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Virgil
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 Posted: Fri Oct 10, 2008 4:07 am Post subject: Re: Why "meta diagonals" are irrelevant |
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In article
<7fe9cec8-d5de-4c39-8757-f4dd79b6418a@z6g2000pre.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 9 Okt., 16:46, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
In article
ec757a57-7954-4593-808a-873ada048...@p31g2000prf.googlegroups.com> WM
mueck...@rz.fh-augsburg.de> writes:
> On 8 Okt., 16:43, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > In article
0dfb7977-a24d-4a7f-933a-c139c616f...@b30g2000prf.googlegroups=
> .com> WM <mueck...@rz.fh-augsburg.de> writes:
> > > On 8 Okt., 14:54, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > > > in a list, but on the other hand there is no
> > > > list that contains *all* reals.
> >
> > > Each meta-diagonal can be put in a list, but on the other hand
there
> > > is no list that contains *all* meta-diagonals.
> > All meta-diagonals of what? All meta-diagonals of all possible lists?
Of
> > course not, because the set of possible lists is uncountable.
> Wrong. A list is a construction.
[/quote]
WRONG! A list is function, but need not be a construction.
I
[quote]
You cannot take more than a countable number of single reals (even if
they were there), because there is no chance to do so (finite
alphabet). And you cannot construct uncountably many lists.
[/quote]
Mathematics is not constrained by the self-mutilation of constructionism. |
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Virgil
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| Posted: Fri Oct 10, 2008 4:19 am Post subject: Re: Why "meta diagonals" are irrelevant |
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In article
<6119fc84-34e3-488f-9b2f-025570d7ffa8@a70g2000hsh.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]
No, the definition is not in error. The definition defines something that
does not exist.
That means it is in error, because the definition includes the
statement of existence.
[/quote]
Then when someone defines a Kris Kringle, or a leprechaun or a unicorn,
one must be simultaneously stating one exists?
In English, specifically including English mathematics, definition is
divorced from instantiation.
One can, for instance, define a field having 6 elements, but one cannot
instantiate it.
[quote]
> If you define M is an empty set that is larger than the set of all
> sets, then this definition is in error.
No. The definition defines something that does not exist, that does not
make the definition erroneous.
Consider the standard way to prove that sqrt(2) is irrational. It starts
with:
Assume p/q = sqrt(2) with p and q integer.
That does not define sqrt(2) as being a rational.
Here is some form of definition, namely of a pair p, q. That is not an
error, but later it is shown that the definition defined something that
does not exist.
The definition, if it was a definition, would have defined sqrt(2).
That exists (at least according to most mathematicians). So you will
eliminate sqrt(2) from the universe of existing numbers? A commendable
approach!
[/quote]
If one defines in English the square root of 2 to be a positive RATIONAL
number whose square is two, there is no instantiation.
If one defines in English the square root of 2 to be a positive REAL
number whose square is two, there is a unique instantiation.
If one defines in English the square root of 2 to be only a REAL number
whose square is two, there are two distinct instantiations.
If other languages work differently, that is their problem, no ours. |
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Virgil
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| Posted: Fri Oct 10, 2008 4:27 am Post subject: Re: Why "meta diagonals" are irrelevant |
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In article
<c2b0f6b0-8769-4fc4-a706-7c69282f3f73@2g2000hsn.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 9 Okt., 20:30, William Hughes <wpihug...@hotmail.com> wrote:
This claim is pure nonsense, and is never uttered. William gave a
construction with the claim that there is a list that contains B. You
say "no" to that with an argument that is irrelevant.
William claimed so. Why not claim that all reals are in a list? I
showed that it is impossible to have all antidiagonals of
constructable lists in one list. That argument cannot be refuted.
So what? No one said it could.
This set is countable.
[/quote]
Can YOU *construct* a list of all anti-diagonals of all constructable
lists?
Or construct a proof showing that such a construction is possible?
Absent those, you have made yet a n o t h e r unproven claim. |
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Virgil
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| Posted: Fri Oct 10, 2008 4:51 am Post subject: Re: Why "meta diagonals" are irrelevant |
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In article
<c452e6d3-3621-4ad9-8047-70e70228338d@w1g2000prk.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 9 Okt., 20:39, William Hughes <wpihug...@hotmail.com> wrote:
On Oct 9, 2:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 00:36, William Hughes <wpihug...@hotmail.com> wrote:
snip
You started this with
I have shown that the standard definition is wrong. A set can have
cardinality aleph_0 and simultaneously cannot be put in a sequence.
Such a set is, for instance, the set of all entries and all
metadiagonals of an infinite sequence of sequences.
"Such a set" is the set I called B. My claim is that B has
cardinality aleph_0 and can be put in sequence C.
That is impossible. In order to do so, you must construct all lists of
the sequence, but you cannot because there is no last one.
Nope. Incredibly you do not seem to know the simple construction
by which you can get one list from an infinite sequence of lists.
Have you never seen a construction by which the rationals
are shown to be in a single list?
Of course. But there we did not consider the question concerning the
meta-diagonals. If you consider all, i.e., also those of your list C,
then you have a countable set that (by indexing the numbers with the
indexes of the lists, see my discussion with Virgil) cannot be in a
single sequence.
[/quote]
Then WM has managed to prove the uncountability of the set of all
infinite binary strings. |
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Virgil
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| Posted: Fri Oct 10, 2008 4:52 am Post subject: Re: Why "meta diagonals" are irrelevant |
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In article
<d827145a-006c-4508-96a7-1bbc4d3e6575@f37g2000pri.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 9 Okt., 20:24, Virgil <Vir...@gmale.com> wrote:
In article
But the principle can also be seen by means of the anti-diagonals.
Then construct for us any list of infinite binary strings containing any
of its anti-diagonals, or concede the impossibility of doing so.
I just argued that it is impossible to construct a list of all
antidiagonals of all constructed lists. Nevertheless the number of
constructed lists and, therefore, also the number of all those
antidiagonals is countable.
You may as well argue that the moon is made of green cheese.
Your mathematical argument fails for (1) lack of proof and
(2) proof of falsehood.
Sorry, I do not understand your sibylline words.
[/quote]
That hardly scratches the surface of what you do not understand. |
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Virgil
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| Posted: Fri Oct 10, 2008 4:55 am Post subject: Re: Why "meta diagonals" are irrelevant |
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In article
<95c20cb3-f7a4-44ce-b539-87b494ed8790@w39g2000prb.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 9 Okt., 20:42, Virgil <Vir...@gmale.com> wrote:
No matter how many times you repeat the process, you will still
only have countably many elements, and you can still combine them
into a single list.
Of course, that is what my proof needs.
Except that at each stage, there are anti-diagonals still not listed.
That is what my proof needs.
[/quote]
It needs a lot more than that, and most of what it needs does not exist.
[quote]
It is possible to have all the meta-diagonals from a countable set of
lists all in one list.
Unless they were indexed and the meta-diagonals of the final list were
included. But that is no longer relevant, because it is sufficient to
consider only the countable uncountable set of all anti-diagonals.
[/quote]
Which, since WM allows it to be uncountable, settles the issue on
whether uncountable sets exist. |
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Virgil
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| Posted: Fri Oct 10, 2008 5:05 am Post subject: Re: Why "meta diagonals" are irrelevant |
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In article
<859e087a-0bb1-4269-969a-bf8167cf25f9@t18g2000prt.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 9 Okt., 21:16, Virgil <Vir...@gmale.com> wrote:
Another definition is "can be injected into a countable set".
Does WM claim that any set which "can be injected into a countable set"
is incapable of being listed? Or that the two definitions are not
logically equivalent (logically equivalent in the sense that when either
holds the other must also)?
The countable set of all anti-diagonals (produced by a unique
replacement rule) of constructed Cantor-lists is countable but not
listable
[/quote]
Into which "countable set" is it injectable (WM-countable)?
And what non-listable but countable set is your ur-countable set into
which all other countable but not listable sets inject?
Unless you can come up with one such set, your definition fails.
[quote]
According to that definition of "countable", the reals are provably
not countable. Even you say so.
According to the other definition the reals are provably countable.
Provably false by any definition of the reals that I am aware of.
See the paths in the binary tree.
[/quote]
Which binary tree? The infinite complete binary tree in which every node
has two child nodes and no path terminates has uncountably many paths
corresponding to the uncountably many infinite binary strings of
branchings, {left,right} which make those paths.
[quote]
So you have some other notion of what "countable" means? What is
it?
The set of paths of the binary tree can be injected into the set of
its nodes.
[/quote]
Not outside of WM>s weird weird private world. |
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Michael Press
Guest
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| Posted: Fri Oct 10, 2008 7:50 am Post subject: Re: A consideration concerning the diagonal argument of G. C |
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In article <gcl34n01t8i@drn.newsguy.com>,
stevendaryl3016@yahoo.com (Daryl McCullough) wrote:
[quote]In article <87r66pzzwp.fsf@alatheia.dsl.inet.fi>, Aatu Koskensilta says...
georgie <geo_cant@yahoo.com> writes:
People who object to things like the diagonal argument simply
disagree with the fanatical belief in certain axioms. Those
"disbelievers" aren>t inferrior idiots just because the mathematical
fanatics can>t stand to see someone disagree with their holy axioms.
So which of the axioms used in the diagonal argument do you object to?
The Axiom of Idiocy, which says that anyone who doesn>t believe is
ZF is an iferrior idiot. The question of whether this axiom is independent
of the others has never been settled, although Cohen was able to show
that it follows from ZF + "there exists a ridiculously large cardinal".
[/quote]
The argument has come to be known as farcing.
--
Michael Press |
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WM
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| Posted: Fri Oct 10, 2008 8:35 am Post subject: Re: Why "meta diagonals" are irrelevant |
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On 9 Okt., 22:11, William Hughes <wpihug...@hotmail.com> wrote:
[quote]On Oct 9, 3:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 20:30, William Hughes <wpihug...@hotmail.com> wrote:
This claim is pure nonsense, and is never uttered. William gave a
construction with the claim that there is a list that contains B. You
say "no" to that with an argument that is irrelevant.
William claimed so. Why not claim that all reals are in a list? I
showed that it is impossible to have all antidiagonals of
constructable lists in one list. That argument cannot be refuted.
So what? No one said it could.
This set is countable.
Nope. It is not a sequence of sequences. You
have no argument that it is countable
[/quote]
It is a sequence of numbers. (A constructable item belongs to a
countable set.)
Regards, WM |
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WM
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| Posted: Fri Oct 10, 2008 10:53 am Post subject: Re: Why "meta diagonals" are irrelevant |
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On 9 Okt., 22:22, William Hughes <wpihug...@hotmail.com> wrote:
[quote]On Oct 9, 3:28 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 20:39, William Hughes <wpihug...@hotmail.com> wrote:
On Oct 9, 2:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 00:36, William Hughes <wpihug...@hotmail.com> wrote:
snip
You started this with
I have shown that the standard definition is wrong. A set can have
cardinality aleph_0 and simultaneously cannot be put in a sequence.
Such a set is, for instance, the set of all entries and all
metadiagonals of an infinite sequence of sequences.
"Such a set" is the set I called B. My claim is that B has
cardinality aleph_0 and can be put in sequence C.
That is impossible. In order to do so, you must construct all lists of
the sequence, but you cannot because there is no last one.
Nope. Incredibly you do not seem to know the simple construction
by which you can get one list from an infinite sequence of lists.
Have you never seen a construction by which the rationals
are shown to be in a single list?
Of course.
So we know know that a sequence of sequences is listable.
Consider the set B, a sequence of sequences, and the metadiagonals.
Your original claim is that B is not listable.
(Note that B only contains the metadiagonals of sequences in B.)
B consists of the union of the B_i, where B_i is a sequence
and its metadiagonals. Define D_i to be a sequence which contains
B_i and its metadiagonals. Define D to be the sequence of sequences
D_i. D is listable. D contains B. B is listable. No reference
to a last sequence.
[/quote]
But a reference to exclude the metadiagonals of D.
With your arguing you can also set up a sequence S of all anti
diagonals of constructed lists. You just refrain from consisdering the
antidiagonal of S.
With your arguing you can also set up a sequence of all real numbers.
You just refrain from considering the antidiagonal.
So in fact your arguing includes a last list. Above this finla list is
called D.
Regards, WM |
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WM
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| Posted: Fri Oct 10, 2008 11:14 am Post subject: Re: Why "meta diagonals" are irrelevant |
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On 9 Okt., 22:22, stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
[quote]WM says...
On 9 Okt., 21:16, Virgil <Vir...@gmale.com> wrote:
Another definition is "can be injected into a countable set".
Does WM claim that any set which "can be injected into a countable set"
is incapable of being listed? Or that the two definitions are not
logically equivalent (logically equivalent in the sense that when either
holds the other must also)?
The countable set of all anti-diagonals (produced by a unique
replacement rule) of constructed Cantor-lists is countable but not
listable
If it>s not listable, then it>s not injectable.
[/quote]
But every item, constructed or only defined as an individual
(requiring a finite string over a finite alpabet) belongs to a
contable set.
That>s an easy
[quote]proof:
[/quote]
It is not appropriate to put on the blinders of formal logic here.
They might look like a jester>s cap.
Regards, WM |
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WM
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| Posted: Fri Oct 10, 2008 11:18 am Post subject: Re: Why "meta diagonals" are irrelevant |
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On 10 Okt., 01:00, Virgil <Vir...@gmale.com> wrote:
[quote]In article
6f40d13d-0e17-4fe0-998a-86a0abfc7...@v16g2000prc.googlegroups.com>,
WM <mueck...@rz.fh-augsburg.de> wrote:
There is no final stage. At every stage that exists, my assertion is
correct. The meta-diagonals are never in one list.
Any set of "meta-diagonals" or anything else whose members cannot be put
into one list is uncountable by definition.
[/quote]
Any set of items that have been constructed or at least defined as
individuals is coutable by the fact that the set finite strings over a
finite alphabet is countable.
The set of all antidiagonals of lists that can be defined or
constructed is countable but not countable.
The logical expression "A and not A" is called contradiction.
Regards, WM
Regards, WM
Regards, WM |
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WM
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| Posted: Fri Oct 10, 2008 11:20 am Post subject: Re: Why "meta diagonals" are irrelevant |
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On 10 Okt., 01:07, Virgil <Vir...@gmale.com> wrote:
[quote]You cannot take more than a countable number of single reals (even if
they were there), because there is no chance to do so (finite
alphabet). And you cannot construct uncountably many lists.
Mathematics is not constrained by the self-mutilation of constructionism
[/quote]
But all constructive sets belong to your mathematics and can be
treated by your mathematics? Or are there exceptions?
Regards, WM |
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WM
Guest
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| Posted: Fri Oct 10, 2008 11:44 am Post subject: Re: Why "meta diagonals" are irrelevant |
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On 10 Okt., 01:19, Virgil <Vir...@gmale.com> wrote:
[quote]In article
6119fc84-34e3-488f-9b2f-025570d7f...@a70g2000hsh.googlegroups.com>,
WM <mueck...@rz.fh-augsburg.de> wrote:
No, the definition is not in error. The definition defines something that
does not exist.
That means it is in error, because the definition includes the
statement of existence.
Then when someone defines a Kris Kringle, or a leprechaun or a unicorn,
one must be simultaneously stating one exists?
[/quote]
He should be aware that the blinders of his logic might resemble a
fools cap.
A definition starts: X is ...
This "is" includes the existence of X: "X is". And the definition is
wrong, in error, mistaken, nonsense, foolish (according to the degree
of erring) if X is not.
Regards, WM |
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WM
Guest
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| Posted: Fri Oct 10, 2008 11:48 am Post subject: Re: Why "meta diagonals" are irrelevant |
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On 10 Okt., 01:27, Virgil <Vir...@gmale.com> wrote:
[quote]In article
c2b0f6b0-8769-4fc4-a706-7c69282f3...@2g2000hsn.googlegroups.com>,
WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 20:30, William Hughes <wpihug...@hotmail.com> wrote:
This claim is pure nonsense, and is never uttered. William gave a
construction with the claim that there is a list that contains B. You
say "no" to that with an argument that is irrelevant.
William claimed so. Why not claim that all reals are in a list? I
showed that it is impossible to have all antidiagonals of
constructable lists in one list. That argument cannot be refuted.
So what? No one said it could.
This set is countable.
Can YOU *construct* a list of all anti-diagonals of all constructable
lists?
[/quote]
I just said that that is impossible.
[quote]
Or construct a proof showing that such a construction is possible?
[/quote]
Such a proof would contradict my proof. I usually try to avoid such
proofs.
[quote]
Absent those, you have made yet a n o t h e r unproven claim.
[/quote]
Every constructable item belongs to a countable set. Therefore I need
not construct all constructable lists to see that they and their
antidiagonals belong to a countable set.
Regards, WM |
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