Re: A consideration concerning the diagonal argument of G. Cantor
WM
> On Oct 3, 4:04 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > No, I do not note that. I note that the definition of countability by
> > a single seqeunce and of countability by an infinite union of
> > sequences contradict each other.
>
> No, there is no such contradition as it is easy
> to show that if there is a bijection from set A to
> a sequence of sequences, then there is a bijection from
> A to a sequence and visa-versa.
Not for an infinite sequence of infinite sequences. The simple proof
is that what evr you construct as a "list", there is a again a
sequence of meta-diagonals not in the list.
>
> The contradiction you think you have found is your
> usual mistake of not taking into account that
> something can change. In this case it is the
> real number you construct. It can change if
> the set you are looking at changes.
Of course it changes. But the actual (!) cannot be excluded to be a
meta-diagonal of the actual list.
>
> More formally:
>
> Given any set or real numbers A with cardinality aleph_0,
> it is possible to contruct a real number,
> *that may depend on the set A*, call it r(A),
> such that r(A) is not an element of A.
>
> You consider a sequence of real numbers,
> call it S.
You misunderstood the argument. Here it is again:
For every line of Cantor's list it is true that this line does not
contain the diagonal number. Nevertheless the diagonal number may be
in the infinite list yet.
Consider the infinite Cantor-list containing the entries a_m1, a_m2,
a_m3, ...
a_11, a_12, a_13, ...
a_21, a_22, a_23, ...
a_31, a_32, a_33, ...
...
The diagonal number
a'_11, a'_22, a'_33, ... with a'_ii =/= a_ii (and a'_ii =/= 9)
cannot be excluded to be identical to the meta-diagonal
a_21, a_ 32, a_43, ...
for example, or generally, to be identical to a meta-diagonal
a_m1, a_ (m+1)2, a_(m+2)3, ... with m =/= 1.
or to a truncated (meta-) diagonal
a_mn, a_ (m+1)(n+1), a_(m+2)(n+2), ... with n =/= 1.
Therefore the diagonal number can exist (infinitely often) in the
Cantor-list containing countably many meta-diagonals like
xxxxxx...
dxxxxx...
xdxxxx...
xxdxxx...
xxxdxx...
...
and truncated meta-diagonals like
xxxxxx...
xxxxxx...
xxxxxx...
xxdxxx...
xxxdxx...
xxxxdx...
...
This construction holds for every list you may construct. None can
exclude it.
Therefore Cantor's diagonal proof does not supply evidence for the
existence of uncountable sets. It cannot exclude the existence of the
diagonal number in the countable list of meta-diagonals and truncated
(meta-) diagonals.
Regards, WM
Jan Andres
> On 3 Okt., 14:48, William Hughes <wpihug...@hotmail.com> wrote:
>> No, there is no such contradition as it is easy
>> to show that if there is a bijection from set A to
>> a sequence of sequences, then there is a bijection from
>> A to a sequence and visa-versa.
>
> Not for an infinite sequence of infinite sequences.
Sure it is.
http://en.wikipedia.org/wiki/Cantor_pairing_function
Jan
Virgil
<e36c1b05-c49e-42cb...@d45g2000hsc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 3 Okt., 14:48, William Hughes <wpihug...@hotmail.com> wrote:
> > On Oct 3, 4:04 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > No, I do not note that. I note that the definition of countability by
> > > a single seqeunce and of countability by an infinite union of
> > > sequences contradict each other.
> >
> > No, there is no such contradition as it is easy
> > to show that if there is a bijection from set A to
> > a sequence of sequences, then there is a bijection from
> > A to a sequence and visa-versa.
>
> Not for an infinite sequence of infinite sequences.
You are wrong again. Any such sequence of sequences can be rearranged
into a single sequence.
> The simple proof
> is that what evr you construct as a "list", there is a again a
> sequence of meta-diagonals not in the list.
IT only takes one to prove Cantor.
> >
> > The contradiction you think you have found is your
> > usual mistake of not taking into account that
> > something can change. In this case it is the
> > real number you construct. It can change if
> > the set you are looking at changes.
>
> Of course it changes. But the actual (!) cannot be excluded to be a
> meta-diagonal of the actual list.
Is this supposed to make sense?
> >
> > More formally:
> >
> > Given any set or real numbers A with cardinality aleph_0,
> > it is possible to contruct a real number,
> > *that may depend on the set A*, call it r(A),
> > such that r(A) is not an element of A.
> >
> > You consider a sequence of real numbers,
> > call it S.
>
> You misunderstood the argument. Here it is again:
> For every line of Cantor's list it is true that this line does not
> contain the diagonal number. Nevertheless the diagonal number may be
> in the infinite list yet.
All Cantor claims is that it is not a LINE in that list of lines, which
you now concede.
Ergo Cantor's proof is vindicated!!!!
William Hughes
> For every line of Cantor's list it is true that this line does not
> contain the diagonal number.
To rephrase, if B is any list of real numbers,
then given B, we can contruct d(B), such that
d(B) is not a line in B.
Let C be any set of real numbers with cardinality aleph_0.
Then we can construct r(C), a real number that is
not in C.
Proof:
C has cardinality aleph_0, so there is
a list D, such that every element of C is
a line in D. We construct d(D). Note that
d(D) is not a line of D (it may be a metadiagonal
of D, we neither know nor care). d(D) is not an element
of C (all elements of C are lines of D).
set r(C) equal to d(D).
- William Hughes
WM
> On Oct 4, 7:15 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > For every line of Cantor's list it is true that this line does not
> > contain the diagonal number.
>
> To rephrase, if B is any list of real numbers,
> then given B, we can contruct d(B), such that
> d(B) is not a line in B.
That is irrelevant for the question whether R can be given as a
countable set of lists, i.e., has cardinal number aleph_0.
>
> Let C be any set of real numbers with cardinality aleph_0.
> Then we can construct r(C), a real number that is
> not in C.
>
> Proof:
is irrelevant for the question whether R has cardinal number aleph_0 *
aleph_0 = aleph_0.
Regards, WM
William Hughes
> On 6 Okt., 02:32, William Hughes <wpihug...@hotmail.com> wrote:
<snip>
>
> > Let C be any set of real numbers with cardinality aleph_0.
> > Then we can construct r(C), a real number that is
> > not in C.
In other words C is not R.
So R does not have cardinality aleph_0
>
> > Proof:
>
> is irrelevant for the question whether R has cardinal number aleph_0 *
> aleph_0 = aleph_0.
Nope. Since aleph_0 * aleph_0 = aleph_0, showing that R does not have
cardinality
aleph_0 also shows that R does not have cardinality aleph_0 * aleph_0.
- William Hughes
WM
> On Oct 6, 9:08 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 6 Okt., 02:32, William Hughes <wpihug...@hotmail.com> wrote:
>
> <snip>
>
>
>
> > > Let C be any set of real numbers with cardinality aleph_0.
> > > Then we can construct r(C), a real number that is
> > > not in C.
The premise is wrong. You cannot take "any set of reals with
cardinality aleph_0". If you think you have taken the largest possible
one, then there is automatically another set of reals with cardinality
aleph_0. It is not so easy to deal with infinity.
That's the simple solution of one of the greatest mathematical
problems of the last 130 years.
>
> In other words C is not R.
> So R does not have cardinality aleph_0
In other words, C is not the largest possible countable set.
>
>
>
> > > Proof:
>
> > is irrelevant for the question whether R has cardinal number aleph_0 *
> > aleph_0 = aleph_0.
>
> Nope. Since aleph_0 * aleph_0 = aleph_0, showing that R does not have
> cardinality
> aleph_0 also shows that R does not have cardinality aleph_0 * aleph_0.
Non sequitur. (Wrong direction of arguing.)
Regards, WM
William Hughes
> On 6 Okt., 16:58, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Oct 6, 9:08 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 6 Okt., 02:32, William Hughes <wpihug...@hotmail.com> wrote:
>
> > <snip>
>
> > > > Let C be any set of real numbers with cardinality aleph_0.
> > > > Then we can construct r(C), a real number that is
> > > > not in C.
>
> The premise is wrong. You cannot take "any set of reals with
> cardinality aleph_0".
OK rephrase
Let C be a set of real numbers with cardinality aleph_0.
Then we can construct r(C), a real number that is
not in C.
Claim:
R does not have cardinality aleph_0.
Assume that R has cardinality aleph_0.
Then R is a set of real numbers with cardinality aleph_0
Then we can construct r(R) a real number that
is not in R. Contradition.
Therefore R does not have cardinality aleph_0.
- William Hughes
Herbert Newman
>
> OK rephrase
>
> Let C be a set of real numbers with cardinality aleph_0.
>
You mean
Let C be an arbitrary set of real numbers with cardinality aleph_0.
right?
Herb
Herbert Newman
>>
>> To rephrase, if B is any list of real numbers,
>> then given B, we can contruct d(B), such that
>> d(B) is not a line in B.
>
> That is irrelevant for the question whether R can be given as a
> countable set of lists, i.e., has cardinal number aleph_0. [WM]
>
That's only irrelevant for that question in Mückenheim's rather weird
anti-mathematical world.
Herb
William Hughes
What I mean in that the only things I care about are that C is a set
of real numbers, and C has cardinality aleph_0. We may or may
not know more about C.
Phrases used to mean this are
i: Let C be any set of real numbers with cardinality aleph_0.
ii: Let C be an arbitrary set of real numbers with cardinality
aleph_0.
iii: Let C be a set of real numbers with cardinality aleph_0.
WM does not like i or ii because he thinks that "any set" or an
"arbitrary set" is a special kind of set. Hence my use of iii.
[iii is a bit ambiguous as it could mean, "there exists at least
on set C" rather than "for every set C" i or ii are preferable]
- William Hughes
Herbert Newman
>
> What I mean in that the only things I care about are that C is a set
> of real numbers, and C has cardinality aleph_0. We may or may
> not know more about C.
>
> Phrases used to mean this are
>
> i: Let C be any set of real numbers with cardinality aleph_0.
> ii: Let C be an arbitrary set of real numbers with cardinality
> aleph_0.
> iii: Let C be a set of real numbers with cardinality aleph_0.
>
> WM does not like i or ii because he thinks that "any set" or an
> "arbitrary set" is a special kind of set. Hence my use of iii.
>
The problem (or rather ONE of the problems) when discussing "mathematical
questions" with WM is that he knows shit about the practice of mathematics.
So IMHO it's a rather pointless task.
Herb
WM
> On Oct 6, 11:19 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > Let C be any set of real numbers with cardinality aleph_0.
> > > > > Then we can construct r(C), a real number that is
> > > > > not in C.
>
> > The premise is wrong. You cannot take "any set of reals with
> > cardinality aleph_0".
>
> OK rephrase
>
> Let C be a set of real numbers with cardinality aleph_0.
You should say: Let C be a set of real numbers that can be in a
sequence.
> Then we can construct r(C), a real number that is
> not in C.
No. You cannot construct a diagonal of the whole set R, although R has
cardinal number aleph_0, i.e., it belongs to an infinite sequence of
sequences.
>
> Claim:
>
> R does not have cardinality aleph_0.
False. R (and any other set) cannot be proven to have cardinality
larger than aleph_0.
>
> Assume that R has cardinality aleph_0.
> Then R is a set of real numbers with cardinality aleph_0
> Then we can construct r(R) a real number that
> is not in R. Contradition.
False. Falsity is easy to see by the fact that the set taken and any
of the diagonal numbers constructed belong to a set with cardinality
aleph_0.
>
> Therefore R does not have cardinality aleph_0.
Non sequitur.
Regards, WM
Herbert Newman
>>
>> Let C be a set of real numbers with cardinality aleph_0.
>> Then we can construct r(C), a real number that is
>> not in C. [William Hughes]
>>
> No. You cannot construct a diagonal of the whole set R, [...]
>
Huh?! What the hell are you talking about, man?! Can't you READ? No one
talked about R here. William Hughes considers an arbitrary set C of real
numbers with cardinality aleph_0. (For example the set {1, 2, 3, ...} would
be one such a set.)
>
> although R has cardinal number aleph_0
>
If "R" denotes the set of real numbers, then your claim is FALSE. R does
_not_ have the cardinality aleph_0.
>>
>> Claim:
>>
>> R does not have cardinality aleph_0.
>>
> False.
>
Nonsense. Of course the claim is true; at least in the context of classical
mathematics (comprising set theory): it can be proved in the context of,
say, ZFC.
>
> R (and any other set) cannot be proven to have cardinality
> larger than aleph_0.
>
Nonsense. Of course it can be proved. Maybe reading a textbook might help.
>>
>> Proof:
>>
>> Assume that R has cardinality aleph_0.
>> Then R is a set of real numbers with cardinality aleph_0
>> Then we can construct r(R) a real number that
>> is not in R. Contradiction.
>>
> False.
>
Huh?! Do you have any mental problems, man?
Herb
William Hughes
> On 6 Okt., 17:38, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Oct 6, 11:19 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > > Let C be any set of real numbers with cardinality aleph_0.
> > > > > > Then we can construct r(C), a real number that is
> > > > > > not in C.
>
> > > The premise is wrong. You cannot take "any set of reals with
> > > cardinality aleph_0".
>
> > OK rephrase
>
> > Let C be a set of real numbers with cardinality aleph_0.
>
> You should say: Let C be a set of real numbers that can be in a
> sequence.
I just did. "C is a set of real numbers that can be in a sequence"
and "C is a set of real numbers with cardinality aleph_0" mean
exactly the same thing.
>
> > Then we can construct r(C), a real number that is
> > not in C.
>
> No. You cannot construct a diagonal of the whole set R,
Neither R nor a diagonal was mentioned.
The claim is that you can construct a real
number that is not in C (this real number may be a diagonal
of some list, but this is not claimed).
- William Hughes
WM
> On Oct 6, 1:18 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
> > On 6 Okt., 17:38, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > On Oct 6, 11:19 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > > > Let C be any set of real numbers with cardinality aleph_0.
> > > > > > > Then we can construct r(C), a real number that is
> > > > > > > not in C.
>
> > > > The premise is wrong. You cannot take "any set of reals with
> > > > cardinality aleph_0".
>
> > > OK rephrase
>
> > > Let C be a set of real numbers with cardinality aleph_0.
>
> > You should say: Let C be a set of real numbers that can be in a
> > sequence.
>
> I just did. "C is a set of real numbers that can be in a sequence"
> and "C is a set of real numbers with cardinality aleph_0" mean
> exactly the same thing.
That shows the inconsistency of these notions.
As I have proven there are sets of cardinality aleph_0 that cannot be
given the form of a sequence.
>
>
>
> > > Then we can construct r(C), a real number that is
> > > not in C.
>
> > No. You cannot construct a diagonal of the whole set R,
>
> Neither R nor a diagonal was mentioned.
> The claim is that you can construct a real
> number that is not in C (this real number may be a diagonal
> of some list, but this is not claimed).
From a sequence you can construct a diagonal that has not been in the
sequence. Nevertheless this diagonal is in a set of cardinality
aleph_0 together with all elements of the sequence. And, of course, it
can be put in a sequence together with the other numbers, after it has
been constructed.
This does not show the existence of cardinals larger than aleph_0 but
it shows only that not all real numbers do exist but have to be
constructed, according to Dedekind.
Regards, WM
Herbert Newman
>
> I have proven there are sets of cardinality aleph_0 that cannot be
> given the form of a sequence.
>
Oh my God!
William Hughes
> On 6 Okt., 20:27, William Hughes <wpihug...@hotmail.com> wrote:
> > I just did. "C is a set of real numbers that can be in a sequence"
> > and "C is a set of real numbers with cardinality aleph_0" mean
> > exactly the same thing.
>
> That shows the inconsistency of these notions.
> As I have proven there are sets of cardinality aleph_0 that cannot be
> given the form of a sequence.
>
Hmm. this is a strange new use of aleph_0 that I was not previously
aware of. Let us call a set that cannot be given the form of a
sequence
an unsequencable set. Can we agree that R is an unsequencable set?
- William Hughes
WM
Yes.
Regards, WM
Herbert Newman
>
> Can we agree that R is an unsequencable set?
>
Great. Now WM claims that he has "proven" that IR has the cardinality
aleph_0. With other words, there is a set with aleph_0 (namely IR, the set
of all real numbers) which is unsequencable. Are YOU glad now? :-o
Herb
William Hughes
Good. Now I say that _by definition_ a set has cardinality
aleph_0 iff it can be put in a sequence. So I say that
R does not have cardinality aleph_0.
Your claim is that R can be put in a sequence of sequences
and thus has cardinality aleph_0.
(Note that you have not shown this, you have only shown
that the standard diagonal proof does not (directly) disprove
this). You refuse to define aleph_0, you just use
aleph_0 * aleph_0 = aleph_0
Note, however, that any set that can be put in a sequence
of sequences can also be put in a sequence. So the fact
that R is unsequencable means that R cannot be put in
a sequence of sequences.
- William Hughes
WM
> On Oct 6, 3:11 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 6 Okt., 21:06, William Hughes <wpihug...@hotmail.com> wrote:
> >Can we agree that R is an unsequencable set?
>
> > Yes.
>
> Good. Now I say that _by definition_ a set has cardinality
> aleph_0 iff it can be put in a sequence. So I say that
> R does not have cardinality aleph_0.
This definition is not new. But the result aleph_0 * aleph_0 =
aleph_0 then is false.
>
> Your claim is that R can be put in a sequence of sequences
> and thus has cardinality aleph_0.
> (Note that you have not shown this, you have only shown
> that the standard diagonal proof does not (directly) disprove
> this). You refuse to define aleph_0, you just use
> aleph_0 * aleph_0 = aleph_0
>
> Note, however, that any set that can be put in a sequence
> of sequences can also be put in a sequence.
False. That would be true only if there was a last sequence of meta-
diagonals. But there isn't.
Regards, WM
Herbert Newman
>>
>>Can we agree that R is an unsequencable set?
>>
>> Yes.
>>
> Good. Now I say that _by definition_ a set has cardinality
> aleph_0 iff it can be put in a sequence. So I say that
> R does not have cardinality aleph_0.
>
> Your claim is that R can be put in a sequence of sequences
> and thus has cardinality aleph_0.
> (Note that you have not shown this, you have only shown
> that the standard diagonal proof does not (directly) disprove
> this). You refuse to define aleph_0, you just use
> aleph_0 * aleph_0 = aleph_0
>
> Note, however, that any set that can be put in a sequence
> of sequences can also be put in a sequence. So the fact
> that R is unsequencable means that R cannot be put in
> a sequence of sequences.
>
Oh, may poor brain. How confusing. Is this math? :-)
Herb
William Hughes
> On 6 Okt., 21:25, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Oct 6, 3:11 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 6 Okt., 21:06, William Hughes <wpihug...@hotmail.com> wrote:
> > >Can we agree that R is an unsequencable set?
>
> > > Yes.
>
> > Good. Now I say that _by definition_ a set has cardinality
> > aleph_0 iff it can be put in a sequence. So I say that
> > R does not have cardinality aleph_0.
>
> This definition is not new.
Indeed as it is the standard definition of a set having cardinality
aleph_0. You have neither shown it is inconsistent (Demonstrating
that
a particular proof does not show that R is not contained by a
sequence of sequences is a long way from showing that there
is a sequence of sequences that contains R) nor have
you have presented an alternate defintion.
- William Hughes
MoeBlee
> For every line of Cantor's list it is true that this line does not
> contain the diagonal number. Nevertheless the diagonal number may be
> in the infinite list yet.
Not if "in the infinite list" means 'in the range of the list', which
is all that matters for the diagonal argument.
But that is not the sense of 'in the list' that the diagonal argument
refers to. All you've done is add another sense of the for the phrase
'in the list'. It's irrelevent to refuting the diagonal argument.
Remove your amphiboly as to what we mean by "in the list" and use
strict mathematical terminology without the word 'list' and the
diagonal argument stands:
Suppose f is a function from w into the set of denumerable binary
sequences. Then the denumerable binary sequence d defined by:
d(n) = 0 if f(n)(n) = 1 and d(n) = 1 if f(n)(n) = 0
is not in the range of f. Therefore there is no function from w onto
the set of denumerable binary sequences. And a set is uncoutnable iff
there is no function from w onto the set. Therefore, the set of
sequences is uncountable.
Why you can't understand that is, of course, a matter of psychology
not mathematics.
MoeBlee
Virgil
<714c4506-d119-4c39...@e2g2000hsh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 6 Okt., 02:32, William Hughes <wpihug...@hotmail.com> wrote:
> > On Oct 4, 7:15 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > For every line of Cantor's list it is true that this line does not
> > > contain the diagonal number.
> >
> > To rephrase, if B is any list of real numbers,
> > then given B, we can contruct d(B), such that
> > d(B) is not a line in B.
>
> That is irrelevant for the question whether R can be given as a
> countable set of lists
If one can show a construction of a bijection between the set of decimal
strings and the set of reals, which one can, it becomes relevant again.
> >
> > Let C be any set of real numbers with cardinality aleph_0.
> > Then we can construct r(C), a real number that is
> > not in C.
> >
> > Proof:
>
> is irrelevant
Proofs, at least logically valid ones, seem always to be held irrelevant
by WM, as clearly evidenced in all those un-refereed papers of his he
keeps referring to.
Virgil
<8857f8a4-6e84-4bcd...@a70g2000hsh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 6 Okt., 16:58, William Hughes <wpihug...@hotmail.com> wrote:
> > On Oct 6, 9:08 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 6 Okt., 02:32, William Hughes <wpihug...@hotmail.com> wrote:
> >
> > <snip>
> >
> >
> >
> > > > Let C be any set of real numbers with cardinality aleph_0.
> > > > Then we can construct r(C), a real number that is
> > > > not in C.
>
> The premise is wrong. You cannot take "any set of reals with
> cardinality aleph_0".
What WM cannot do many others can.
> If you think you have taken the largest possible
> one,
There is no such thing as a /largest possible countable set of reals/,
and William Hughes has not suggested that there was.
So that WM is again off on a tangent unrelated to the issue.
> >
> > In other words C is not R.
> > So R does not have cardinality aleph_0
>
> In other words, C is not the largest possible countable set.
Of reals, precisely!
There is no countable set of reals which is not a proper subset of
another countable set of reals. That is precisely what Cantor is saying
in those theorems that WM does not comprehend.
Virgil
<7028f537-6bf2-410f...@l64g2000hse.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 6 Okt., 17:38, William Hughes <wpihug...@hotmail.com> wrote:
> > On Oct 6, 11:19 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > > > Let C be any set of real numbers with cardinality aleph_0.
> > > > > > Then we can construct r(C), a real number that is
> > > > > > not in C.
> >
> > > The premise is wrong. You cannot take "any set of reals with
> > > cardinality aleph_0".
> >
> > OK rephrase
> >
> > Let C be a set of real numbers with cardinality aleph_0.
>
> You should say: Let C be a set of real numbers that can be in a
> sequence.
He should say it the way he wants to say it, not the way tyo want him to.
Doing things your way leads to anarchy.
>
> > Then we can construct r(C), a real number that is
> > not in C.
>
> No. You cannot construct a diagonal of the whole set R
Who ever said "the whole set R"? Only WM, and he is wrong!
> >
> > R does not have cardinality aleph_0.
>
> False. R (and any other set) cannot be proven to have cardinality
> larger than aleph_0.
R already has been proven to have cardinality geater that aleph_0.
That WM is too dense to grasp it is not our fault.
> >
> > Assume that R has cardinality aleph_0.
> > Then R is a set of real numbers with cardinality aleph_0
> > Then we can construct r(R) a real number that
> > is not in R. Contradition.
>
> False.
A theorem of Cantor's, plus a bit of polishing , does it nicely, even if
WM fails to grasp it.
> Falsity is easy to see by the fact that the set taken and any
> of the diagonal numbers constructed belong to a set with cardinality
> aleph_0.
It is not whether that one item (the anti-diagonal) belongs to some
other set, but whether it belongs to the particular set from which it is
derived, which it does not.
> >
> > Therefore R does not have cardinality aleph_0.
Wrong again, WM.
Virgil
<13a14724-510e-483d...@v28g2000hsv.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 6 Okt., 20:27, William Hughes <wpihug...@hotmail.com> wrote:
> > On Oct 6, 1:18 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 6 Okt., 17:38, William Hughes <wpihug...@hotmail.com> wrote:
> >
> > > > On Oct 6, 11:19 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > > > > Let C be any set of real numbers with cardinality aleph_0.
> > > > > > > > Then we can construct r(C), a real number that is
> > > > > > > > not in C.
> >
> > > > > The premise is wrong. You cannot take "any set of reals with
> > > > > cardinality aleph_0".
> >
> > > > OK rephrase
> >
> > > > Let C be a set of real numbers with cardinality aleph_0.
> >
> > > You should say: Let C be a set of real numbers that can be in a
> > > sequence.
> >
> > I just did. "C is a set of real numbers that can be in a sequence"
> > and "C is a set of real numbers with cardinality aleph_0" mean
> > exactly the same thing.
>
> That shows the inconsistency of these notions.
> As I have proven there are sets of cardinality aleph_0 that cannot be
> given the form of a sequence.
Since having cardinality aleph_0 is defined as capable of being put into
a sequence, WM is playing the fool again.
> >
> >
> >
> > > > Then we can construct r(C), a real number that is
> > > > not in C.
> >
> > > No. You cannot construct a diagonal of the whole set R,
> >
> > Neither R nor a diagonal was mentioned.
> > The claim is that you can construct a real
> > number that is not in C (this real number may be a diagonal
> > of some list, but this is not claimed).
>
> From a sequence you can construct a diagonal that has not been in the
> sequence. Nevertheless this diagonal is in a set of cardinality
> aleph_0 together with all elements of the sequence.
When the critical test is whether element a is in set B, having it at
most e in some superset of B is not sufficient.
AS WM's grasp of logic is perpetually not sufficient.
Virgil
<13a14724-510e-483d...@v28g2000hsv.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> I have proven there are sets of cardinality aleph_0 that cannot be
> given the form of a sequence.
For his next trick, WM will "prove" 2 = 1.
Virgil
<71093608-1e10-498a...@34g2000hsh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 6 Okt., 21:25, William Hughes <wpihug...@hotmail.com> wrote:
> > On Oct 6, 3:11 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 6 Okt., 21:06, William Hughes <wpihug...@hotmail.com> wrote:
> > >Can we agree that R is an unsequencable set?
> >
> > > Yes.
> >
> > Good. Now I say that _by definition_ a set has cardinality
> > aleph_0 iff it can be put in a sequence. So I say that
> > R does not have cardinality aleph_0.
>
> This definition is not new. But the result aleph_0 * aleph_0 =
> aleph_0 then is false.
> >
> > Your claim is that R can be put in a sequence of sequences
> > and thus has cardinality aleph_0.
> > (Note that you have not shown this, you have only shown
> > that the standard diagonal proof does not (directly) disprove
> > this). You refuse to define aleph_0, you just use
> > aleph_0 * aleph_0 = aleph_0
> >
> > Note, however, that any set that can be put in a sequence
> > of sequences can also be put in a sequence.
>
> False.
On the contrary, it is almost trivially true.
The objects in sequence of sequences are essentially indexed by an
ordered pair of natural number indices, one for outer sequence and one
for the inner sequence.
It is then the problem defining a sequential ordering on the set of
ordered pairs of naturals, {(n,m): n in N and M in N}.
We do it by these two simple rules for distinct pairs (n1,m1) and (n2,m2)
(1) if n1 + m1 < n2 + m2, then (n1,m1) < (n2,m2)
(2) if n1 + m1 = n2 + m2 and n1 < n2 then (n1,m1) < (n2,m2).
This results is a sequential ordering of the ordered pairs of naturals.
It is omething obvious to mathematicians, but beyond WM's ken.
WM
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.
Regards, WM
WM
> > That shows the inconsistency of these notions.
> > As I have proven there are sets of cardinality aleph_0 that cannot be
> > given the form of a sequence.
>
> Since having cardinality aleph_0 is defined as capable of being put into
> a sequence, WM is playing the fool again.
>
Consider a Cantor-list. Put its meta-diagonals in the list, for
instance put one meta-diagonal between any pair of regular entries.
Then you have a new list. Repeat the same procedure and repeat and
repeat ...
The set of all entries and all metadiagonals of the infinte sequence
of sequences is countable but cannot be put in one sequence.
Regards, WM
William Hughes
> On 6 Okt., 22:07, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
> > On Oct 6, 3:41 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 6 Okt., 21:25, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > On Oct 6, 3:11 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > > On 6 Okt., 21:06, William Hughes <wpihug...@hotmail.com> wrote:
> > > > >Can we agree that R is an unsequencable set?
>
> > > > > Yes.
>
> > > > Good. Now I say that _by definition_ a set has cardinality
> > > > aleph_0 iff it can be put in a sequence. So I say that
> > > > R does not have cardinality aleph_0.
>
> > > This definition is not new.
>
> > Indeed as it is the standard definition of a set having cardinality
> > aleph_0. You have neither shown it is inconsistent (Demonstrating
> > that
> > a particular proof does not show that R is not contained by a
> > sequence of sequences is a long way from showing that there
> > is a sequence of sequences that contains R) nor have
> > you have presented an alternate defintion.
>
> I have shown that the standard definition is wrong.
No
>A set
Well at least you have abandoned any pretence that
you have shown something about R.
> can have
> cardinality aleph_0
using the standard definition or some private
definition that you refuse to share?
>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.
No, Call your set B. B consists of an infinite
number of sets B_i.
[Is it possible that you do not know the standard simple
contruction that shows that an infinite sequence of sequences
can be put in a sequence? ]
B_i consisists of a sequence and its metadiagonals.
Put this in a sequence C_i.
The C_i are no an infinite sequence of sequences,
Put this into a sequence, call it D
- William Hughes
William Hughes
> On 7 Okt., 03:47, Virgil <Vir...@gmale.com> wrote:
>
> > > That shows the inconsistency of these notions.
> > > As I have proven there are sets of cardinality aleph_0 that cannot be
> > > given the form of a sequence.
>
> > Since having cardinality aleph_0 is defined as capable of being put into
> > a sequence, WM is playing the fool again.
>
> Consider a Cantor-list. Put its meta-diagonals in the list, for
> instance put one meta-diagonal between any pair of regular entries.
> Then you have a new list.
Correct, the procedure produces a list
> Repeat the same procedure and repeat and
> repeat ...
>
And you always have a list
> The set of all entries and all metadiagonals of the infinte sequence
> of sequences is
a list.
- William Hughes
Dik T. Winter
> On 7 Okt., 03:47, Virgil <Vir...@gmale.com> wrote:
>
> > > That shows the inconsistency of these notions.
> > > As I have proven there are sets of cardinality aleph_0 that cannot be
> > > given the form of a sequence.
> >
> > Since having cardinality aleph_0 is defined as capable of being put into
> > a sequence, WM is playing the fool again.
>
> Consider a Cantor-list. Put its meta-diagonals in the list, for
> instance put one meta-diagonal between any pair of regular entries.
> Then you have a new list. Repeat the same procedure and repeat and
> repeat ...
Yes, and each of the sets you get is incomplete.
> The set of all entries and all metadiagonals of the infinte sequence
> of sequences is countable but cannot be put in one sequence.
It can if it is countable. That is the *definition* of countable!
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
WM
that has meta-diagonals which are not in the list.
Regards, WM
Virgil
<f70f6e17-000a-43da...@u28g2000hsc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> I have shown that the standard definition is wrong.
Not to the satisfaction of anyone competent.
> A set can have cardinality aleph_0 and simultaneously cannot be put
> in a sequence.
Since having cardinality aleph_0 MEANS capable of being arranged into an
endless sequence, at least in mathematics, WM is off in his
anti-mathematical dreamland again.
>
> Such a set is, for instance, the set of all entries and all
> metadiagonals of an infinite sequence of sequences.
A countable set of countable sets is countable (of cardinality <=
aleph_0), and so is any finite iteration of such a countable set of
countable sets construction.
Virgil
<1e091648-b08d-46e4...@t41g2000hsc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Yes such a set can be put into one sequence quite easily.
Each list naturally has a natural number index in the sequence of lists,
and in each entry in a listed sequence naturally has an index in that
listed sequence, so each such entry is locatable by a double index in
the form of an ordered pair of natural number indices:
(list_index, entry_index), one for each entry in each list.
Then one can order them by the following rules:
(1) If m1 = m2 and n1 = n2 then (m1,n1) = (m2,n2)
(2) If M1 = M2 and n1 < n2 then (m1,n1) < (m2,n2)
(3) If m1 <> m2 and m1 + n1 < m2 + n2 then (m1,n1) < (m2,n2)
The resulting ordering puts all the aleph_0 entries of all the aleph_0
lists into a single sequence/list, as anyone with even minor
mathematical competence in mathematics can verify.
Thus WM falls flat on his face again by claiming falsehoods.
William Hughes
so you put these meta-diagonals in the list and
you get a new list with meta-diagonals, so you put
these meta-diagonals in the list, so you get a new
list with meta-diagonals.
At the last step there is a contradiction.
Guess What?
- William Hughes
Virgil
<52a615cf-2906-4725...@i76g2000hsf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Since your 'meta-diagonals' are not members of the list, they are
totally irrelevant anyway, as Cantor's argument is only about what are,
or are not, members of the list. They can, at most, give examples of
other non-members of the list, which reinforces Cantor's result.
WM
> >A set
>
> Well at least you have abandoned any pretence that
> you have shown something about R.
The set in question might well be R. It cannot be excluded.
>
> > can have
> > cardinality aleph_0
>
> using the standard definition or some private
> definition that you refuse to share?
>
A sequence of sequences contains only countably many elements.
> >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.
>
> No, Call your set B. B consists of an infinite
> number of sets B_i.
>
> [Is it possible that you do not know the standard simple
> contruction that shows that an infinite sequence of sequences
> can be put in a sequence? ]
>
> B_i consisists of a sequence and its metadiagonals.
> Put this in a sequence C_i.
> The C_i are no an infinite sequence of sequences,
> Put this into a sequence, call it D
and continue. At *every* stage there is a set of meta-diagonals that
is not in the sequence. In order to maintain the actual infinity of
set theory, you must assert that the complete sequence has no meta-
diagonals. Obviously set theorists believe so, obviously that is
nonsense.
Regards, WM
WM
No. At *every* step there is a contradiction of the assumption that
all meta-diagonals are in a sequence.
Regards, WM
WM
> In article <1e091648-b08d-46e4-9ce7-f374f3f7e...@t41g2000hsc.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 7 Okt., 03:47, Virgil <Vir...@gmale.com> wrote:
> >
> > > > That shows the inconsistency of these notions.
> > > > As I have proven there are sets of cardinality aleph_0 that cannot be
> > > > given the form of a sequence.
> > >
> > > Since having cardinality aleph_0 is defined as capable of being put into
> > > a sequence, WM is playing the fool again.
> >
> > Consider a Cantor-list. Put its meta-diagonals in the list, for
> > instance put one meta-diagonal between any pair of regular entries.
> > Then you have a new list. Repeat the same procedure and repeat and
> > repeat ...
>
> Yes, and each of the sets you get is incomplete.
So it is. Nevertheless all the elements (regular entries and meta-
diagonals) belong to a countable set.
>
> > The set of all entries and all metadiagonals of the infinte sequence
> > of sequences is countable but cannot be put in one sequence.
>
> It can if it is countable. That is the *definition* of countable!
That is like a definition of negative numbers as being positive
numbers.
Regards, WM
William Hughes
Nope, at each step you have a sequence C_i and its metadiagonals.
C_i does not contain the metadiagonals of C_i, but sequence
C_(i+1) does. Of course C_(i+1) does not contain its metadiagonals
but the sequence C_(i+2) does and so forth
At no step do you get a sequence that contains its own metadiagonals.
So what? At no step do you get a set of metadiagonals that are
not contained in a sequence. The only way you could get a
contradiction
is at the last step, where you cannot create a new sequence.
Virgil
<0f0637df-b046-45fc...@j22g2000hsf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 7 Okt., 13:44, William Hughes <wpihug...@hotmail.com> wrote:
>
> > >A set
> >
> > Well at least you have abandoned any pretence that
> > you have shown something about R.
>
> The set in question might well be R. It cannot be excluded.
If the set in question is to be countable, then R CAN be excluded.
> >
> > > can have
> > > cardinality aleph_0
> >
> > using the standard definition or some private
> > definition that you refuse to share?
> >
>
> A sequence of sequences contains only countably many elements.
But R cannot be so represented.
>
> > >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.
> >
> > No, Call your set B. B consists of an infinite
> > number of sets B_i.
> >
> > [Is it possible that you do not know the standard simple
> > contruction that shows that an infinite sequence of sequences
> > can be put in a sequence? ]
> >
> > B_i consisists of a sequence and its metadiagonals.
> > Put this in a sequence C_i.
> > The C_i are no an infinite sequence of sequences,
> > Put this into a sequence, call it D
>
> and continue. At *every* stage there is a set of meta-diagonals that
> is not in the sequence.
At NO stage does any meta_diagonal have to be listed (though some may
be) in the listing from which they were constructed, so that they are
all totally irrelvant to the only relevant issue:
Is an anti-diagonal ever listed in the list
from which it is constructed?
If "YES" then Cantor's proof fails.
If "NO" the Cantor's proof is valid, and the Set of all binary
sequences is NOT countable!.
> In order to maintain the actual infinity of
> set theory, you must assert that the complete sequence has no meta-
> diagonals.
On the contrary, WE do not have to make any such foolish assertion.
If WM chooses to claim it is at all relevant, it is he who makes foolish
assertions, such as the above.
William Hughes
However, at no stage is there a set of meta-diagonals that is
not in _some_ sequence. Not that the claim is that there
is some sequence which contains the meta-diagonals, not that
the meta-diagonals are contained in the sequence that created them.
You could only have
a "complete sequence" at the last step. Guess What?
> In order to maintain the actual infinity of
> set theory, you must assert that the complete sequence has no meta-
> diagonals.
Correct, since there is no such thing
as "the complete sequence" there is no
such thing as "the meta-diagonals of the complete sequence"
- William Hughes
WM
> > > At the last step there is a contradiction.
>
> > No. At *every* step there is a contradiction of the assumption that
> > all meta-diagonals are in a sequence.
>
> Nope, at each step you have a sequence C_i and its metadiagonals.
Yes it is, therefore your "Nope" is false.
> C_i does not contain the metadiagonals of C_i, but sequence
> C_(i+1) does. Of course C_(i+1) does not contain its metadiagonals
> but the sequence C_(i+2) does and so forth
Yes. But that does not help, because every C(n+m) creates an infinite
but countable set of meta-diagonals that is not in the list.
>
> At no step do you get a sequence that contains its own metadiagonals.
> So what? At no step do you get a set of metadiagonals that are
> not contained in a sequence. The only way you could get a
> contradiction
> is at the last step, where you cannot create a new sequence.
Wrong. There is no last step required to see that at *every* step
there are metadiagonals that are not in the sequence.
You can even construct a sequence by means of the cardinal number of
the set of metadiagonals that are not in their list:
aleph_0, aleph_0, aleph_0, aleph_0, aleph_0, aleph_0, ...
Obviously the limit is zero. A virtuoso treatment of mathematics.
Regards, WM
Virgil
<0f0637df-b046-45fc...@j22g2000hsf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 7 Okt., 13:44, William Hughes <wpihug...@hotmail.com> wrote:
>
> > >A set
> >
> > Well at least you have abandoned any pretence that
> > you have shown something about R.
>
> The set in question might well be R. It cannot be excluded.
> >
> > > can have
> > > cardinality aleph_0
> >
> > using the standard definition or some private
> > definition that you refuse to share?
> >
>
> A sequence of sequences contains only countably many elements.
But no sequence of sequences contains all reals.
>
> At *every* stage there is a set of meta-diagonals that
> is not in the sequence.
At no stage does any meta-diagonal have any relevance to the issue of
whether any list can list all reals.
Cantor's proof, as modified by others to apply to decimal expansions,
shows that no sequence of reals can list all reals.
Since none of the 'meta-diagonals' need be listed, they are irrelevant.
> In order to maintain the actual infinity of
> set theory, you must assert that the complete sequence has no meta-
> diagonals.
In the matter of whether a list lists infinitely many objects,
meta-diagonals are irrelevant.
> Obviously set theorists believe so, obviously that is nonsense.
It is a good deal saner and more logical a form of nonsense than the
contra-logical nonsense that WM purveys.
Virgil
<4e68f397-670f-4576...@17g2000hsk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> No. At *every* step there is a contradiction of the assumption that
> all meta-diagonals are in a sequence.
AS no one except possibly WM assumes that all meta-diagonals of a given
sequence are in THAT sequence, any such contradictions are all in WM's
Virgil
<c1c229ef-059b-452c...@k7g2000hsd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 7 Okt., 15:03, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > In article
> > <1e091648-b08d-46e4-9ce7-f374f3f7e...@t41g2000hsc.googlegroups.com> WM
> > <mueck...@rz.fh-augsburg.de> writes:
> > > On 7 Okt., 03:47, Virgil <Vir...@gmale.com> wrote:
> > >
> > > > > That shows the inconsistency of these notions.
> > > > > As I have proven there are sets of cardinality aleph_0 that cannot
> > be
> > > > > given the form of a sequence.
> > > >
> > > > Since having cardinality aleph_0 is defined as capable of being put
> > into
> > > > a sequence, WM is playing the fool again.
> > >
> > > Consider a Cantor-list. Put its meta-diagonals in the list, for
> > > instance put one meta-diagonal between any pair of regular entries.
> > > Then you have a new list. Repeat the same procedure and repeat and
> > > repeat ...
> >
> > Yes, and each of the sets you get is incomplete.
>
> So it is. Nevertheless all the elements (regular entries and meta-
> diagonals) belong to a countable set.
If you mean that interspersing the meta-diagaonals of a list with the
members of a list to form a new list may include members previously not
listed, so what?
Cantor does no say that there is any binary string which cannot be
listed, he only says that for any such list there will be at least one
such binary string not listed in that list.
To show that false, WM must produce a list containing as a listed member
its own anti-diagonal.
Since WM cannot do that, he keeps trying to bring in all sorts of
irrelevancies, such as his present concentrations on the irrelevant
"meta-diagonals".
If any meta-diagonal turns out not to be a listed member of the list,
that just confirms Cantor's theorem even more solidly.
> >
> > > The set of all entries and all metadiagonals of the infinte sequence
> > > of sequences is countable but cannot be put in one sequence.
> >
> > It can if it is countable. That is the *definition* of countable!
>
> That is like a definition of negative numbers as being positive
> numbers.
My definition of a set being countable is that there exists a surjection
from N to that set.
If yours is different, you should at least tell us what it is, because
according to my definition, you are dead wrong.
William Hughes
> On 7 Okt., 19:26, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > At the last step there is a contradiction.
>
> > > No. At *every* step there is a contradiction of the assumption that
> > > all meta-diagonals are in a sequence.
>
> > Nope, at each step you have a sequence C_i and its metadiagonals.
>
> Yes it is, therefore your "Nope" is false.
Lean to distinguish between
i: This set of meta-diagonals is not in a specific list
and
ii This set of meta-diagonals is not in any list.
You claim is that there is a set of meta-diagonals that
is not in any list.
>
> > C_i does not contain the metadiagonals of C_i, but sequence
> > C_(i+1) does. Of course C_(i+1) does not contain its metadiagonals
> > but the sequence C_(i+2) does and so forth
>
> Yes. But that does not help, because every C(n+m) creates an infinite
> but countable set of meta-diagonals that is not in the list.
True, these metadiagonals are not in C(n+m). So what?
C(n+m) does not create a set of meta-diagonals that is
not in any list.
>
>
>
> > At no step do you get a sequence that contains its own metadiagonals.
> > So what? At no step do you get a set of metadiagonals that are
> > not contained in a sequence. The only way you could get a
> > contradiction
> > is at the last step, where you cannot create a new sequence.
>
> Wrong. There is no last step required to see that at *every* step
> there are metadiagonals that are not in the sequence.
This is true, but hardly a problem. At *every* step the meta-
diagonals
are in _some_ list.
>
> You can even construct a sequence by means of the cardinal number of
> the set of metadiagonals that are not in their list:
Since we are looking for a set of metadiagonals that are not in _any_
list
this is irrelevant.
-William Hughes
Virgil
<0eb587bd-8881-4de6...@n38g2000prl.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 7 Okt., 19:26, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > At the last step there is a contradiction.
> >
> > > No. At *every* step there is a contradiction of the assumption that
> > > all meta-diagonals are in a sequence.
> >
> > Nope, at each step you have a sequence C_i and its metadiagonals.
>
> Yes it is, therefore your "Nope" is false.
>
> > C_i does not contain the metadiagonals of C_i, but sequence
> > C_(i+1) does. Of course C_(i+1) does not contain its metadiagonals
> > but the sequence C_(i+2) does and so forth
>
> Yes. But that does not help, because every C(n+m) creates an infinite
> but countable set of meta-diagonals that is not in the list.
One has no guarantee the a meta-diagonal is not contained in the list
from which it is constructed unless it is really like an anti-diagonal
in structure.
>
>
> >
> > At no step do you get a sequence that contains its own metadiagonals.
> > So what? At no step do you get a set of metadiagonals that are
> > not contained in a sequence. The only way you could get a
> > contradiction
> > is at the last step, where you cannot create a new sequence.
Does Cantor's diagonal proof claim that his anti-diagonal cannot be
listed in another list? NO! so that only the original list counts.
It is like a game in which Cantor challenges all comers to come up with
a listing of all binary sequences.
If you wish to play, YOU have one chance to find a list congtaining all
binary sequences, and if you fail, Cantor wins.
You have played and failed, as everyone must.
>
> Wrong. There is no last step required to see that at *every* step
> there are metadiagonals that are not in the sequence.
Each step is a "last" step, unless you find a list for which an
anti-diagaonal is actually listed in THAT list. Which it never is.
WM keeps trying to make out that the Cantor proof says things it does
not say.
When he finally realizes his error, he may finally be on the road to
sanity.
WM
> On Oct 7, 1:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
> However, at no stage is there a set of meta-diagonals that is
> not in _some_ sequence.
At every step there is a set of meta-diagonals that is not in our
sequence.
> Not that the claim is that there
> is some sequence which contains the meta-diagonals, not that
> the meta-diagonals are contained in the sequence that created them.
>
> You could only have
> a "complete sequence" at the last step. Guess What?
You can only have the complete sequence of natural numbers at a last
number???
>
> > In order to maintain the actual infinity of
> > set theory, you must assert that the complete sequence has no meta-
> > diagonals.
>
> Correct, since there is no such thing
> as "the complete sequence" there is no
> such thing as "the meta-diagonals of the complete sequence"
No? There is no complete infinite set? Not even N, for instance? Not
even N lists?
Regards, WM
WM
> On Oct 7, 2:12 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 7 Okt., 19:26, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > At the last step there is a contradiction.
>
> > > > No. At *every* step there is a contradiction of the assumption that
> > > > all meta-diagonals are in a sequence.
>
> > > Nope, at each step you have a sequence C_i and its metadiagonals.
>
> > Yes it is, therefore your "Nope" is false.
>
> Lean to distinguish between
>
> i: This set of meta-diagonals is not in a specific list
>
> and
>
> ii This set of meta-diagonals is not in any list.
>
> You claim is that there is a set of meta-diagonals that
> is not in any list.
I claim that not all meta-diagonals can be in in one list.
Nevertheless all belong to a countable set.
Cantor claims that not all real numers can be in one list.
Nevertheless all can belong to a countable set.
>
>
>
> > > C_i does not contain the metadiagonals of C_i, but sequence
> > > C_(i+1) does. Of course C_(i+1) does not contain its metadiagonals
> > > but the sequence C_(i+2) does and so forth
>
> > Yes. But that does not help, because every C(n+m) creates an infinite
> > but countable set of meta-diagonals that is not in the list.
>
> True, these metadiagonals are not in C(n+m). So what?
> C(n+m) does not create a set of meta-diagonals that is
> not in any list.
>
Cantor's diagonal argument does not create any anti-diagonal that is
not in any list.
>
>
> > > At no step do you get a sequence that contains its own metadiagonals.
> > > So what? At no step do you get a set of metadiagonals that are
> > > not contained in a sequence. The only way you could get a
> > > contradiction
> > > is at the last step, where you cannot create a new sequence.
>
> > Wrong. There is no last step required to see that at *every* step
> > there are metadiagonals that are not in the sequence.
>
> This is true, but hardly a problem. At *every* step the meta-
> diagonals
> are in _some_ list.
Same holds for real numbers. At every step all real numbers are in
some list. (There is no real number that can be excluded from every
list.)
>
>
>
> > You can even construct a sequence by means of the cardinal number of
> > the set of metadiagonals that are not in their list:
>
> Since we are looking for a set of metadiagonals that are not in _any_
> list
> this is irrelevant.
So you see that your recent arguing was false? However: Since we are
looking for real numbers that are not in any list (and cannot find
them), Cantor's argument is false.
Regards, WM
WM
> In article
> <0eb587bd-8881-4de6-840c-a7a6c9393...@n38g2000prl.googlegroups.com>,
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 7 Okt., 19:26, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > At the last step there is a contradiction.
>
> > > > No. At *every* step there is a contradiction of the assumption that
> > > > all meta-diagonals are in a sequence.
>
> > > Nope, at each step you have a sequence C_i and its metadiagonals.
>
> > Yes it is, therefore your "Nope" is false.
>
> > > C_i does not contain the metadiagonals of C_i, but sequence
> > > C_(i+1) does. Of course C_(i+1) does not contain its metadiagonals
> > > but the sequence C_(i+2) does and so forth
>
> > Yes. But that does not help, because every C(n+m) creates an infinite
> > but countable set of meta-diagonals that is not in the list.
>
> One has no guarantee the a meta-diagonal is not contained in the list
> from which it is constructed unless it is really like an anti-diagonal
> in structure.
>
But the probability is very low, for an infinite number of digits.
>
>
> > > At no step do you get a sequence that contains its own metadiagonals.
> > > So what? At no step do you get a set of metadiagonals that are
> > > not contained in a sequence. The only way you could get a
> > > contradiction
> > > is at the last step, where you cannot create a new sequence.
>
> Does Cantor's diagonal proof claim that his anti-diagonal cannot be
> listed in another list? NO! so that only the original list counts.
Please tell that William Hughes.
>
> It is like a game in which Cantor challenges all comers to come up with
> a listing of all binary sequences.
And I have revealed that this game has not the result Cantorists
assume.
>
> If you wish to play, YOU have one chance to find a list congtaining all
> binary sequences, and if you fail, Cantor wins.
You have the chance to find one list containing all the metadiagonals.
And if you fail, then it is clear that Cantor's game does not prove
the existence of some aleph > aleph_0.
>
> You have played and failed, as everyone must.
Intriguing how wooden-headed Cantorists can be. Try to see: Game is
over. There is no infinity larger than infinity. The emporer is naked.
Regards, WM
William Hughes
> On 7 Okt., 19:36, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Oct 7, 1:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > However, at no stage is there a set of meta-diagonals that is
> > not in _some_ sequence.
>
> At every step there is a set of meta-diagonals that is not in our
> sequence.
So what? Your claim is not that there is a set of metadiagonals
that is not in _our_ sequence. Your claim is that there is a set of
meta-diagonals
that is not in _some_ sequence.
- William Hughes
William Hughes
> On 7 Okt., 21:24, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
> > On Oct 7, 2:12 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 7 Okt., 19:26, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > > At the last step there is a contradiction.
>
> > > > > No. At *every* step there is a contradiction of the assumption that
> > > > > all meta-diagonals are in a sequence.
>
> > > > Nope, at each step you have a sequence C_i and its metadiagonals.
>
> > > Yes it is, therefore your "Nope" is false.
>
> > Lean to distinguish between
>
> > i: This set of meta-diagonals is not in a specific list
>
> > and
>
> > ii This set of meta-diagonals is not in any list.
>
> > You claim is that there is a set of meta-diagonals that
> > is not in any list.
>
> I claim that not all meta-diagonals can be in in one list.
"all meta diagonals" is a set of meta-diagonals.
not in one list = not in any list
So you claim that there is a set of meta-diagonals
that is not in any list.
- William Hughes
WM
I never said so. (I know that there is nothing that could not be in
some sequence.) All meta-diagonals are in a sequence of sequences,
hence, belong to a set of cardinal number aleph_0. That is my premise!
Hence every meta-diagonal is in some sequenceIt is impossible,
however, to put them in one sequence.
Every real number is in some sequence too.
Regards, WM
Virgil
<9cbb4ecb-b5e7-4e75...@p10g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 7 Okt., 19:36, William Hughes <wpihug...@hotmail.com> wrote:
> > On Oct 7, 1:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> > However, at no stage is there a set of meta-diagonals that is
> > not in _some_ sequence.
>
> At every step there is a set of meta-diagonals that is not in our
> sequence.
Anti-diagonals can not be listed in the list from which they are
constructed., but meta-diagonals may be.
>
> > Not that the claim is that there
> > is some sequence which contains the meta-diagonals, not that
> > the meta-diagonals are contained in the sequence that created them.
> >
> > You could only have
> > a "complete sequence" at the last step. Guess What?
>
> You can only have the complete sequence of natural numbers at a last
> number???
WM guessed wrong again! The correct answer is that there is no last step
at which any "completion" occurs.
> >
> > > In order to maintain the actual infinity of
> > > set theory, you must assert that the complete sequence has no meta-
> > > diagonals.
> >
> > Correct, since there is no such thing
> > as "the complete sequence" there is no
> > such thing as "the meta-diagonals of the complete sequence"
>
> No? There is no complete infinite set?
There are certainly infinite sets, but there is no such thing as an
infinite sequence being completed by having a last element.
So that any completion by WM's anti-mathematical standards is impossible.
WM
There are meta-diagonals that are not in that list that containing the
entries and the meta-diagonals of the previous lists.
>
> So you claim that there is a set of meta-diagonals
> that is not in any list.
Why should I?
Regards, WM
William Hughes
"All Metadiagonals" = the set of metadiagonals in the sequence of
sequences.
So your claim is that it is imposssible to put the set of
all metadiagonals from a seqence of sequences in one
sequence.
That is your claim is that there is a set of meta-diagonals
that is not in some sequence.
- William Hughes
Virgil
<21caaf4e-0891-4785...@k37g2000hsf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> I claim that not all meta-diagonals can be in in one list.
> Nevertheless all belong to a countable set.
Since one can easily have more that countably many lists with at least
potentially at least one distinct meta-diagonal from each, it is
difficult to see how.
> Cantor claims that not all real numers can be in one list.
> Nevertheless all can belong to a countable set.
You keep claiming that but never produce such a counting of the reals.
In mathematics, listability and countability are equivalent notions so
that a set cannot be one without being the other, nor fail to be one
without failing to be the other.
In WM's weird weird world they seem not to be equivalent, but WM has
never managed to make that alleged distinction clear to anyone but
himself.
WM has been asked repeatedly to explain *his* distinction between them
but seems determined to keep it secret.
Virgil
<21caaf4e-0891-4785...@k37g2000hsf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
>
> I claim that not all meta-diagonals can be in in one list.
> Nevertheless all belong to a countable set.
That is a claim that something is true and simultaneously false.
No wonder WM has so much trouble getting things straight.
Virgil
<5f502c55-f0b0-4791...@c36g2000prc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
WH seems a good deal more aware of it thatn you do.
> >
> > It is like a game in which Cantor challenges all comers to come up with
> > a listing of all binary sequences.
>
> And I have revealed that this game has not the result Cantorists
> assume.
Well, WM has never beaten Cantor at it. Nor has anyone else, or they
would have posted their win in headlines all over the world, just as
Wiles proof of FLT was headlined.
> >
> > If you wish to play, YOU have one chance to find a list congtaining all
> > binary sequences, and if you fail, Cantor wins.
>
> You have the chance to find one list containing all the metadiagonals.
A meta-diagonal's presence in a list for which it is a meta-diagonal is
irrelevant, and even less relevant re any other list.
> And if you fail, then it is clear that Cantor's game does not prove
> the existence of some aleph > aleph_0.
It does prove specifically the existence of a set, R, which satisfies
Cantor's definition of card(R) > card(N).
> >
> > You have played and failed, as everyone must.
>
> Intriguing how wooden-headed Cantorists can be. Try to see: Game is
> over.
The game IS over, and you lost!
> The emporer is naked.
And WM thinks he is the emperor.
Virgil
<42089daf-062d-4c35...@v30g2000hsa.googlegroups.com>,
William Hughes <wpih...@hotmail.com> wrote:
WM has it wrong, as usual. What Hughes claims is that no meta-diagonal
need be in the list that spawns it. And as every real will be a
meta-diagonal for SOME list, it is the set of all reals which Hughes is
claiming are not all in any one list, which is just a reiteration of the
Cantor based result.
Virgil
<44f539c3-c96a-4b3b...@e17g2000hsg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 7 Okt., 22:05, William Hughes <wpihug...@hotmail.com> wrote:
> > On Oct 7, 3:47 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 7 Okt., 19:36, William Hughes <wpihug...@hotmail.com> wrote:
> >
> > > > On Oct 7, 1:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > However, at no stage is there a set of meta-diagonals that is
> > > > not in _some_ sequence.
> >
> > > At every step there is a set of meta-diagonals that is not in our
> > > sequence.
> >
> > So what? Your claim is not that there is a set of metadiagonals
> > that is not in _our_ sequence. Your claim is that there is a set of
> > meta-diagonals
> > that is not in _some_ sequence.
>
> I never said so. (I know that there is nothing that could not be in
> some sequence.) All meta-diagonals are in a sequence of sequences,
> hence, belong to a set of cardinal number aleph_0. That is my premise!
And it is, as usual, wrong. There can be different sequences of
sequences which generate different sets of meta-diagonals, neither being
a subset of the other.
William Hughes
> On 7 Okt., 22:11, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
> > On Oct 7, 3:55 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > I claim that not all meta-diagonals can be in in one list.
>
A simple rephasing using the fact that
"all meta diagonals" is a set of meta-diagonals means that your claim
is:
I claim that there is a set of meta-diagonals that can not be in
one list.
euivalently
I claim that there is a set of meta-diagonals that is not in any
list.
- William Hughes
Virgil
<5d639d1d-7e0a-4580...@o40g2000prn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> There are meta-diagonals that are not in that list that containing the
> entries and the meta-diagonals of the previous lists.
You "meta-diagonals" are totally irrelevant to the validity of the
Cantor diagonal proof.
So that no matter how long you go on mumbling about them, that Cantor
proof will remain totally unaffected by your mumblings.
WM
So it is. And that is clear because, by definition, the meta-diagonals
of a list are not in the sequence represented by that list.
>
> That is your claim is that there is a set of meta-diagonals
> that is not in some sequence.
No. Every meta-diagonal is in *some* sequence. The meta-diagonals of a
list form a sequence. So they are in that sequence. Further you can
define as many sequences as you like, each of which may start with a
meta-diagonal of your choice. Same is true with every anti-diagonal of
the set of Cantor- lists.
Regards, WM
WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > I claim that not all meta-diagonals can be in in one list.
> > Nevertheless all belong to a countable set.
>
> Since one can easily have more that countably many lists with at least
> potentially at least one distinct meta-diagonal from each, it is
> difficult to see how.
You should look up how those lists have been constructed: Take some
list, include its meta-diagonals, take that list, include its meta-
diagonals, and so on. The set of them is countable.
>
> > Cantor claims that not all real numers can be in one list.
> > Nevertheless all can belong to a countable set.
>
> You keep claiming that but never produce such a counting of the reals.
I produce it by means of the binary tree. Look there.
>
> WM has been asked repeatedly to explain *his* distinction between them
> but seems determined to keep it secret.
I did it more than once, but you did not understand.
A set is countable, by definition, if it can be written in the form of
a list or sequence.
According to a theorem of set theory, a set has cardinal number
aleph_0, if its elements can be shown to be in a sequence of
sequences.
The meta-diagonals of a sequence can never be in that sequence.
Therefore it is impossible to have all meta-diagonals of my
construction in a single sequence (this sequence has meta-diagonals).
Nevertheless the set of all those meta-diagonals has cardinal number
aleph_0, because they all are in a sequence of seqeunces.
Regards, WM
WM
> In article
> <21caaf4e-0891-4785-88e2-c52292bd2...@k37g2000hsf.googlegroups.com>,
>
>
> > I claim that not all meta-diagonals can be in in one list.
> > Nevertheless all belong to a countable set.
>
> That is a claim that something is true and simultaneously false.
The claim for instance, of finished infinity (vollendete
Unendlichkeit).
> No wonder WM has so much trouble getting things straight.
But after all, I think I got it.
Regards, WM
William Hughes
<snip>
> > That is your claim is that there is a set of meta-diagonals
> > that is not in some sequence.
>
> No.
-
So it is not possible to put the _set_ of meta-diagonals
in "one sequence", but it is possible to put the _set_
of meta-diagonals in "some sequence" ?
- William Hughes
WM
There is no list that contains all meta-diagonals as its entries
(because it does not contain its own meta-diagonals).
Nevertheless the set of all meta-diagonals has cardinal number
aleph_0.
Therefore Cantor's proof does not prove the uncountability of
anything.
Regards, WM
Dik T. Winter
> On 7 Okt., 15:03, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > > Consider a Cantor-list. Put its meta-diagonals in the list, for
> > > instance put one meta-diagonal between any pair of regular entries.
> > > Then you have a new list. Repeat the same procedure and repeat and
> > > repeat ...
> >
> > Yes, and each of the sets you get is incomplete.
>
> So it is. Nevertheless all the elements (regular entries and meta-
Of course. Of every set each element also belongs to a countable set, take
for instance the singleton set with that element only.
> > > The set of all entries and all metadiagonals of the infinte sequence
> > > of sequences is countable but cannot be put in one sequence.
> >
> > It can if it is countable. That is the *definition* of countable!
>
> That is like a definition of negative numbers as being positive
> numbers.
What definition of "countable" do *you* use? Or do you not use *any*
definition?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
William Hughes
>
> There is no list that contains all meta-diagonals as its entries
> (because it does not contain its own meta-diagonals).
Correct. What does this have to do with B, the set of meta-diagonals
in a sequence of sequences? This is certainly not all possible
metadiagonals. My claim is that there is a list that contains
B.
- William Hughes
Dik T. Winter
...
> I claim that not all meta-diagonals can be in in one list.
> Nevertheless all belong to a countable set.
No, each time you add the newly created meta-diagonals you get a new list,
and you keep on extending. At *no* stage will the list contain all
meta-diagonals. So you have to show proof that they all belong to a countable
set.
> Cantor claims that not all real numers can be in one list.
> Nevertheless all can belong to a countable set.
You have to provide proof for that.
And you will fail, because the mathematical definition of countable is that
the elements can be in a single list.
> > True, these metadiagonals are not in C(n+m). So what?
> > C(n+m) does not create a set of meta-diagonals that is
> > not in any list.
> >
> Cantor's diagonal argument does not create any anti-diagonal that is
> not in any list.
Right. Each real can be put in a list, but on the other hand there is no
list that contains *all* reals.
> Same holds for real numbers. At every step all real numbers are in
> some list. (There is no real number that can be excluded from every
> list.)
Yes, but that is irrelevant.
> So you see that your recent arguing was false? However: Since we are
> looking for real numbers that are not in any list (and cannot find
> them), Cantor's argument is false.
Nobody except you is looking for such non-existent numbers. That not
all reals can be put in a single list does *not* mean that there is a
real that can not be put in any list. Arguing that the two mean the
same is like arguing that because there are no natural numbers that
can not be put in a set of two elements, *all* natural numbers can be
out in a set of two elements.
WM
> In article <c1c229ef-059b-452c-99af-005c396ce...@k7g2000hsd.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 7 Okt., 15:03, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> ...
> > > > Consider a Cantor-list. Put its meta-diagonals in the list, for
> > > > instance put one meta-diagonal between any pair of regular entries.
> > > > Then you have a new list. Repeat the same procedure and repeat and
> > > > repeat ...
> > >
> > > Yes, and each of the sets you get is incomplete.
> >
> > So it is. Nevertheless all the elements (regular entries and meta-
> > diagonals) belong to a countable set.
>
> Of course. Of every set each element also belongs to a countable set, take
> for instance the singleton set with that element only.
All meta-diagonals however belong to one and the same set that has
cardinal number aleph_0 but cannot be given the form of a single
sequence.
>
> > > > The set of all entries and all metadiagonals of the infinte sequence
> > > > of sequences is countable but cannot be put in one sequence.
> > >
> > > It can if it is countable. That is the *definition* of countable!
> >
> > That is like a definition of negative numbers as being positive
> > numbers.
>
> What definition of "countable" do *you* use? Or do you not use *any*
> definition?
A countable set can be given the form of a single sequence. A set of
cardinality aleph_0 * aleph_0 = aleph_0 is countable by a theorem of
set theory. But there are sets of cardinal number aleph_0 that cannot
be given the form of a single sequence. Therefore the notion of
countability is ill-defined. A set can be countable and uncountable
simultaneously.
Regards, WM
WM
> In article <21caaf4e-0891-4785-88e2-c52292bd2...@k37g2000hsf.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> ...
> > I claim that not all meta-diagonals can be in in one list.
> > Nevertheless all belong to a countable set.
>
> No, each time you add the newly created meta-diagonals you get a new list,
> and you keep on extending. At *no* stage will the list contain all
> meta-diagonals.
That is what I say.
> So you have to show proof that they all belong to a countable
> set.
There are as many meta-diagonals as list entries. (Each one starts at
the first digit of a line, with exception of the first line, similar
to the regular entries.) The number of list entries of an infinite
list of lists in aleph_0.
>
> > Cantor claims that not all real numers can be in one list.
> > Nevertheless all can belong to a countable set.
>
> You have to provide proof for that.
>
> And you will fail, because the mathematical definition of countable is that
> the elements can be in a single list.
That is an error.
Regards, WM
WM
> 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.
>
> > Same holds for real numbers. At every step all real numbers are in
> > some list. (There is no real number that can be excluded from every
> > list.)
>
> Yes, but that is irrelevant.
Why?
>
> > So you see that your recent arguing was false? However: Since we are
> > looking for real numbers that are not in any list (and cannot find
> > them), Cantor's argument is false.
>
> Nobody except you is looking for such non-existent numbers. That not
> all reals can be put in a single list does *not* mean that there is a
> real that can not be put in any list.
If all reals can be put in lists, then there must be an uncountable
number of lists. That, however, is impossible. Guess why.
Regards, WM
WM
No. It does not contain its own meta-diagonals.
Regards, WM
Dik T. Winter
> On 8 Okt., 14:48, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
...
> > > > Yes, and each of the sets you get is incomplete.
> > >
> > > So it is. Nevertheless all the elements (regular entries and meta-
> > > diagonals) belong to a countable set.
> >
> > Of course. Of every set each element also belongs to a countable set,
> > take for instance the singleton set with that element only.
>
> All meta-diagonals however belong to one and the same set that has
> cardinal number aleph_0 but cannot be given the form of a single
> sequence.
That is all meta-diagonals that you encounter belong to a single list,
indeed. But the meta-diagonals that belong to that single list you do
*not* encounter. So they are not in that single list. They again are
in a new list.
> > > > It can if it is countable. That is the *definition* of countable!
> > >
> > > That is like a definition of negative numbers as being positive
> > > numbers.
> >
> > What definition of "countable" do *you* use? Or do you not use *any*
> > definition?
>
> A countable set can be given the form of a single sequence. A set of
> cardinality aleph_0 * aleph_0 = aleph_0 is countable by a theorem of
> set theory. But there are sets of cardinal number aleph_0 that cannot
> be given the form of a single sequence.
You state so without proof. Provide a set of cardinal number aleph_0 that
cannot be given in that form.
Dik T. Winter
> On 8 Okt., 14:55, William Hughes <wpihug...@hotmail.com> wrote:
> > Correct. What does this have to do with B, the set of meta-diagonals
> > in a sequence of sequences? This is certainly not all possible
> > metadiagonals. My claim is that there is a list that contains
> > B.
>
> No. It does not contain its own meta-diagonals.
Why "no"? It does indeed not contain its own meta-diagonals. But they are
also not in B. And the claim was *not* that it did contain its own
meta-diagonals but only B.
Dik T. Winter
> 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. And
specifying "meta-diagonal" without specifying where it comes from is
nonsense.
> > > Same holds for real numbers. At every step all real numbers are in
> > > some list. (There is no real number that can be excluded from every
> > > list.)
> >
> > Yes, but that is irrelevant.
>
> Why?
Because from each set, be it countable or not, each element can be put in
a list. That is quite trivial.
> > > So you see that your recent arguing was false? However: Since we are
> > > looking for real numbers that are not in any list (and cannot find
> > > them), Cantor's argument is false.
> >
> > Nobody except you is looking for such non-existent numbers. That not
> > all reals can be put in a single list does *not* mean that there is a
> > real that can not be put in any list.
>
> If all reals can be put in lists, then there must be an uncountable
> number of lists. That, however, is impossible. Guess why.
There is an uncountable number of lists. Guess why.
Dik T. Winter
> On 8 Okt., 14:54, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> .com> WM <mueck...@rz.fh-augsburg.de> writes:
> > ...
> > > I claim that not all meta-diagonals can be in in one list.
> > > Nevertheless all belong to a countable set.
> >
> > No, each time you add the newly created meta-diagonals you get a new list,
> > and you keep on extending. At *no* stage will the list contain all
> > meta-diagonals.
>
> That is what I say.
>
> > So you have to show proof that they all belong to a countable
> > set.
>
> There are as many meta-diagonals as list entries. (Each one starts at
> the first digit of a line, with exception of the first line, similar
> to the regular entries.) The number of list entries of an infinite
> list of lists in aleph_0.
Yes, so you have a collection of entries that consists of the entries of
the lists and the meta-diagonals of that that you encounter during the
process. That collection is countable again. And it can be put in a
list. But again that collection does *not* contain all real numbers.
You can continue that way with a second set of aleph_0 lists, and still
are not complete. And you can go on and on, never getting more than
aleph_0 entries and never getting all the reals.
> > > Cantor claims that not all real numers can be in one list.
> > > Nevertheless all can belong to a countable set.
> >
> > You have to provide proof for that.
> >
> > And you will fail, because the mathematical definition of countable is
> > that the elements can be in a single list.
>
> That is an error.
A definition is *never* an error.
Daryl McCullough
> WM <muec...@rz.fh-augsburg.de> writes:
> > A countable set can be given the form of a single sequence. A set of
> > cardinality aleph_0 * aleph_0 = aleph_0 is countable by a theorem of
> > set theory. But there are sets of cardinal number aleph_0 that cannot
> > be given the form of a single sequence.
>
>You state so without proof.
It's not just a matter of proof. The *definition* of "having cardinality
aleph_0" is "having a bijection with the naturals", which is the *same*
as the definition of "can be given the form of a single sequence".
There are three different phrases that mean exactly the same thing:
(at least when S is an infinite set)
1. S is countable
2. S has cardinality of aleph_0
3. There is a sequence containing all the elements of S
WM is saying, in essence: There are countable sets that are not
countable.
--
Daryl McCullough
Ithaca, NY
William Hughes
The list that contains B is not B. My claim
is that there is a list, call it C, that contains B.
C does not contain the meta-diagonals of C, but I
do not claim this.
- William Hughes
WM
> On Oct 8, 9:39 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 8 Okt., 14:55, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > On Oct 8, 8:19 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > There is no list that contains all meta-diagonals as its entries
> > > > (because it does not contain its own meta-diagonals).
>
> > > Correct. What does this have to do with B, the set of meta-diagonals
> > > in a sequence of sequences? This is certainly not all possible
> > > metadiagonals. My claim is that there is a list that contains
> > > B.
>
> > No. It does not contain its own meta-diagonals.
>
> The list that contains B is not B. My claim
> is that there is a list, call it C, that contains B.
And my claim is that the set of meta-diagonals of all the lists
*including the meta-diagonals of C* (or whatever you construct to
prove B being in a list) has cardinality aleph_0 - but cannot be in a
list.
> C does not contain the meta-diagonals of C, but I
> do not claim this.
Then your construction is uninteresting with respect to the original
discussion. Further it is false. Your list C is a sub-list of my
infinite sequence of lists containing all meta-diagonals. Therefore C
cannot exist as a single list.
Regards, WM
WM
> > > A countable set can be given the form of a single sequence. A set of
> > > cardinality aleph_0 * aleph_0 = aleph_0 is countable by a theorem of
> > > set theory. But there are sets of cardinal number aleph_0 that cannot
> > > be given the form of a single sequence.
>
> >You state so without proof.
The set of meta-diagonals of all lists that you construct is countable
but cannot be put in a single list.
>
> It's not just a matter of proof. The *definition* of "having cardinality
> aleph_0" is "having a bijection with the naturals", which is the *same*
> as the definition of "can be given the form of a single sequence".
A related definition is that of a square without corners.
>
> There are three different phrases that mean exactly the same thing:
> (at least when S is an infinite set)
>
> 1. S is countable
> 2. S has cardinality of aleph_0
> 3. There is a sequence containing all the elements of S
Which sequence contains all meta-diagonals of all lists that can be
constructed? (Their entries and meta-diagonals are counatable.) All
those lists exist, they are denumerable.
>
> WM is saying, in essence: There are countable sets that are not
> countable.
In essence, I have shown that set theory contains an ill-defined
notion, namely a self-contradition.
Regards, WM
WM
> In article <2c3e3d41-9092-4489-9cdf-06d813bdd...@w1g2000prk.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 8 Okt., 14:48, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> ...
> > > > > Yes, and each of the sets you get is incomplete.
> > > >
> > > > So it is. Nevertheless all the elements (regular entries and meta-
> > > > diagonals) belong to a countable set.
> > >
> > > Of course. Of every set each element also belongs to a countable set,
> > > take for instance the singleton set with that element only.
> >
> > All meta-diagonals however belong to one and the same set that has
> > cardinal number aleph_0 but cannot be given the form of a single
> > sequence.
>
> That is all meta-diagonals that you encounter belong to a single list,
> indeed. But the meta-diagonals that belong to that single list you do
> *not* encounter. So they are not in that single list. They again are
> in a new list.
Correct. And that new list has also meta-diagonals. All metas are
countable but not in a single list.
>
> > > > > It can if it is countable. That is the *definition* of countable!
> > > >
> > > > That is like a definition of negative numbers as being positive
> > > > numbers.
> > >
> > > What definition of "countable" do *you* use? Or do you not use *any*
> > > definition?
> >
> > A countable set can be given the form of a single sequence. A set of
> > cardinality aleph_0 * aleph_0 = aleph_0 is countable by a theorem of
> > set theory. But there are sets of cardinal number aleph_0 that cannot
> > be given the form of a single sequence.
>
> You state so without proof. Provide a set of cardinal number aleph_0 that
> cannot be given in that form.
You just agreed to the result, see above.
Regards, WM
WM
> In article <23a85ab9-95b1-4bc6-8162-32afd7f4a...@a29g2000pra.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 8 Okt., 14:55, William Hughes <wpihug...@hotmail.com> wrote:
> ...
> > > Correct. What does this have to do with B, the set of meta-diagonals
> > > in a sequence of sequences? This is certainly not all possible
> > > metadiagonals. My claim is that there is a list that contains
> > > B.
> >
> > No. It does not contain its own meta-diagonals.
>
> Why "no"? It does indeed not contain its own meta-diagonals. But they are
> also not in B. And the claim was *not* that it did contain its own
> meta-diagonals but only B.
The claim was (and is): The set of all meta-diagonals that are in
lists are not in one list.
Regards, WM
WM
> 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. Lists do not exist in Platonist
heaven. Therefore the set of lists is countable. But if you should
disagree, then take all lists that ever will be constructed.
>
> > > > Same holds for real numbers. At every step all real numbers are in
> > > > some list. (There is no real number that can be excluded from every
> > > > list.)
> > >
> > > Yes, but that is irrelevant.
> >
> > Why?
>
> Because from each set, be it countable or not, each element can be put in
> a list. That is quite trivial.
and quite wrong.
>
> > > > So you see that your recent arguing was false? However: Since we are
> > > > looking for real numbers that are not in any list (and cannot find
> > > > them), Cantor's argument is false.
> > >
> > > Nobody except you is looking for such non-existent numbers. That not
> > > all reals can be put in a single list does *not* mean that there is a
> > > real that can not be put in any list.
> >
> > If all reals can be put in lists, then there must be an uncountable
> > number of lists. That, however, is impossible. Guess why.
>
> There is an uncountable number of lists.
There? Where can I find it?
Regards, WM
WM
> In article <c3a2d403-4e2d-4ba2-9e50-5b6d572d8...@s20g2000prd.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 8 Okt., 14:54, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > > In article <21caaf4e-0891-4785-88e2-c52292bd2...@k37g2000hsf.googlegroups=
> > .com> WM <mueck...@rz.fh-augsburg.de> writes:
> > > ...
> > > > I claim that not all meta-diagonals can be in in one list.
> > > > Nevertheless all belong to a countable set.
> > >
> > > No, each time you add the newly created meta-diagonals you get a new list,
> > > and you keep on extending. At *no* stage will the list contain all
> > > meta-diagonals.
> >
> > That is what I say.
> >
> > > So you have to show proof that they all belong to a countable
> > > set.
> >
> > There are as many meta-diagonals as list entries. (Each one starts at
> > the first digit of a line, with exception of the first line, similar
> > to the regular entries.) The number of list entries of an infinite
> > list of lists in aleph_0.
>
> Yes, so you have a collection of entries that consists of the entries of
> the lists and the meta-diagonals of that that you encounter during the
> process. That collection is countable again. And it can be put in a
> list. But again that collection does *not* contain all real numbers.
> You can continue that way with a second set of aleph_0 lists, and still
> are not complete. And you can go on and on, never getting more than
> aleph_0 entries and never getting all the reals.
Same is true with all meta-diagonals that will ever be constructed.
>
> > > > Cantor claims that not all real numers can be in one list.
> > > > Nevertheless all can belong to a countable set.
> > >
> > > You have to provide proof for that.
> > >
> > > And you will fail, because the mathematical definition of countable is
> > > that the elements can be in a single list.
> >
> > That is an error.
>
> A definition is *never* an error.
If you define A is B and A is not B, then this definition is an error.
If you define M is an empty set that is larger than the set of all
sets, then this definition is in error.
Regards, WM
Daryl McCullough
>> It's not just a matter of proof. The *definition* of "having cardinality
>> aleph_0" is "having a bijection with the naturals", which is the *same*
>> as the definition of "can be given the form of a single sequence".
>
>A related definition is that of a square without corners.
So now you are saying that there *are* no countable sets (in
the same sense that there are no squares without corners)?
You don't believe that the naturals are countable?
You are not making any sense whatsoever.
>> There are three different phrases that mean exactly the same thing:
>> (at least when S is an infinite set)
>>
>> 1. S is countable
>> 2. S has cardinality of aleph_0
>> 3. There is a sequence containing all the elements of S
>
>Which sequence contains all meta-diagonals of all lists that can be
>constructed?
There is no such sequence. That's why we say that the set of
all reals is uncountable. That's what "uncountable" means---a
set S is uncountable if there is no sequence containing every
element of S.
>> WM is saying, in essence: There are countable sets that are not
>> countable.
>
>In essence, I have shown that set theory contains an ill-defined
>notion, namely a self-contradition.
We've been through this many times before. You were asked to
find a contradiction from the axioms of ZF, and you were unable
to. What is contradictory is *your* understanding of set theory.
But your understanding of set theory is not standard set theory.
Once again, to say that a theory T is inconsistent is to say that
there is a finite sequence of statements such that each statement
is either an axiom of T, or follows from previous statements by
logical deduction, such that the last statement in the sequence
is of the form A&~A.
As far as anyone knows, there is no such sequence for ZF. So
your statements about set theory being inconsistent are just
blather and nonsense. You don't know what you are talking about.
Herbert Newman
>
> You are not making any sense whatsoever.
>
Really?! When was the first time you realized that?
Herb
WM
> WM says...
>
> >> It's not just a matter of proof. The *definition* of "having cardinality
> >> aleph_0" is "having a bijection with the naturals", which is the *same*
> >> as the definition of "can be given the form of a single sequence".
>
> >A related definition is that of a square without corners.
>
> So now you are saying that there *are* no countable sets (in
> the same sense that there are no squares without corners)?
> You don't believe that the naturals are countable?
Are you really too limited to understand? Or do you wilfully
misunderstand?
The notion countable is ill-defined.
>
> >> There are three different phrases that mean exactly the same thing:
> >> (at least when S is an infinite set)
>
> >> 1. S is countable
> >> 2. S has cardinality of aleph_0
> >> 3. There is a sequence containing all the elements of S
>
> >Which sequence contains all meta-diagonals of all lists that can be
> >constructed?
>
> There is no such sequence. That's why we say that the set of
> all reals is uncountable.
The set of meta-diagonals is in question.
> That's what "uncountable" means---a
> set S is uncountable if there is no sequence containing every
> element of S.
The set of meta-diagonals of all lists that can be constructed is
countable.
>
> >> WM is saying, in essence: There are countable sets that are not
> >> countable.
>
> >In essence, I have shown that set theory contains an ill-defined
> >notion, namely a self-contradition.
>
> We've been through this many times before. You were asked to
> find a contradiction from the axioms of ZF, and you were unable
> to.
I have given many. See for instance: Binary Tree and Pairs of Nodes
> What is contradictory is *your* understanding of set theory.
> But your understanding of set theory is not standard set theory.
Cantor's understanding of a set consisting of distinguishable elements
was ok. When it became clear that most real numbers cannot be
distuinguished, there was the first big contradiction, because two
numbers are not numbers unless they can be distinguished.
>
> Once again, to say that a theory T is inconsistent is to say that
> there is a finite sequence of statements such that each statement
> is either an axiom of T, or follows from previous statements by
> logical deduction, such that the last statement in the sequence
> is of the form A&~A.
>
> As far as anyone knows,
You mean a group of ignorants who cannot see that undistinguishable
numbers cannot be numbers?
> there is no such sequence for ZF. So
> your statements about set theory being inconsistent are just
> blather and nonsense. You don't know what you are talking about.
Try to improve your understanding of lists, constructable, meta-
diagonal and so on. Then you will understand at least why countable
and aleph_0 are not the same.
Regards, WM
Daryl McCullough
>On 8 Okt., 19:13, stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
>The notion countable is ill-defined.
On the contrary, it has a perfectly straight-forward definition:
A set S is countable if there is a function f whose domain is N
and whose image is all of S.
>> That's what "uncountable" means---a
>> set S is uncountable if there is no sequence containing every
>> element of S.
>
>The set of meta-diagonals of all lists that can be constructed is
>countable.
No, it is not. A countable set is one that can be put into
a list. That's what "countable" means.
>> We've been through this many times before. You were asked to
>> find a contradiction from the axioms of ZF, and you were unable
>> to.
>
>I have given many.
No, you haven't.
>See for instance: Binary Tree and Pairs of Nodes
That thread does not contain a sequence of statements
such that each statement is either an axiom of ZF or
follows from previous statements by the rules of logical
deduction.
You've never given a proof of a contradiction from the
axioms of ZF. I don't believe that you've ever given a
proof of anything in your life.
Why don't you start off with something simple? How about
proving that addition is associative, starting from the
axioms of Peano Arithmetic? Can you do that?
WM
> WM says...
>
> >On 8 Okt., 19:13, stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
> >The notion countable is ill-defined.
>
> On the contrary, it has a perfectly straight-forward definition:
> A set S is countable if there is a function f whose domain is N
> and whose image is all of S.
>
> >> That's what "uncountable" means---a
> >> set S is uncountable if there is no sequence containing every
> >> element of S.
>
> >The set of meta-diagonals of all lists that can be constructed is
> >countable.
>
> No, it is not. A countable set is one that can be put into
> a list. That's what "countable" means.
So the theorem aleph_0 * aleph_0 = aleph_0 is wrong?
>
> >> We've been through this many times before. You were asked to
> >> find a contradiction from the axioms of ZF, and you were unable
> >> to.
>
> >I have given many.
>
> No, you haven't.
Which sequence contains all meta-diagonals of all lists that can be
constructed? (Their entries and meta-diagonals are counatable.) All
those lists exist, they are denumerable.
>
> >See for instance: Binary Tree and Pairs of Nodes
>
> That thread does not contain a sequence of statements
> such that each statement is either an axiom of ZF or
> follows from previous statements by the rules of logical
> deduction.
>
No. It uses the mathematical foundations of real numbers.
>
> Why don't you start off with something simple? How about
> proving that addition is associative, starting from the
> axioms of Peano Arithmetic? Can you do that?
The foundations of mathematics are counting, adding and drawing lines.
To prove something from artificial axioms that have been culled from
those real foundations is nonsense. Addition is associative and
commutative as has been found by observation and can be proved by
experiment (abacus). That is fact. Fine for the axioms if they can
reproduce it. A deplorable mathematician who does not know it without
axioms. Superfluous to prove that. You see what foolish results you
get when some articfical axioms are stated that are not in agreement
with mathematics, like ZFC and the like. Numbers that cannot be
distinguished! Unbelievable that mathematicians are boasting to
believe in such nonsense. With respect to these results one could
translate ZFC as Zermelo-Fraenkel-Crimes.
Regards, WM
Regards, WM
Herbert Newman
>
> The notion countable is ill-defined.
>
Only in your dreamland. In mathematics /countable/ has a precise and proper
definition. (See http://en.wikipedia.org/wiki/Countable_set)
>
> Cantor's understanding of a set consisting of distinguishable elements
> was ok. When it became clear that most real numbers cannot be
> distinguished, there was the first big contradiction, because two
> numbers are not numbers unless they can be distinguished.
>
Huh?! Any two different real numbers are "distinguishable". For simplicity
let's consider two different real numbers a,b e ]0,1[. If [a]_i and [b]_i
are the i-th digits of the decimal representations of a and b resp. (which
do not end in all 9s) then there is an i e IN such that [a]_i =/= [b]_i.
With other words, for all a,b e ]0,1[:
if a =/= b then Ei e IN: [a]_i =/= [b]_i.
You see: a and b can be "distinguished" if a =/= b. I.e. there is an index
n where the decimal representations (not ending in all 9s) of a and b
differ.
Did you ever attend a lecture in math?
Herb
Daryl McCullough
>
>On 8 Okt., 20:18, stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
>> A countable set is one that can be put into
>> a list. That's what "countable" means.
>
>So the theorem aleph_0 * aleph_0 = aleph_0 is wrong?
No, that conclusion *follows* from the definition of "countable".
There are two different operations that you might be referring
to with the * in your statement "aleph_0 * aleph_0 = aleph_0".
What I think you probably mean is cardinal multiplication.
If A is a set with cardinality alpha, and B is a set with
cardinality beta, then AxB (the set of all pairs <x,y> such
that x is an element of A and y is an element of B) has
cardinality alpha * beta. By this definition of cardinal
multiplication, it follows that aleph_0 * aleph_0 = aleph_0.
>> >> We've been through this many times before. You were asked to
>> >> find a contradiction from the axioms of ZF, and you were unable
>> >> to.
>>
>> >I have given many.
>>
>> No, you haven't.
>
>Which sequence contains all meta-diagonals of all lists that can be
>constructed?
By the fact that you've changed the subject, does that mean
that you agree, that you have never written down a sequence
of statements such that each statement is either an axiom of ZF,
or follows from previous statements by logical deduction?
>> >See for instance: Binary Tree and Pairs of Nodes
>>
>> That thread does not contain a sequence of statements
>> such that each statement is either an axiom of ZF or
>> follows from previous statements by the rules of logical
>> deduction.
>>
>No.
So you agree that you've never proved that ZF is inconsistent?