the only explanation of this and other paradoxes. At
least you cannot deny that it would resolve he problem.
Here is another one: Let {q_1, q_2, q_3, ...} the well-ordered set of
all rational numbers.
If you say that it exists, then I can prove that it can be ordered by
magnitude without destroying the well-order. Obviously that is
impossible. Hence {q_1, q_2, q_3, ...} does not exist.
[quote:f91363e5d5]Let's see. Let S be the set of finite subsets of N. And let us have
a bijection f: N -> S (they do exist). Now obviously f is not a
bijection from N to M(f), that is clear by the construction.
[/quote:f91363e5d5]
Hence by constuction, the set M(f) does not exist, if f is required to
be surjective.
[quote:f91363e5d5]Note again: M depends on f, a fact that you leave out every time. M in
itself is not a set, but for every given f, M(f) is a set.
[/quote:f91363e5d5]
And for every given f, this set is not in bijection with |N.
Regards, WM |
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| Guest |
| Posted: Thu Jun 22, 2006 5:47 am |
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Virgil schrieb:
[quote:f0297c7a88]No more than it proves Muecken does not exist.
[/quote:f0297c7a88]
Please note: My name is Mueckenheim or short WM. It is fairly long,
but so much time must be for polite people. And I am not going to talk
to impolite people.
Regards, WM |
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| Daryl McCullough |
| Posted: Thu Jun 22, 2006 7:21 am |
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Guest
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mueckenh@rz.fh-augsburg.de says...
[quote:3685625aa6]Maybe. It is, however, the same approach as yours. If it is impossible
to find a direct surjective mapping, then first define the set and then
try to find a surjection. I can proudly declare in your words:
A *surjective* mapping does exist, but it is not f. There exists a
surjective mapping g -> M(f).
[/quote:3685625aa6]
And those words are perfectly correct: f is not a surjective mapping
from N to M(f), but g *is* a surjective mapping from N to M(f). So
M(f) is countable.
In contrast, there is no surjection from N to P(N). So P(N) is *not*
countable.
--
Daryl McCullough
Ithaca, NY |
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| Daryl McCullough |
| Posted: Thu Jun 22, 2006 7:23 am |
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Guest
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In article <1150975362.277573.39860@b68g2000cwa.googlegroups.com>,
mueckenh@rz.fh-augsburg.de says...
[quote:c95d7d05d5]
Virgil schrieb:
So the resolution is that either f, K_f and M_f are self-contradictory
and do not exist at all
if f is defined to be a surjection. That is right. And why do you
believe that a self contradictory approach should prove anything (for
instance about the surjectivity of certain mappings)?
[/quote:c95d7d05d5]
If an assumption leads to a contradiction, then that assumption must
be false. The assumption that there is a surjection from N to P(N) leads
to a contradiction. Therefore, it is false that there is a surjection
from N to P(N). Therefore, P(N) is uncountable.
--
Daryl McCullough
Ithaca, NY |
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| Jens Kruse Andersen |
| Posted: Thu Jun 22, 2006 1:14 pm |
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Guest
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mueckenh wrote:
[quote:4363133ea3]Jens Kruse Andersen schrieb:
By definition, an infinite set M is uncountable if and only if:
There is no bijective mapping f from N to M.
Note that M is given first, and *then* it is said that no bijective f
exists for that M, with no other condition on f than it has to be
bijective.
Oh, I see. Let's apply this new insight:
2) 0.12324389
3) 0.23123123
4) 0.85348714
5) 0.11133333
6) 0.31415161
..
1) 0.24446...
So this is the proof, that the proof that a surjective mapping f: |N
--> [0,1] does not exist, does not exist, isn't it?
[/quote:4363133ea3]
As others have said: Nonsense. A small list of numbers with no definitions,
assumptioms, explanations or anything else is, well, a small list of numbers.
Your final claim is hard to parse but if you think your small list of numbers
with no comments proves that Cantor's diagonal proof is wrong, then: Nonsense.
After seeing your replies here and elsewhere, I have decided to not spend more
time on this discussion.
Others have already said several times what I would say but you appear to
ignore it. I suggest rereading the replies carefully.
[quote:4363133ea3]Please note: My name is Mueckenheim or short WM.
[/quote:4363133ea3]
If you want to be referred to by some name then use it as poster name instead
of your e-mail address.
Many people reply with a program which automatically inserts your poster name.
--
Jens Kruse Andersen |
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| Guest |
| Posted: Thu Jun 22, 2006 2:52 pm |
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Jens Kruse Andersen schrieb:
[quote:45607a1f42]mueckenh wrote:
Jens Kruse Andersen schrieb:
By definition, an infinite set M is uncountable if and only if:
There is no bijective mapping f from N to M.
Note that M is given first, and *then* it is said that no bijective f
exists for that M, with no other condition on f than it has to be
bijective.
Oh, I see. Let's apply this new insight:
2) 0.12324389
3) 0.23123123
4) 0.85348714
5) 0.11133333
6) 0.31415161
..
1) 0.24446...
So this is the proof, that the proof that a surjective mapping f: |N
--> [0,1] does not exist, does not exist, isn't it?
As others have said: Nonsense. A small list of numbers with no definitions,
assumptioms, explanations or anything else is, well, a small list of numbers.
[/quote:45607a1f42]
I think everyone her does know this list of numbers. It is not a small
list, but a countably infinte one. (One point was missing, it should
read "...".)
[quote:45607a1f42]Your final claim is hard to parse but if you think your small list of numbers
with no comments proves that Cantor's diagonal proof is wrong, then: Nonsense.
[/quote:45607a1f42]
Comment: The first mapping read f = 1), 2), 3), .... The second mapping
reads 2), 3), 4), ..., 1). Isn't that enough?
[quote:45607a1f42]
After seeing your replies here and elsewhere, I have decided to not spend more
time on this discussion.
Others have already said several times what I would say but you appear to
ignore it. I suggest rereading the replies carefully.
[/quote:45607a1f42]
I have never stated the following problem. So you might be interested:
Cantor said: A well-ordered set remains well-ordered, if finitely many
or infinitely many transpositions are executed. Let's see what happens.
Let {q_1, q_2, q_3, ...} be the well-ordered set of all positive
rationals.
1) Consider for n e |N the rationals q_n and q_n+1. If q_n > q_n+1,
exchange q_n and q_n+1, otherwise let the succession unaltered. This
requires at most aleph_0 transpositions.
2) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2,
exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This
requires at most aleph_0 transpositions.
3) Consider for n e |N the rationals q_n and q_n+1. If q_n > q_n+1,
exchange q_n and q_n+1, otherwise let the succession unaltered. This
requires at most aleph_0 transpositions.
4) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2,
exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This
requires at most aleph_0 transpositions.
Of course, this all is defined in zero time, like the definition of
Cantor's diagonal.
After aleph_0 * aleph_0 = aleph_0 transpositions, the set {q_k_1,
q_k_2, q_k_3, ...} of all positive rationals is well-ordered and
simultaneously ordered by magnitude, starting with the smallest
positive rational and ending with the largest.
This can happen, if the well ordered set of all positive rational does
exist. But we know, it cannot happen. Hence, the well ordered set of
all positive rational does not exist.
Regards, WM |
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| Guest |
| Posted: Thu Jun 22, 2006 2:57 pm |
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Daryl McCullough schrieb:
[quote:f2a05970ca]
That case is impossible. To prove a case impossible, you only
need to show that it leads to a logical contradiction.
[/quote:f2a05970ca]
1) Consider a bjective mapping from the set {a, 1} --> P({1}) = {{},
{1}}, where a is *not a natural number*. This mapping is arbitrary but
has to satisfy only one condition, namely that the set K = {k e |N & k
/e f(k)} is in the image. This mapping is impossible. (See my article
"On Cantor's Theorem", arXiv, math.GM/0505648, 30 May 2005.)
2) Consider a mapping |N --> P(|N) which need not be surjective but has
to satisfy only one condition, namely that the set K = {k e |N & k /e
f(k)} is in the image. This mapping is impossible.
In both cases there is this impredicable request {f, k, K} which is
impossible to satisfy. But in the proof by Hessenberg, you insist, it
would prove non-surjectivity?
In a bijective mapping |N --> P(|N) there is every element of |N and
every subset of |N. But no set is defined as K above, because this
very *definition* leads to an impossible result as can be seen by the
examples above.
[quote:f2a05970ca]This well-ordering gives the following first few terms of the sequence:
[/quote:f2a05970ca]
It will be enough to use the positive rationals only.
[quote:f2a05970ca]
0/1,
-1/1, +1/1,
-2/1, -1/2, +1/2, +2/1
-3/1, -1/3, +1/3, +3/1
-4/1, -3/2, -2/3, -1/4, +2/3, +3/2, +4/1
-5/1, -1/5, +1/5, +5/1
etc.
(The first row has |p| + q = 1, the next row has |p| + q = 2, etc.)
But notice that this is not ordered by magnitude, since 1/2 < 1/1, but
1/1 comes before 1/2 in the ordering.
The ordering by magnitude in not a well-ordering.
Here is how we obtain it:[/quote:f2a05970ca]
Cantor said: A well-ordered set remains well-ordered, if finitely many
or infinitely many transpositions are executed. Let's see what happens.
Let {q_1, q_2, q_3, ...} be the well-ordered set of all positive
rationals.
1) Consider for n e |N the rationals q_n and q_n+1. If q_n > q_n+1,
exchange q_n and q_n+1, otherwise let the succession unaltered. This
requires at most aleph_0 transpositions.
2) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2,
exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This
requires at most aleph_0 transpositions.
3) Consider for n e |N the rationals q_n and q_n+1. If q_n > q_n+1,
exchange q_n and q_n+1, otherwise let the succession unaltered. This
requires at most aleph_0 transpositions.
4) Consider for n e |N the rationals q_n+1 and q_n+2. If q_n+1 > q_n+2,
exchange q_n+1 and q_n+2, otherwise let the succession unaltered. This
requires at most aleph_0 transpositions.
Of course, this all is defined in zero time, like the definition of
Cantor's diagonal.
After aleph_0 * aleph_0 = aleph_0 transpositions, the set {q_k_1,
q_k_2, q_k_3, ...} of all positive rationals is well-ordered and
simultaneously ordered by magnitude, starting with the smallest
positive rational and ending with the largest.
This can happen, if the well ordered set of all positive rational does
exist. But we know, it cannot happen.
Hence, the well ordered set of all positive rational does not exist.
Regards, WM |
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| Guest |
| Posted: Thu Jun 22, 2006 2:59 pm |
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Daryl McCullough schrieb:
[quote:5882eb05d8]In article <1150975362.277573.39860@b68g2000cwa.googlegroups.com>,
mueckenh@rz.fh-augsburg.de says...
Virgil schrieb:
So the resolution is that either f, K_f and M_f are self-contradictory
and do not exist at all
if f is defined to be a surjection. That is right. And why do you
believe that a self contradictory approach should prove anything (for
instance about the surjectivity of certain mappings)?
If an assumption leads to a contradiction, then that assumption must
be false. The assumption that there is a surjection from N to P(N) leads
to a contradiction. Therefore, it is false that there is a surjection
from N to P(N). Therefore, P(N) is uncountable.
Consider a mapping |N --> P(|N) which need not be surjective but has to[/quote:5882eb05d8]
satisfy only one condition, namely that the set K = {k e |N & k /e
f(k)} is in the image. This mapping is impossible.
{f, k, K} is impossible to satisfy. But in the proof by Hessenberg,
you insist, it would prove non-surjectivity?
In a bijective mapping |N --> P(|N) there is every element of |N and
every subset of |N. But no set is defined as K above, because this very
*definition* leads to an impossible result as can be seen by the
examples above.
Regards, WM |
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| Guest |
| Posted: Thu Jun 22, 2006 3:12 pm |
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Daryl McCullough schrieb:
[quote:b315456ead]mueckenh@rz.fh-augsburg.de says...
Maybe. It is, however, the same approach as yours. If it is impossible
to find a direct surjective mapping, then first define the set and then
try to find a surjection. I can proudly declare in your words:
A *surjective* mapping does exist, but it is not f. There exists a
surjective mapping g -> M(f).
And those words are perfectly correct: f is not a surjective mapping
from N to M(f), but g *is* a surjective mapping from N to M(f). So
M(f) is countable.
In contrast, there is no surjection from N to P(N). So P(N) is *not*
countable.
[/quote:b315456ead]
This is proven by three invalid proofs. Do you know Canto's first proof
of 1874? Probably not, so we need not discuss it. You certainly know
his second one. It is wrong, because all it shows is the construction
of a number of a set of countably many constructible numbers. It is
ridiculous to believe this to be a proof of uncountability. There are
simply not more than countably many constructible and individualizable
real numbers, because there are not more than countably many different
symbols or names in any language. And I pointed out already that your
arguing invalidates this proof completely:
[quote:b315456ead]2) 0.12324389
3) 0.23123123
4) 0.85348714
...
1) 0.244...
[/quote:b315456ead]
(I hope, you understand, what I mean.)
The third proof is that by Hessenberg, which utiizes an impredicable
definition. This set K does *exist*, if every subset of |N does exist,
but it cannot be *defined* as it is done. But I discussed that with you
in a previous posting.
Uncountably many sets, however, cannot be distinguished, because of
laking names.
Regards, WM |
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