A consideration concerning the diagonal argument of G. Cantor
Albrecht
if there is an infinite sequence (list) of all rational numbers in [0,
1] and an antidiagonal number of this list (according to G. Cantor's
diagonal argument) and this antidiagonal number is a rational number
in [0, 1]
would destroy this fact the diagonal argument of G. Cantor which
proves that the set of reals is more infinite than the set of rational
numbers?
If you comprehend german language you may consult my little homepage,
especially the page "Weiterungen":
http://www.albrecht-storz.homepage.t-online.de/
I will provide an english version, but this may need a little bit
time.
But my general question may be considered without these details.
Thank you and with kind regards
Albrecht S. Storz
Baphomet
--
I bring the chain
1% - give to law what law gives to me
1% - give to law what law gives to me
1% - give to law what law gives to me
1% - give to law what law gives to me
narcs, chain breakers are not LAW
neither are jocks or dainties
in fact narcs are at many odds with law
repeat, I am not talking about giving narcs much
let 1%(not narcs) eat what they narc on, unless they are femme,
then give them as much string as they want, the land of corn is purse-i-
on, Persian, the chains can take care of purse string problems,
much more fun to do it the hardest way
give narcs the length of string yu can afford
and the dainties then jocks will fall
rises should happen before raises
?-The Devil
?kid? - Dale
prospect? - Dale Kelly
healer? - Dale Richard Kelly
soothsayer? - Dale Kelly
seer? - Mr. Kelly
magi? - Don Dale Kelly
gaul(one) - Dale Kelly
celt (oNe) - O'Cellaigh
druID (oNe) - THE DRUID
druId (oNe) - Dale
tripper (OnE) - Dale Kelly
trip dealer (OnE) - Satan
tripper wheeler (OnE) - Dale
wheeler (OnE) - The Rider
billy' GOAT (One)- Beelzubub
hilly' GOAT (One) - Leader Pagan's MC
SEEKER (ONE) - Baphomet (Captain) Pagans MC
1% - give to law what law gives to me
1% - give to law what law gives to me
1% - give to law what law gives to me
1% - give to law what law gives to me
http://sheilafeistyshy.blogspot.com/
http://barphomet.blog.com/
Aatu Koskensilta
> if there is an infinite sequence (list) of all rational numbers in [0,
> 1] and an antidiagonal number of this list (according to G. Cantor's
> diagonal argument) and this antidiagonal number is a rational number
> in [0, 1]
What makes you think it's rational? This is boring old claptrap. On a
more constructive note, if it is absolutely necessary to post inane
rubbish in sci.logic and sci.math, could one not at least put in the
effort to come up with some novel inane rubbish?
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Albrecht
Aatu Koskensilta schrieb:
> On 2008-01-26, in sci.logic, Albrecht wrote:
> > if there is an infinite sequence (list) of all rational numbers in [0,
> > 1] and an antidiagonal number of this list (according to G. Cantor's
> > diagonal argument) and this antidiagonal number is a rational number
> > in [0, 1]
>
> What makes you think it's rational?
I posed a question which is interesting for me and, may be, some other
people. The question is: if there would be a list of all rational
numbers, e.g. in [0, 1], and an antidiagonal number of this list in
[0, 1], would this fact destroys Cantors diagonal argument?
If you are not interested in this question, why do you have to answer?
Do you react to all posts in sci.math and sci.logic which are rubbish
in your opinion?
What makes you think, a number is not rational?
> This is boring old claptrap. On a
> more constructive note, if it is absolutely necessary to post inane
> rubbish in sci.logic and sci.math, could one not at least put in the
> effort to come up with some novel inane rubbish?
>
> --
> Aatu Koskensilta (aatu.kos...@xortec.fi)
>
> "Wovon man nicht sprechen kann, daruber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
So, why don't you?
Regards
Albrecht S. Storz
Aatu Koskensilta
> I posed a question which is interesting for me and, may be, some other
> people. The question is: if there would be a list of all rational
> numbers, e.g. in [0, 1], and an antidiagonal number of this list in
> [0, 1], would this fact destroys Cantors diagonal argument?
No.
> Do you react to all posts in sci.math and sci.logic which are rubbish
> in your opinion?
Only to some. Do you think this time someone will have something novel
to say about your confused, tired tirades about the diagonal argument?
> What makes you think, a number is not rational?
Diagonalising a list of rationals provably results in an irrational.
WM
wrote:
> On 2008-01-26, in sci.logic, Albrecht wrote:
>
> > I posed a question which is interesting for me and, may be, some other
> > people. The question is: if there would be a list of all rational
> > numbers, e.g. in [0, 1], and an antidiagonal number of this list in
> > [0, 1], would this fact destroys Cantors diagonal argument?
>
> No.
>
> > Do you react to all posts in sci.math and sci.logic which are rubbish
> > in your opinion?
>
> Only to some. Do you think this time someone will have something novel
> to say about your confused, tired tirades about the diagonal argument?
>
> > What makes you think, a number is not rational?
>
> Diagonalising a list of rationals provably results in an irrational.
Counter example:
0.000...
0.1000...
0.11000...
0.111000...
...
is a list of rationals. Substituting 0 by 1 results in 0.111...
Regards, WM
lwa...@lausd.net
> Aatu Koskensilta schrieb:
> > What makes you think it's rational?
> Do you react to all posts in sci.math and sci.logic which are rubbish
> in your opinion?
The reason Aatu was offended by the OP's post is that in these
newsgroups, Cantor's theories are challenged very often. Those
who are not regular members, like the OP, may believe that he
is the first person to come up with such a challenge to Cantor,
but the regulars are tired of seeing arguments against Cantor,
especially when its the same argument repeated by several
different people.
Then again, the fact that so many people challenge Cantor at
all shows just how nonintuitive standard set theory is. The OP
is asking why the diagonal proof shows that card(N) < card(R),
but doesn't prove that card(N) < card(Q).
> > On a constructive note, if it is absolutely necessary to post inane
> > rubbish in sci.logic and sci.math, could one not at least put in the
> > effort to come up with some novel inane rubbish?
I believe that Aatu's use of the term "rubbish" here is a bit
too harsh. Typically, most adherents of a standard set
theory, such as ZFC, believe that standard set theory is
the only theory worth studying, and label all nonstandard
set theories as "rubbish."
The OP, by arguing that card(N) < card(Q), is probably
thinking about a set theory where proper subsets always
have strictly smaller cardinalities. This is, of course, not
provable in standard ZFC, but I don't believe that a set
theory (other than ZFC) in which all proper subsets have
strictly smaller cardinalities is "rubbish," that their own
set theory is the only possible theory and that all other
set theories are "rubbish."
I would never consider a post "rubbish" merely because it
raises the possibly of a set theory other than the one that
Cantor once imagined, but, as the OP is finding out, that
is the typical attitude of standard mathematicians.
Albrecht
Aatu Koskensilta schrieb:
> On 2008-01-26, in sci.logic, Albrecht wrote:
> > I posed a question which is interesting for me and, may be, some other
> > people. The question is: if there would be a list of all rational
> > numbers, e.g. in [0, 1], and an antidiagonal number of this list in
> > [0, 1], would this fact destroys Cantors diagonal argument?
>
> No.
Why do you think so. I'm interested in more than a individual vote.
>
> > Do you react to all posts in sci.math and sci.logic which are rubbish
> > in your opinion?
>
> Only to some. Do you think this time someone will have something novel
> to say about your confused, tired tirades about the diagonal argument?
>
> > What makes you think, a number is not rational?
>
> Diagonalising a list of rationals provably results in an irrational.
Wrong. All rationals have infinite extensions. Some have only infinite
extensions.
Best regards
Albrecht S. Storz
lwa...@lausd.net
> > Diagonalising a list of rationals provably results in an irrational.
> Counter example:
> 0.000...
> 0.1000...
> 0.11000...
> 0.111000...
> ...
> is a list of rationals. Substituting 0 by 1 results in 0.111...
If we assume that WM's list is in binary, then it's correct
that the diagonal is rational. Still, it is distinct from all
of the other numbers on the list, which is all the
diagonalization is supposed to accomplish.
More to the point would be if we switched the first two:
> 0.1000...
> 0.000...
> 0.11000...
> 0.111000...
> ...
Then the diagonal would be 0.0111... -- which equals 0.1,
which actually is on the list.
The usual solution is to group the digits in pairs:
> 0.(10)(00)...
> 0.(00)(00)...
> 0.(11)(00)(00)...
> 0.(11)(10)(00)...
> ...
and use only (01) or (10) on the diagonal. In this case,
we can use (01), so the diagonal number is 0.010101...,
which, while still rational (it's actually 1/3), will not be on
the list. This is the equivalent of working in quaternary
rather than binary, and only using the digits 1 and 2.
No matter what base the numbers are in, it still doesn't
include all the rationals, such as 0.01. What Aatu is
arguing is that if the list contains _all_ the rationals,
then the diagonal must be _irrational_.
Then again, I know that WM doesn't really want to prove
the rationals to be uncountable, since his earlier set
theories are based on the existence of only countably
many reals. So WM and the OP are actually attacking
Cantor from opposite sides:
WM: card(N) = card(Q) = card(R)
ZFC: card(N) = card(Q) < card(R)
AS: card(N) < card(Q) < card(R)
Albrecht
lwal...@lausd.net schrieb:
Thanks for your appeasing posting.
But you are wrong with your supposition, I might believe that card(N)
< card(Q). I think that the diagonal argument is not stringent. So I'm
analysing the problem from different viewpoints.
One of them is the approach to try to construct lists of rationals
which might have rational antidiagonals which should be part of the
list. So I question if there is an analog proof on rationals what
might mean that to the diagonal argument of G. Cantor.
Best regards
Albrecht S. Storz
>
Peter Webb
"Albrecht" <albs...@gmx.de> wrote in message
news:6545ba30-2eaf-409c...@u10g2000prn.googlegroups.com...
> Dear members of sci.math and sci.logic,
>
> if there is an infinite sequence (list) of all rational numbers in [0,
> 1] and an antidiagonal number of this list (according to G. Cantor's
> diagonal argument) and this antidiagonal number is a rational number
> in [0, 1]
>
> would destroy this fact the diagonal argument of G. Cantor which
> proves that the set of reals is more infinite than the set of rational
> numbers?
>
No, it would destroy the proof that the rationals are countable. It would
say nothing as to whether Reals are countable, because your argument would
concern only the rationals.
http://en.wikipedia.org/wiki/Countable provides a means of listing all
Rational numbers, which proves they are in fact countable. If you can apply
some version of Cantor's proof to the decimal representation of the Rational
numbers listed in this way, and it produces a Rational not on the list, then
we would have to chuck out a fair percentage of the mathematics done over
the last 100 years. That's not going to happen, but have a go anyway if you
want.
G. Frege
>
> Then again, the fact that so many people challenge Cantor at
> all shows just how nonintuitive standard set theory is.
>
First: "so many" is wrong, "some" is right.
Second: It does NOT show "how nonintuitive standard set theory is" but
that (1) some people are not able to comprehend simple mathematical
arguments and (2) that there are mathematical cranks out there who are
determined to falsify Cantor and "transfinite" set theory.
F.
--
E-mail: info<at>simple-line<dot>de
Mike Terry
news:6545ba30-2eaf-409c...@u10g2000prn.googlegroups.com...
>
> if there is an infinite sequence (list) of all rational numbers in [0,
> 1] and an antidiagonal number of this list (according to G. Cantor's
> diagonal argument)
..the antidiagonal number is not on the list, and so by your hypothesis is
irrational..
> and this antidiagonal number is a rational number
> in [0, 1]
..and by your additional hypothesis the antidiagonal number is also
rational..
>
> would destroy this fact the diagonal argument of G. Cantor which
> proves that the set of reals is more infinite than the set of rational
> numbers?
>
So you are asking "if there is [a certain number] that is both irrational
and rational, would this destroy Cantor's diagonal argument?".
A fascinating question!... (not)
Mike.
G. Frege
<aatu.kos...@xortec.fi> wrote:
To avoid misunderstandings.
>>
>> What makes you think, a number is not rational?
>>
> Diagonalising a list of [all] rationals provably results in an irrational.
> ~~~
F.
William Elliot
> On 2008-01-26, in sci.logic, Albrecht wrote:
> > if there is an infinite sequence (list) of all rational numbers in [0,
> > 1] and an antidiagonal number of this list (according to G. Cantor's
> > diagonal argument) and this antidiagonal number is a rational number
> > in [0, 1]
>
> What makes you think it's rational? This is boring old claptrap. On a
> more constructive note, if it is absolutely necessary to post inane
> rubbish in sci.logic and sci.math, could one not at least put in the
> effort to come up with some novel inane rubbish?
>
Oh come on Aatu, isn't my rubbish in the sci.logic thread "Second order
Logic" novel? Why I nary heard a peep from you, while for this
Baphomet neophyte you lavish upon him your 'irrational' observations.
WM
> On Jan 26, 1:50 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > Diagonalising a list of rationals provably results in an irrational.
> > Counter example:
> > 0.000...
> > 0.1000...
> > 0.11000...
> > 0.111000...
> > ...
> > is a list of rationals. Substituting 0 by 1 results in 0.111...
>
> If we assume that WM's list is in binary, then it's correct
> that the diagonal is rational.
In binaries the simple substitution would not be sufficient for the
argument. Let my list be in decimals. The diagonal is nevertheless
rational, namely 1/9.
> What Aatu is
> arguing is that if the list contains _all_ the rationals,
> then the diagonal must be _irrational_.
Then he should say so.
Regards, WM
Albrecht
Peter Webb schrieb:
> "Albrecht" <albs...@gmx.de> wrote in message
> news:6545ba30-2eaf-409c...@u10g2000prn.googlegroups.com...
> > Dear members of sci.math and sci.logic,
> >
> > if there is an infinite sequence (list) of all rational numbers in [0,
> > 1] and an antidiagonal number of this list (according to G. Cantor's
> > diagonal argument) and this antidiagonal number is a rational number
> > in [0, 1]
> >
> > would destroy this fact the diagonal argument of G. Cantor which
> > proves that the set of reals is more infinite than the set of rational
> > numbers?
> >
>
> No, it would destroy the proof that the rationals are countable. It would
> say nothing as to whether Reals are countable, because your argument would
> concern only the rationals.
I don't think so. A proof which shows that the rationals are
uncountable would distroy the diagonal argument - not the proof that
the rationals are countable, I think.
The diagonal argument of G. Cantor "measures with two measures": The
real numbers in Cantor's lists are actual inifinite in the sense of
being completely defined by given an infinite sequence of digits. The
real numbers symbolise their limits.
The list itself isn't actual infinite since it lacks of its limit.
E.g. the infinite list
0.011111...
0.101111...
0.110111...
0.111011...
0.111101...
...
doesn't contain its limit 0.111111111.... which is the antidiagonal of
the list. But it is thought that any number of the list _is_ the limit
of the infinite sum which the number respresents.
This small gap between diffenrent understandings of infinity leads to
the senseless view of different infinites, I think.
Best regards
Albrecht S. Storz
>
Dave Seaman
> Peter Webb schrieb:
>> "Albrecht" <albs...@gmx.de> wrote in message
>> news:6545ba30-2eaf-409c...@u10g2000prn.googlegroups.com...
>> > Dear members of sci.math and sci.logic,
>> >
>> > if there is an infinite sequence (list) of all rational numbers in [0,
>> > 1] and an antidiagonal number of this list (according to G. Cantor's
>> > diagonal argument) and this antidiagonal number is a rational number
>> > in [0, 1]
>> >
>> > would destroy this fact the diagonal argument of G. Cantor which
>> > proves that the set of reals is more infinite than the set of rational
>> > numbers?
>> >
>>
>> No, it would destroy the proof that the rationals are countable. It would
>> say nothing as to whether Reals are countable, because your argument would
>> concern only the rationals.
> I don't think so. A proof which shows that the rationals are
> uncountable would distroy the diagonal argument - not the proof that
> the rationals are countable, I think.
A proof that the rationals are uncountable would demonstrate a contradiction
in mathematics, just the same as if you proved that 1+1=3. All of mathematics
would be affected.
> The diagonal argument of G. Cantor "measures with two measures": The
> real numbers in Cantor's lists are actual inifinite in the sense of
> being completely defined by given an infinite sequence of digits. The
> real numbers symbolise their limits.
> The list itself isn't actual infinite since it lacks of its limit.
The list is a mapping f : N -> R, where N is the natural numbers and R is
the reals. It is infinite because N is infinite.
> E.g. the infinite list
> 0.011111...
> 0.101111...
> 0.110111...
> 0.111011...
> 0.111101...
> ...
> doesn't contain its limit 0.111111111.... which is the antidiagonal of
> the list. But it is thought that any number of the list _is_ the limit
> of the infinite sum which the number respresents.
> This small gap between diffenrent understandings of infinity leads to
> the senseless view of different infinites, I think.
Those are not "different understandings of infinity". The notion of
limit is exactly the same in both cases.
Instead of getting bogged down in considerations of the real numbers,
which are only confusing you, you should look at the discussion of
Cantor's theorem at <http://en.wikipedia.org/wiki/Cantor's_theorem>,
which shows that P(N), the power set of the integers, is uncountable. In
order to understand this proof, you don't need to know anything about
real numbers or limits of decimal representations. If you can rephrase
your objections without resorting to talk of limits, it will greatly
simplify the discussion.
It turns out that P(N) has the same cardinality as R.
--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
Peter Webb
>
>
> Peter Webb schrieb:
>> "Albrecht" <albs...@gmx.de> wrote in message
>> news:6545ba30-2eaf-409c...@u10g2000prn.googlegroups.com...
>> > Dear members of sci.math and sci.logic,
>> >
>> > if there is an infinite sequence (list) of all rational numbers in [0,
>> > 1] and an antidiagonal number of this list (according to G. Cantor's
>> > diagonal argument) and this antidiagonal number is a rational number
>> > in [0, 1]
>> >
>> > would destroy this fact the diagonal argument of G. Cantor which
>> > proves that the set of reals is more infinite than the set of rational
>> > numbers?
>> >
>>
>> No, it would destroy the proof that the rationals are countable. It would
>> say nothing as to whether Reals are countable, because your argument
>> would
>> concern only the rationals.
>
> I don't think so. A proof which shows that the rationals are
> uncountable would distroy the diagonal argument - not the proof that
> the rationals are countable, I think.
>
A proof based on anything that proved the Rationals are uncountable would be
big news, as its very simple to show they are countable.
> The diagonal argument of G. Cantor "measures with two measures": The
> real numbers in Cantor's lists are actual inifinite in the sense of
> being completely defined by given an infinite sequence of digits. The
> real numbers symbolise their limits.
> The list itself isn't actual infinite since it lacks of its limit.
>
> E.g. the infinite list
>
> 0.011111...
> 0.101111...
> 0.110111...
> 0.111011...
> 0.111101...
> ...
>
> doesn't contain its limit 0.111111111.... which is the antidiagonal of
> the list. But it is thought that any number of the list _is_ the limit
> of the infinite sum which the number respresents.
>
No, when you use this notation:
0.0111...
with three dots ... after the 1, you are explicity saying you mean the sum
of the infinite series 1/4 + 1/8 + 1/16 ... the notation 0.0111... is a
shorthand for the actual infinite sum.
> This small gap between diffenrent understandings of infinity leads to
> the senseless view of different infinites, I think.
>
> Best regards
> Albrecht S. Storz
>
Well, good luck proving the Rationals are uncountable.
G. Frege
wrote:
>
> A proof which shows that the rationals are uncountable would distroy
> the diagonal argument [...]
>
Actually, it would "destroy" a good deal of mathematics, especially set
theory, since we can prove in our standard set theories that the
rationals are countable. Actually, the argument (that proves that the
rationals are countable) is quite simple/natural, hence it's hard to see
how to avoid the resulting /contradiction/. Indeed "A proof that the
rationals are uncountable would demonstrate a contradiction in
mathematics, just the same as if you proved that 1+1=3. All of
mathematics would be affected." (Dave Seaman)
F.
Daryl McCullough
>I posed a question which is interesting for me and, may be, some other
>people. The question is: if there would be a list of all rational
>numbers, e.g. in [0, 1], and an antidiagonal number of this list in
>[0, 1], would this fact destroys Cantors diagonal argument?
No. If that happens, you have proved both a statement and its negation.
I doubt that that is Cantor's fault. We would have to look at the
axioms you are using to figure out exactly which one is at fault.
If you have a list of all rationals, the the antidiagonal is *not*
a rational. The antidiagonal is constructed to be a number that is
*not* on the list (if you do it correctly, so as to avoid reals
with multiple decimal representations).
If you have a list of reals in [0,1], the numbers with multiple
representations all have decimal representations that end in all
0s or all 9s. To avoid these numbers, you can do one of several things:
1. Add all such numbers (in both representations) to your list.
Then diagonalization is guaranteed to produce a real that has only
one decimal representation.
2. Alternatively, when diagonalizing, always choose 1-8 as the
anti-diagonal, not 0 or 9.
3. Etc.
--
Daryl McCullough
Ithaca, NY
herbzet
Peter Webb wrote:
> "Albrecht" wrote
> >
> > Dear members of sci.math and sci.logic,
> >
> > if there is an infinite sequence (list) of all rational numbers in [0,
> > 1] and an antidiagonal number of this list (according to G. Cantor's
> > diagonal argument) and this antidiagonal number is a rational number
> > in [0, 1]
> >
> > would destroy this fact the diagonal argument of G. Cantor which
> > proves that the set of reals is more infinite than the set of rational
> > numbers?
> >
>
> No, it would destroy the proof that the rationals are countable. It would
> say nothing as to whether Reals are countable, because your argument would
> concern only the rationals.
Very good answer.
> http://en.wikipedia.org/wiki/Countable provides a means of listing all
> Rational numbers,
Easy to do, and there are yet easier ways than the one in Wikipedia.
A rational number is just an ordered pair <a, b> of integers, with
b =/= 0. Just systematically list every such ordered pair.
> which proves they are in fact countable.
Right.
> If you can apply
> some version of Cantor's proof to the decimal representation of the
> Rational numbers listed in this way, and it produces a Rational not
> on the list,
by hypothesis, all the rationals in [0, 1] are on the list.
> then
Either:
a) You have a reductio argument showing that the rationals are
not, in fact, countable, i.e, the hypothesis cannot be true,
or
b) The diagonal argument produces a number already on the list
of all rationals in [0,1] and thus does not produce what it
is designed to produce: a number not already on the list.
This latter situation would, as the OP supposes, throw doubt on
any proof that uses a similar diagonal argument.
So, what would be called for here is a listing of ... well,
whatever, and an application of the diagonal argument that
fails to produce an item not already on the list.
> we would have to chuck out a fair percentage of the mathematics done over
> the last 100 years. That's not going to happen, but have a go anyway if you
> want.
--
hz
herbzet
WM wrote:
> On 26 Jan., 23:58, lwal...@lausd.net wrote:
> > On Jan 26, 1:50 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > Aatu wroteL
> > > > Diagonalising a list of rationals provably results in an irrational.
> > > Counter example:
> > > 0.000...
> > > 0.1000...
> > > 0.11000...
> > > 0.111000...
> > > ...
> > > is a list of rationals. Substituting 0 by 1 results in 0.111...
> >
> > If we assume that WM's list is in binary, then it's correct
> > that the diagonal is rational.
>
> In binaries the simple substitution would not be sufficient for the
> argument. Let my list be in decimals. The diagonal is nevertheless
> rational, namely 1/9.
>
> > What Aatu is
> > arguing is that if the list contains _all_ the rationals,
> > then the diagonal must be _irrational_.
>
> Then he should say so.
Correct. It might have been more charitable to point out that
Aatu may have merely spoken with some slight ambiguity.
Personally, I assumed that by "a list of rationals" he meant,
in this context, "a list of all rationals".
Thanks for catching the ambiguous locution.
--
hz
herbzet
herbzet wrote:
> So, what would be called for here is a listing of ... well,
> whatever, and an application of the diagonal argument that
> fails to produce an item not already on the list.
Hard to see how this might be done, since a "diagonal argument"
is the production of an object which, by construction, is
different from each item already on a given list.
--
hz
WM
> On Sun, 27 Jan 2008 04:19:25 -0800 (PST), Albrecht wrote:
> > Peter Webb schrieb:
> >>news:6545ba30-2eaf-409c...@u10g2000prn.googlegroups.com...
> >> > Dear members of sci.math and sci.logic,
>
> >> > if there is an infinite sequence (list) of all rational numbers in [0,
> >> > 1] and an antidiagonal number of this list (according to G. Cantor's
> >> > diagonal argument) and this antidiagonal number is a rational number
> >> > in [0, 1]
>
> >> > would destroy this fact the diagonal argument of G. Cantor which
> >> > proves that the set of reals is more infinite than the set of rational
> >> > numbers?
>
> >> No, it would destroy the proof that the rationals are countable. It would
> >> say nothing as to whether Reals are countable, because your argument would
> >> concern only the rationals.
> > I don't think so. A proof which shows that the rationals are
> > uncountable would distroy the diagonal argument - not the proof that
> > the rationals are countable, I think.
>
> A proof that the rationals are uncountable would demonstrate a contradiction
> in mathematics, just the same as if you proved that 1+1=3. All of mathematics
> would be affected.
Not at all. It has been attempted during the last 100 years or so to
base mathematics on set theory. But in fact mathematics is completely
independent of set theory. It existed before and will remain after. In
mathematics there is at most one infinity. Everything else is
matheology.
Regards, WM
herbzet
in a given set. More general.
WM
This was Cantor's original opinion which however turned out wrong. A
simple example is well known:
Take the list of binaries
1.000...
0.1000...
0.11000...
0.111000...
...
When substituting 0 by 1 the diagonal 0.111... produces the first
entry. This case is always explicitly excluded (since the case has
been known). But why should this be the only ambiguity? Because 1000
mathematicians have not yet been able to figure out others? There are
more. You can see it by
0.121
0.11211
0.1112111
...
0.111...[n times]...1112111...[n times]...111
which has the same limit as
0.111...[n times]...1113111...[n times]...111.
lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
= lim {n-->oo} 0.111...[n times]...1113111...[n times]...111.
These are not real numbers? Why not?
Regards, WM
herbzet
WM wrote:
> herbzet wrote:
> > herbzet wrote:
Yes: for both sequences lim {n --> oo} = .1111 ... = 1/9.
> These are not real numbers? Why not?
Of course 1/9 is a real number. It's not, however, on either list.
You could put .1111 ... on either list, and both lists would still
have lim {n --> oo) = 1/9, but now 1/9 would be on both lists.
I'm afraid I'm missing your point.
--
hz
WM
That is understandable, because I was very short. My point is the
following:
lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
and
lim {n-->oo} 0.111...[n times]...1113111...[n times]...111
both are the real number 1/9 although they differ by one digit.
This shows that real numbers differing by one (or more) digit(s) are
not necessarily different in value, if infinities are involved. This
invalidates Cantor's argument.
Regards, WM
Tonico
> Regards, WM-
***************************************************
Herbzet, meet Mueckenheim, our resident german crank. The funniest
(!?) part of it all is that he's an actual teacher (??) in a rather
obscure, shy german "university".
He loves writing lots of 1's and before them he writes lim (n-->oo),
and for some reason he believes that makes some actual sense in
mathematics.
Oh, well: enjoy!
Regards
Tonio
Eric Schmidt
> lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
> and
> lim {n-->oo} 0.111...[n times]...1113111...[n times]...111
> both are the real number 1/9 although they differ by one digit.
>
> This shows that real numbers differing by one (or more) digit(s) are
> not necessarily different in value, if infinities are involved. This
> invalidates Cantor's argument.
No it doesn't show that at all. All you have are two diferent sequences
converging to 1/9. And Cantor's diagonal argument doesn't involve limits
at all. The *only* double representations of real numbers are found in
an infinite sequence of trailing 9's, equal to another representation
ending in 0's. (This obviously generalizes to bases other than 10.) In
particular, 0.111... is the only decimal representation of 1/9. In
Cantor's argument, the double representations can easily be avoided by
never producing a 0 or 9 on the anti-diagonal.
--
Eric Schmidt
--
Posted via a free Usenet account from http://www.teranews.com
herbzet
WM wrote:
> herbzet wrote:
> > WM wrote:
> > > herbzet wrote:
> > > > herbzet wrote:
> > > > > So, what would be called for here is a listing of ... well,
> > > > > whatever, and an application of the diagonal argument that
> > > > > fails to produce an item not already on the list.
> >
> > > > Hard to see how this might be done, since a "diagonal argument"
> > > > is the production of an object which, by construction, is
> > > > different from each item already on a given list.
Just want to call attention to the above point, so it doesn't
get lost.
???
In what digit do they differ?
Both are .1111 ... Where do they differ?
> This shows that real numbers differing by one (or more) digit(s) are
> not necessarily different in value, if infinities are involved. This
> invalidates Cantor's argument.
Not following.
--
hz
hagman
> Dear members of sci.math and sci.logic,
>
> if there is an infinite sequence (list) of all rational numbers in [0,
> 1] and an antidiagonal number of this list (according to G. Cantor's
> diagonal argument) and this antidiagonal number is a rational number
> in [0, 1]
>
> would destroy this fact the diagonal argument of G. Cantor which
> proves that the set of reals is more infinite than the set of rational
> numbers?
>
Definitely YES, you are totally right!
The reason is "ex falso quodlibet":
IF
there is an infinite sequence (list) of all rational numbers in [0,
1] and an antidiagonal number of this list (according to G. Cantor's
diagonal argument) and this antidiagonal number is a rational number
in [0, 1] (this is the "falso")
THEN
the diagonal argument would be destroyed (this is the "quodlibet")
You could also say
IF
(the same "falso")
THEN
pigs can fly.
> If you comprehend german language you may consult my little homepage,
> especially the page "Weiterungen":http://www.albrecht-storz.homepage.t-online.de/
> I will provide an english version, but this may need a little bit
> time.
BTW, you write
[...] Wahrscheinlichkeit dass dieser Punkt durch eine unangebbare
reelle Zahl bezeichnet werden würde 1 (es wäre sicher eine
unangebbare Zahl),
die Wahrscheinlich dass dieser Punkt durch eine angebbare reelle
Zahl
bezeichnet werden würde wäre 0 (es wäre sicher keine angebbare Zahl)
[...]
The correct statement uses "fast sicher" instead of "sicher" which
in this context simply means up to an exceptional set of (Lebesgue)
measure 0.
hagman
WM
> WM wrote:
> > lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
> > and
> > lim {n-->oo} 0.111...[n times]...1113111...[n times]...111
> > both are the real number 1/9 although they differ by one digit.
>
> > This shows that real numbers differing by one (or more) digit(s) are
> > not necessarily different in value, if infinities are involved. This
> > invalidates Cantor's argument.
>
> No it doesn't show that at all. All you have are two diferent sequences
> converging to 1/9. And Cantor's diagonal argument doesn't involve limits
> at all.
You are wrong. Cantor's argument involves "all natural numbers".
Therefore it is a limit argument. Otherwise the double representation
would not be a problem.
> The *only* double representations of real numbers are found in
> an infinite sequence of trailing 9's,
Here you object your own statement above.
> equal to another representation
> ending in 0's. (This obviously generalizes to bases other than 10.) In
> particular, 0.111... is the only decimal representation of 1/9. In
> Cantor's argument, the double representations can easily be avoided by
> never producing a 0 or 9 on the anti-diagonal.
But it shows that the argument "da wir sonst vor dem Widerspruch
stehen würden, daß ein Ding E0 sowohl Element von M, wie auch nicht
Elemente von M wäre" is wrong.
Regards, WM
WM
> WM wrote:
> > herbzet wrote:
> > > WM wrote:
> > > > herbzet wrote:
> > > > > herbzet wrote:
> > > > > > So, what would be called for here is a listing of ... well,
> > > > > > whatever, and an application of the diagonal argument that
> > > > > > fails to produce an item not already on the list.
>
> > > > > Hard to see how this might be done, since a "diagonal argument"
> > > > > is the production of an object which, by construction, is
> > > > > different from each item already on a given list.
>
> Just want to call attention to the above point, so it doesn't
> get lost.
We have seen that this argument is wrong by the identity of 1 and
0.999...
>
> > lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
> > and
> > lim {n-->oo} 0.111...[n times]...1113111...[n times]...111
> > both are the real number 1/9 although they differ by one digit.
>
> ???
>
> In what digit do they differ?
in 2 and 3, respectively.
>
> Both are .1111 ... Where do they differ?
Just there where lim {n-->oo} 10^-n and 0 differ.
Regards, WM
Aatu Koskensilta
> Those who are not regular members, like the OP, may believe that he
> is the first person to come up with such a challenge to Cantor, but
> the regulars are tired of seeing arguments against Cantor,
> especially when its the same argument repeated by several different
> people.
In case of Allbrecht it's the same arguments repeated by the same
person over and over for several years by now.
> Then again, the fact that so many people challenge Cantor at
> all shows just how nonintuitive standard set theory is. The OP
> is asking why the diagonal proof shows that card(N) < card(R),
> but doesn't prove that card(N) < card(Q).
Many mathematical notions and results often appear puzzling at first,
and questions such as why the diagonal argument does not show the
uncountability of the reationals, how there can be a countable model
of set theory, etc. naturally occur to students. Such questions are
answered by mathematical explanations, illustrations, explications,
practice and such like. With Allbrecht, Mueckenheim, and their ilk
we're not dealing with such natural puzzlement, and they are on the
whole whole not swayed in the least by any argument or explanation.
> I believe that Aatu's use of the term "rubbish" here is a bit
> too harsh. Typically, most adherents of a standard set
> theory, such as ZFC, believe that standard set theory is
> the only theory worth studying, and label all nonstandard
> set theories as "rubbish."
A peculiar idea. Quine's New Foundations, intuitionistic mathematics,
constructive type theory, ultra-finitism, ultra-intuitionism, Nelson's
predicative arithmetic, and a plethora of other non-mainstream
theories and conceptions of mathematics are not by any means
rubbish. Inane "refutations" of simple mathematical results, on the
other hand, are of no interest and are indeed rubbish.
> The OP, by arguing that card(N) < card(Q), is probably
> thinking about a set theory where proper subsets always
> have strictly smaller cardinalities.
Allbrecht has not presented any coherent theory or conception of the
world of sets, only boring babble about the supposed flaws in the
diagonal argument. Of course there's nothing to stop anyone from
presenting such a theory or conception, which might or might not be of
some interest -- repeating the same boring babble for years certainly
isn't.
> I would never consider a post "rubbish" merely because it
> raises the possibly of a set theory other than the one that
> Cantor once imagined, but, as the OP is finding out, that
> is the typical attitude of standard mathematicians.
You're imagining things. Allbrecht's post being rubbish is not due to
his presenting any alternative set theory or conception of the world
of sets, but rather because it consists of nothing but mathematical
idiocy.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Virgil
<396705d3-a831-48c2...@f10g2000hsf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
It is WM's own views of mathematics that are theological instead of
being merely logical.
What parts of mathematics does WM claim cannot be embedded in set theory?
Aatu Koskensilta
> Then again, the fact that so many people challenge Cantor at
> all shows just how nonintuitive standard set theory is.
Why? People challenge all sorts of perfectly clear stuff.
As to the supposed non-intuitiveness of standard set theory, I doubt
many people have any intuitions at all regarding the technical notions
of set theory. They might of course have all sorts of geometrical
intuitions, or intuitions based on density, or what not, about the
notion of "size". Whatever non-intuitiveness there is to the
elementary results of set theory probably stems from mistakenly
assuming the notion of cardinality corresponds to these intuitive
notions.
Virgil
<1f887394-00cf-4216...@l32g2000hse.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Because they are so ambiguous as to have no meaning.
You do not specify the lengths of the repeating sequences in
"0.111...[n times]...1112111...[n times]...111"
or in
0.111...[n times]...1113111...[n times]...111
so neither has any meaning at all.
Aatu Koskensilta
Here goes: if a list contains all the rationals then diagonalisation
necessarily results in an irrational.
Virgil
<738a944b-2993-4d0f...@i72g2000hsd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Actually, neither is a number at all as the alleged sequences of digits
are not well defined sequences of digits as described.
>
> This shows that real numbers differing by one (or more) digit(s) are
> not necessarily different in value, if infinities are involved.
One can prove that
(1) the set of binary sequences (not numbers) is uncountable.
(2) the set binary reals having multiple binary representations
is countable.
From which it follows that the set of binary numbers is uncountable.
> This
> invalidates Cantor's argument.
Only in WM's World. Elsewhere there are so many alternate proofs of
uncountability that WM's World is overwhelmed.
Virgil
<11e76fa3-3c17-4eb7...@j20g2000hsi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Jan., 19:05, Eric Schmidt <eric41...@comcast.net> wrote:
> > WM wrote:
> > > lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
> > > and
> > > lim {n-->oo} 0.111...[n times]...1113111...[n times]...111
> > > both are the real number 1/9 although they differ by one digit.
> >
> > > This shows that real numbers differing by one (or more) digit(s) are
> > > not necessarily different in value, if infinities are involved. This
> > > invalidates Cantor's argument.
> >
> > No it doesn't show that at all. All you have are two diferent sequences
> > converging to 1/9. And Cantor's diagonal argument doesn't involve limits
> > at all.
>
> You are wrong. Cantor's argument involves "all natural numbers".
> Therefore it is a limit argument. Otherwise the double representation
> would not be a problem.
What delta for what epsilon does the existence of a set of all naturals
imply?
WM's own arguments are too often the limit, but the infiniteness of a
set in no way requires any limits.
Virgil
<39971a6d-38e6-47f8...@u10g2000prn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
The LIMITs, if both exist, do not differ at all.
What WM is apparently arguing that there can only be one sequence
converging to a given limit, which stupidity is typical of him.
> >
> > Both are .1111 ... Where do they differ?
>
> Just there where lim {n-->oo} 10^-n and 0 differ.
Since lim {n-->oo} 10^-n = 0, they do not differ at all in value,
however much they may appear to differ in form.
Ross A. Finlayson
> In article
> <738a944b-2993-4d0f-8dd8-f0cdbfdf4...@i72g2000hsd.googlegroups.com>,
Let's say included among that list of purported results of
uncountability (of reals) that there are these:
antidiagonal argument: antidiagonal of list of continuum of reals is
real
powerset result of N to P(N): bijection of R to P(N), and Cantor-
Bernstein transitivity
nested intervals result: strictly monotone sequences converge to a
real from above and below
Besides those, or arguments using basically the same mechanisms as
those, which others would help comprise "so many"?
(That the antidiagonal argument doesn't follow in unary, binary,
ternary, or representation in an infinite alphabet, antidiagonal's off
the list; N e N nonstandardly of course; and the well-ordering of
reals leads to adjacency, are among arguments for reasonable
sophisticates that alternatives to not having a linear continuum
exist.)
Half of the integers are even.
In consideration of the two different sequences
.000<n-many zeroes>1<n-many zeroes>000
and
.000<n-many zeroes>1<n-many zeroes>000
that is,
.1000...
.0100...
.0010...
and
.0000....
.0000....
.0000....
In the limit the first sequence equals the second, as for no finite
difference a(n)-b(n) is there not a smaller finite difference a(n+m)-
b(n+m) between the two sequences, n >= 0, m > 0. For no element with
finite index n of the first sequence does it actually equal zero.
Anyways my question is as to how many is "so many"? Then, which are
those?
Thanks,
Ross
--
Finlayson Consulting
Albrecht
Peter Webb schrieb:
> "Albrecht" <albs...@gmx.de> wrote in message
> news:a793dcea-240d-4b0b...@s8g2000prg.googlegroups.com...
> >
> >
> > Peter Webb schrieb:
> >> "Albrecht" <albs...@gmx.de> wrote in message
> >> news:6545ba30-2eaf-409c...@u10g2000prn.googlegroups.com...
> >> > Dear members of sci.math and sci.logic,
> >> >
> >> > if there is an infinite sequence (list) of all rational numbers in [0,
> >> > 1] and an antidiagonal number of this list (according to G. Cantor's
> >> > diagonal argument) and this antidiagonal number is a rational number
> >> > in [0, 1]
> >> >
> >> > would destroy this fact the diagonal argument of G. Cantor which
> >> > proves that the set of reals is more infinite than the set of rational
> >> > numbers?
> >> >
> >>
> >> No, it would destroy the proof that the rationals are countable. It would
> >> say nothing as to whether Reals are countable, because your argument
> >> would
> >> concern only the rationals.
> >
> > I don't think so. A proof which shows that the rationals are
> > uncountable would distroy the diagonal argument - not the proof that
> > the rationals are countable, I think.
> >
>
> A proof based on anything that proved the Rationals are uncountable would be
> big news, as its very simple to show they are countable.
Yes, they are countable. And because this fact, a proof which proves
the rationals uncountable disprove this proof itself. And if this
proof works like the diagonal argument of G. Cantor, it would disprove
this proof alike.
>
>
> > The diagonal argument of G. Cantor "measures with two measures": The
> > real numbers in Cantor's lists are actual inifinite in the sense of
> > being completely defined by given an infinite sequence of digits. The
> > real numbers symbolise their limits.
> > The list itself isn't actual infinite since it lacks of its limit.
> >
> > E.g. the infinite list
> >
> > 0.011111...
> > 0.101111...
> > 0.110111...
> > 0.111011...
> > 0.111101...
> > ...
> >
> > doesn't contain its limit 0.111111111.... which is the antidiagonal of
> > the list. But it is thought that any number of the list _is_ the limit
> > of the infinite sum which the number respresents.
> >
>
> No, when you use this notation:
>
> 0.0111...
>
> with three dots ... after the 1, you are explicity saying you mean the sum
> of the infinite series 1/4 + 1/8 + 1/16 ... the notation 0.0111... is a
> shorthand for the actual infinite sum.
Exactly in the definition of infinite sums lays the problem. The
definition of the infinite series as its own limit introduces the
actual infinity in math.
In the usual using it doesn't made any problem. But in the diagonal
argument this definition leads to senseless results. "Senseless" in my
opinion since infinity is the acme of "bigger than all". And there is
no one "bigger than all" bigger than another "bigger than all".
>
>
>
> > This small gap between diffenrent understandings of infinity leads to
> > the senseless view of different infinites, I think.
> >
> > Best regards
> > Albrecht S. Storz
> >
>
> Well, good luck proving the Rationals are uncountable.
Thank you.
Best regards
Albrecht S. Storz
WM
wrote:
> Many mathematical notions and results often appear puzzling at first,
> and questions such as why the diagonal argument does not show the
> uncountability of the reationals, how there can be a countable model
> of set theory, etc. naturally occur to students. Such questions are
> answered by mathematical explanations, illustrations, explications,
> practice and such like.
How can you believe that your explanations, illustrations,
explications, practice (?) and such like could convince any sober mind
when you are unable to explain how you would try to order
indistinguishable elements but claim that you are able to do? Do you
know how often the word verschieden appears in Zermelos "proof" of
1904?
Regards, WM
Aatu Koskensilta
> explications, practice (?) and such like could convince any sober mind
> when you are unable to explain how you would try to order
> indistinguishable elements but claim that you are able to do?
As I have no idea what "ordering indistinguishable elements" means
I've no idea. As to Zermelo's well-ordering theorem, there's no
pretence that any actual human being could in any concrete sense carry
out any ordering. Indeed it's totally obscure what that would mean in
case of, say, the reals or the omega+omega'th level of the cumulative
hierarchy.
It's apparent you don't like the abstractions and notions introduced
in set theory. So why not go on about doing mathematics on whatever
basis you find congenial, instead of carping ineffectually on
perfectly clear mathematics on the grounds that it does not fit your
ideal of how we should conceive sets, functions, reals, what not?
WM
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Jan., 19:05, Eric Schmidt <eric41...@comcast.net> wrote:
> > > WM wrote:
> > > > lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
> > > > and
> > > > lim {n-->oo} 0.111...[n times]...1113111...[n times]...111
> > > > both are the real number 1/9 although they differ by one digit.
>
> > > > This shows that real numbers differing by one (or more) digit(s) are
> > > > not necessarily different in value, if infinities are involved. This
> > > > invalidates Cantor's argument.
>
> > > No it doesn't show that at all. All you have are two diferent sequences
> > > converging to 1/9. And Cantor's diagonal argument doesn't involve limits
> > > at all.
>
> > You are wrong. Cantor's argument involves "all natural numbers".
> > Therefore it is a limit argument. Otherwise the double representation
> > would not be a problem.
>
> What delta for what epsilon does the existence of a set of all naturals
> imply?
>
irrational. Every irrational number is a limit in binary or decimal
representation: lim {n-->oo} sum {1 to n} a_n*10^-n. Two numbers are
identical if the differ by lim {n-->oo} 10^-n. In a Cantor list the
diagonal can be formed by addition of 1 at the n-th place. In the
limit this yields an indistinguishable difference.
Regards, WM
WM
Correct, but the sequences of digits differ. That is why Cantor's
arguing fails.
Regards, WM
Virgil
<eb15e8e0-6a61-4860...@i72g2000hsd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
1904 was more than a century ago, and mthematics has not een entirely in
cold storage during that century, so it makes no matter now how often
"verschieden" appeared then.
Virgil
<75390ff8-8048-483c...@i7g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Jan., 20:28, Virgil <Vir...@com.com> wrote:
> > In article
> > <11e76fa3-3c17-4eb7-8382-b2035c60d...@j20g2000hsi.googlegroups.com>,
> >
> >
> >
> >
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 27 Jan., 19:05, Eric Schmidt <eric41...@comcast.net> wrote:
> > > > WM wrote:
> > > > > lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
> > > > > and
> > > > > lim {n-->oo} 0.111...[n times]...1113111...[n times]...111
> > > > > both are the real number 1/9 although they differ by one digit.
> >
> > > > > This shows that real numbers differing by one (or more) digit(s) are
> > > > > not necessarily different in value, if infinities are involved. This
> > > > > invalidates Cantor's argument.
> >
> > > > No it doesn't show that at all. All you have are two diferent sequences
> > > > converging to 1/9. And Cantor's diagonal argument doesn't involve limits
> > > > at all.
> >
> > > You are wrong. Cantor's argument involves "all natural numbers".
> > > Therefore it is a limit argument. Otherwise the double representation
> > > would not be a problem.
> >
> > What delta for what epsilon does the existence of a set of all naturals
> > imply?
> >
> If a Cantor list contains all reals, then its diagonal must be
> irrational.
If pigs had wings, they could fly.
> Every irrational number is a limit in binary or decimal
> representation: lim {n-->oo} sum {1 to n} a_n*10^-n.
Nope! Only those between 0 and 1, unless your a_n are allowed to take
other than 0 to (base - 1) digit values.
> Two numbers are
> identical if the differ by lim {n-->oo} 10^-n.
Since lim {n-->oo} 10^-n = 0, that says A = B if and only if A - B = 0?
> In a Cantor list the
> diagonal can be formed by addition of 1 at the n-th place.
No it can't. Among other problems, one cannot add 1 to 9 and get a
single decimal digit.
> In the
> limit this yields an indistinguishable difference.
Not in my limits.
WM
wrote:
> On 2008-01-27, in sci.logic, WM wrote:
>
> > How can you believe that your explanations, illustrations,
> > explications, practice (?) and such like could convince any sober mind
> > when you are unable to explain how you would try to order
> > indistinguishable elements but claim that you are able to do?
>
> As I have no idea what "ordering indistinguishable elements" means
> I've no idea. As to Zermelo's well-ordering theorem, there's no
> pretence that any actual human being could in any concrete sense carry
> out any ordering.
That's not the point. An ordering of countably many different elements
would already be impossible in practice. What I object to is the
alleged possibility to well-order an uncountable set of elements which
cannot be distinguished in principle because there are only countably
many labels.
So you and your ilk say: for every subset of the uncountable set I can
find an element which is different from all the other elements. I call
this the first element of that subset. But you know that in most cases
it is impossible to distinguish this element from most of the other
elements. I do not accept that.
>
> It's apparent you don't like the abstractions and notions introduced
> in set theory.
I don't like explicit contradictions as that one explained above.
> So why not go on about doing mathematics on whatever
> basis you find congenial,
That's what I am doing and what most others are doing too. It is
absolutely unnecessary to have transfinite set theory for doing
mathematics. As a philantropist however, I would like to prevent as
many students as possible from getting trapped by this kind of useless
mathelogy.
Regards, WM
Virgil
<fbe35112-e133-4486...@j78g2000hsd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
WRONG, since in any listing of reals in non-terminating decimal format,
no number is listed in which any digit is allowed to move around within
that number, as you argument requires.
One can still construct for any list of proper non-terminating decimals
another which was not listed in it.
From which it follows that, for any listing of such numbers, there are
at least as many unlisted as listed
Virgil
Aatu Koskensilta <aatu.kos...@xortec.fi> wrote:
> On 2008-01-27, in sci.logic, WM wrote:
> > On 26 Jan., 23:58, lwal...@lausd.net wrote:
> >
> >> What Aatu is
> >> arguing is that if the list contains _all_ the rationals,
> >> then the diagonal must be _irrational_.
> >
> > Then he should say so.
>
> Here goes: if a list contains all the rationals then diagonalisation
> necessarily results in an irrational.
A natural, obvious and inevitable corollary of the trivial theorem:
If set A contains all the rationals,
then any number not in A is irrational.
But WM is notoriously unable to see what is natural, what is obvious, or
what is inevitable in mathematics.
lwa...@lausd.net
Let's see what's going on here.
The two numbers with which WM are dealing with are:
a = lim n->oo (1/9 + 1/10^n) for 0.111...1112111...111...
and
b = lim n->oo (1/9 + 2/10^n) for 0.111...1113111...111...
In standard analysis, we prove that a = b = 1/9.
Now what WM is arguing is that suppose that a appears
on a list of reals, and in the process of creating a
Cantor diagonal, the digit "2" is changed to the
digit "3," and the diagonal number is actually b. So
WM concludes that since a = b and a is on the list,
we have that b is on the list. Thus the diagonal
number is not always distinct from the numbers on the
list, and so Cantor's argument would fail.
Why isn't WM's reasoning valid in ZFC? It's because
there are infinitely many 1's between the decimal
point and the "2" in a. In order for the diagonal
to change the "2" to a "3," there must be infinitely
many rows in the list above the number a. But the
rows are indexed by the standard natural numbers,
so each row has only finitely many rows above it.
So in ZFC, if a, which is actually 1/9, is on the
list, then a diagonal can only change the "1"'s --
the diagonal can't reach the "2" at all. In order
for the diagonal to reach the "2," the rows would
have to be indexed by nonstandard natural numbers.
We know that WM wishes to prove card(N)=card(R),
which is false in ZFC. There are two ways to
have card(N)=card(R). One way is to deny that
certain standard reals actually exist -- and that
is what WM usually does with his pi*googleplex and
related arguments.
The other is to add nonstandard naturals, and that
is what WM is doing here. He's not denying that the
diagonal is actually a real number, but adding
extra rows to the list.
G. Frege
wrote:
>>
>> Diagonalising a list of rationals provably results in an irrational.
>>
> Wrong. [bla bla bla]
>
Ok, Aatu omitted a crucial word here. Of course he meant
Diagonalising a list of _all_ rationals provably results
in an irrational.
(Modulo some technical details not essential to the argument as such.)
F.
Aatu Koskensilta
> We know that WM wishes to prove card(N)=card(R),
> which is false in ZFC. There are two ways to
> have card(N)=card(R). One way is to deny that
> certain standard reals actually exist -- and that
> is what WM usually does with his pi*googleplex and
> related arguments.
Could you elaborate on this "way"? Constructive reals are
constructively uncountable, for example, and it seems difficult to
envisage any conception of mathematics on which the reals are
bijectable with the naturals without equivocating, e.g. taking 'real'
to mean 'recursive real' in context of classical reasoning. The
diagonal argument relies on very few and weak principles, and is valid
on all the usual alternative conceptions of mathematics -- how we
choose to interpret what it establishes is another matter of course.
> The other is to add nonstandard naturals, and that
> is what WM is doing here.
What in Mueckenheim's outpouring makes you think he's doing
non-standard analysis or considering non-standard naturals? In any
case, non-standard naturals, non-standard analysis and such like have
nothing to do with the diagonal argument, and it's puzzling to see any
observation regarding them presented as supporting the idea that
there's something wrong with standard set theoretical results,
especially given that as usually conceived they rely on ordinary set
theoretic understanding and machinery.
As already said anyone is of course at liberty to find standard set
theory meaningless, unfounded, ill-justified, repugnant, removed from
the real world, a fantasy, and so on, and go on about doing
mathematics on a basis they find more congenial, happily ignoring any
results derived from principles one finds unacceptable. Marrying this
activity with idiotic "refutations" of the standard conception and
results is pointless.
Aatu Koskensilta
> That's not the point. An ordering of countably many different elements
> would already be impossible in practice. What I object to is the
> alleged possibility to well-order an uncountable set of elements which
> cannot be distinguished in principle because there are only countably
> many labels.
These worries of yours have nothing to do with set theory as usually
understood. They will be effective in dissuading a poor soul from
engaging in the fantasy game of set theory only in case of those who,
for some reason, share your ideas about labels and what not; these
concepts do not enter into any set theoretic arguments in ordinary
mathematics.
> As a philantropist however, I would like to prevent as many students
> as possible from getting trapped by this kind of useless mathelogy.
Yours is a noble calling, protecting the innocent lest they be led
astray! Alas, your mode of presentation and argumentation are not
likely to appeal to but a very few.
Virgil
<f02ca39a-c099-4b3e...@i72g2000hsd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Jan., 22:00, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> wrote:
> > On 2008-01-27, in sci.logic, WM wrote:
> >
> > > How can you believe that your explanations, illustrations,
> > > explications, practice (?) and such like could convince any sober mind
> > > when you are unable to explain how you would try to order
> > > indistinguishable elements but claim that you are able to do?
> >
> > As I have no idea what "ordering indistinguishable elements" means
> > I've no idea. As to Zermelo's well-ordering theorem, there's no
> > pretence that any actual human being could in any concrete sense carry
> > out any ordering.
>
> That's not the point. An ordering of countably many different elements
> would already be impossible in practice. What I object to is the
> alleged possibility to well-order an uncountable set of elements which
> cannot be distinguished in principle because there are only countably
> many labels.
Lots of people have trouble with requiring every set to be well
orderable. So there is no such requirement. The axiom of choice is
optional.
>
> So you and your ilk say: for every subset of the uncountable set I can
> find an element which is different from all the other elements. I call
> this the first element of that subset.
According to what ordering principle?
> But you know that in most cases
> it is impossible to distinguish this element from most of the other
> elements. I do not accept that.
The issue is not whether a single comparison can distinguish one real
simultaneously from all other reals, but whether it is capable of
distinguishing one real from another real.
> >
> > It's apparent you don't like the abstractions and notions introduced
> > in set theory.
>
> I don't like explicit contradictions as that one explained above.
>
> > So why not go on about doing mathematics on whatever
> > basis you find congenial,
>
> That's what I am doing and what most others are doing too. It is
> absolutely unnecessary to have transfinite set theory for doing
> mathematics.
That depends entirely on the sort of mathematics being done.
As a philantropist however, I would like to prevent as
> many students as possible from getting trapped by this kind of useless
> mathelogy.
And as philanthropists, we would like to prevent your sort of egoism
from infecting the innocent.
>
> Regards, WM
Aatu Koskensilta
> Ok, Aatu omitted a crucial word here. Of course he meant
>
> Diagonalising a list of _all_ rationals provably results
> in an irrational.
That's indeed what I meant, but the word I omitted was "the":
Diagonalising a list of [the] rationals results in an irrational.
Aatu Koskensilta
> Lots of people have trouble with requiring every set to be well
> orderable. So there is no such requirement. The axiom of choice is
> optional.
The axiom of choice has historically been subject to debate, but these
days it's an accepted principle in ordinary mathematics, in the sense
that people will happily accept a statement P proved even if the proof
appeals to choice, and not merely say that it has been established
that the axiom of choice implies P. On the whole, it's only logicians
and set theorists who pay any attention to such details.
> And as philanthropists, we would like to prevent your sort of egoism
> from infecting the innocent.
The idea that the innocent need any protection from Mueckenheim's
twaddle is just as silly as his noble quest of exposing the
contradictions in set theory lest some poor soul be led astray, brains
rotting and oozing out of their nostrils.
lwa...@lausd.net
wrote:
> On 2008-01-27, in sci.logic, lwal...@lausd.net wrote:
> > We know that WM wishes to prove card(N)=card(R),
> > which is false in ZFC. There are two ways to
> > have card(N)=card(R). One way is to deny that
> > certain standard reals actually exist -- and that
> > is what WM usually does with his pi*googleplex and
> > related arguments.
> Could you elaborate on this "way"?
Well, one has to read some of WM's arguments from
over the summer to figure out exactly what his
argument was from back then.
But I recall his making several comments about the
natural number floor(pi*googleplex). The problem is
that no one knows whether this number is odd or even,
because no one knows its units digit. If we were to
change its unit digit to a 5, we can't tell whether
we've increased or decreased the number. So then, WM
concludes, the number pi doesn't exist -- and
neither would most transcendentals.
Of course, I'm also curious as to what WM would have
to say about numbers such as Moser's Number and
Graham's Number. Now we know several tricks with
"mod" that will determine the units digits of the
Moser and Graham numbers as "6" and "7" respectively,
but what about the _leftmost_ digits? If we were to
change the leftmost digits of either number to "5"
would we have increased or decreased the number?
Yet even Kronecker would have been forced to accept
the existence of the Moser and Graham numbers, since
they are natural numbers defined using nothing other
than operations on the natural numbers. What about WM?
G. Frege
WM:
" lim {n-->oo} 0.111...[n times]...1112111...[n times]...111
and
lim {n-->oo} 0.111...[n times]...1113111...[n times]...111
both are the real number 1/9 although they differ by one
digit."
An innocent reader: "In what digit do they differ?"
WM: "in 2 and 3, respectively."
Virgil: "The LIMITs, if both exist, do not differ at all."
WM: "Correct, but the sequences of digits differ. That is why Cantor's
arguing fails."
Virgil: "WRONG..."
Indeed. Actually, in decimal representation of 1/9 none of the digits is
2 and/or 3.
But remember, WM allows for digits in the decimal representation of real
numbers with an index >= w. (!)
[WM never realized that the domain of a _sequence_ usually
-and especially in this case- is required to be IN.]
With other words, he _seems_ to think that
lim {n-->oo} 0.0...01
`-.-´
n-times
= 0.000...1 ,
where the digit 1 occurs at the w-th position in the decimal
representation of lim {n-->oo} 0.0...01.
`-.-´
n-times
While WE -of course- would argue that
lim {n-->oo} 0.0...01
`-.-´
n-times
= lim {n-->oo} 1/10^(n+1) = 0 = 0.000...
Again, WM would claim that
lim {n-->oo} 0.0...01 = 0.000...1
`-.-´
n-times
differs from
0.000...(0)
in (the digit) 1.
"Though this be madness, yet there is method in't." (Hamlet)
F.
--
E-mail: info<at>simple-line<dot>de
G. Frege
<muec...@rz.fh-augsburg.de> wrote:
>>>>>
>>>>> lim {n-->oo} 0.111...[n times]...1112
>>>>>
>>>>> and
>>>>>
>>>>> lim {n-->oo} 0.111...[n times]...1113
>>>>>
>>>>> both are the real number 1/9 although they differ by one digit.
>>>>>
>>>> ??? In what digit do they differ?
>>>>
>>> in 2 and 3, respectively.
>>>
>> The LIMITs, if both exist, do not differ at all.
>>
> Correct, but the sequences of digits differ.
>
No, the sequents of digits (of those numbers) do NOT differ here, since
"both" numbers are one and the same number:
1/9 = lim {n-->oo} 0.111...[n times]...1112
and
1/9 = lim {n-->oo} 0.111...[n times]...1113.
You know, by law of IDENTITY 1/9 has the SAME decimal representation as
1/9.
And no, decimal representations of real numbers don't have indexes >= w.
>
> That is why Cantor's arguing fails.
>
That's why you are a mathematical crank.
G. Frege
>
> Again, WM would claim that
>
> lim {n-->oo} 0.0...01 = 0.000...1
> `-.-´
> n-times
> differs from
>
> 0.000...(0)
>
> in (the digit) 1.
>
>
Though he still would maintain that
0 = 0.000...(0) = 0.000...1.
Remember:
Virgil: "The LIMITs, if both exist, do not differ at all."
WM: "Correct, but the sequences of digits differ. That is why Cantor's
arguing fails."
>
>
Tonico
> On Mon, 28 Jan 2008 01:59:48 +0100, G. Frege <nomail@invalid> wrote:
>
> > Again, WM would claim that
>
> > lim {n-->oo} 0.0...01 = 0.000...1
> > `-.-´
> > n-times
> > differs from
>
> > 0.000...(0)
>
> > in (the digit) 1.
>
> Though he still would maintain that
>
> 0 = 0.000...(0) = 0.000...1.
>
> Remember:
>
> Virgil: "The LIMITs, if both exist, do not differ at all."
>
> WM: "Correct, but the sequences of digits differ. That is why Cantor's
> arguing fails."
>
...and over and over and over again. Mueckenheim, Albrecht, T. Orlow
and their ilk are, apparently, a lost cause: they just WON'T
understand, some by their own choice, and some by their own inability.
Again, for the sake of the new ones that may believe this is the first
time these guys crank around: none of them is a mathematician.
Regards
Tonio
Gc
wrote:
> On 2008-01-27, in sci.logic, Virgil wrote:
>
> > Lots of people have trouble with requiring every set to be well
> > orderable. So there is no such requirement. The axiom of choice is
> > optional.
>
> The axiom of choice has historically been subject to debate, but these
> days it's an accepted principle in ordinary mathematics, in the sense
> that people will happily accept a statement P proved even if the proof
> appeals to choice, and not merely say that it has been established
> that the axiom of choice implies P. On the whole, it's only logicians
> and set theorists who pay any attention to such details.
In FOM somebody said that because in algeibrac geometry Groethendick
universe in used so often, then Tarski proved the consistency of the
AC by introducting Tarski`s axiom. To me it seems that the thing is
not so easy that there are ordinary mathematicians who couldn`t care
less and if some thing are done many years lose all their objections
and in the other pole people how are interested in foundations. Some
mathematicians when they encounter things like really weird infinite
groups, just decide to stay away from them and choose to deal with
things they find nice.
Tonico
*********************************************************8
Uh?? What mathematician found, say some really weird infinite group
(like an ugly wreath product, or just an amalgamated free
product...how weird?), and then decided to "stay away" from it?
Where did she/he found it? In the street or while making some
mathematical work? And if it is the latter, how did she/he manage to
"stay away" from it?
Just give, for the fun of it, 1-2 examples, if you please.
Regards
Tonio
Gc
I said that decide not to start to study infinite groups. What is so
weird about this?
> Where did she/he found it? In the street or while making some
> mathematical work? And if it is the latter, how did she/he manage to
> "stay away" from it?
>
> Just give, for the fun of it, 1-2 examples, if you please.
No, I don`t. Your style of writing is insulting and you seem like an
asshole, so I don`t wan`t to continue discuss with you anymore.
Tonico
*********************************************
Wise decision.
And you did NOT write "decide not to start to study infinite groups",
but rather you wrote exactly "Some mathematicians when they encounter
things like really weird infinite groups, just decide to stay away
from them and choose to deal with things they find nice." , as
implying that when they're doing some mathematical work and they
"find" these things they better let go...Of course you may believe
these two things are one and the same. Ok.
Kisses
Tonio
WM
Of course. This is also the case in my example. n --> oo does not mean
that n gets a non-natural number but that the assertion is correct for
every natural number however large it may be.
You agree that irrational numbers are in Cantor's list, at least at
the diagonal. For irrational numbers you accept the limit
lim{n --> oo} sum(1 to n) a_n*10^-n. Changing one diagonal digit by (+
or -) 1 at position n yields a different number. But changing all
diagonal digits includes the limit lim{n --> oo} (+ or -) 10^-n.
Otherwise you have not changed every digit of an irrational (or
periodical rational) number but only an initial segment of digits.
>
> So in ZFC, if a, which is actually 1/9, is on the
> list, then a diagonal can only change the "1"'s --
> the diagonal can't reach the "2" at all. In order
> for the diagonal to reach the "2," the rows would
> have to be indexed by nonstandard natural numbers.
1.000... can never be 0.999...9 for a finite index of the last 9. io
why are there problems?
Regards, WM
WM
wrote:
> On 2008-01-27, in sci.logic, WM wrote:
>
> > That's not the point. An ordering of countably many different elements
> > would already be impossible in practice. What I object to is the
> > alleged possibility to well-order an uncountable set of elements which
> > cannot be distinguished in principle because there are only countably
> > many labels.
>
> These worries of yours have nothing to do with set theory as usually
> understood.
O yes, they have. It can be proved as the most basic principle of any
language of logic that no more than countably many symbols, names,
labels, or numbers (call it as you like) are available. Ordering
elements, however, means in the first step to distinguish them. So set
theory is busy to distinguish what can be proved indistinguishable (or
to prove that it can be distinguished). And that is nonsense under any
aspect, be it from inside or outside mathematics.
Regards, WM
Gc
>
>
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Jan., 14:08, Dave Seaman <dsea...@no.such.host> wrote:
> > > On Sun, 27 Jan 2008 04:19:25 -0800 (PST), Albrecht wrote:
> > > > Peter Webb schrieb:
> > > >> "Albrecht" <albst...@gmx.de> wrote in message
> > > >>news:6545ba30-2eaf-409c...@u10g2000prn.googlegroups.com...
> > > >> > Dear members of sci.math and sci.logic,
>
> > > >> > if there is an infinite sequence (list) of all rational numbers in [0,
> > > >> > 1] and an antidiagonal number of this list (according to G. Cantor's
> > > >> > diagonal argument) and this antidiagonal number is a rational number
> > > >> > in [0, 1]
>
> > > >> > would destroy this fact the diagonal argument of G. Cantor which
> > > >> > proves that the set of reals is more infinite than the set of rational
> > > >> > numbers?
>
> > > >> No, it would destroy the proof that the rationals are countable. It
> > > >> would
> > > >> say nothing as to whether Reals are countable, because your argument
> > > >> would
> > > >> concern only the rationals.
> > > > I don't think so. A proof which shows that the rationals are
> > > > uncountable would distroy the diagonal argument - not the proof that
> > > > the rationals are countable, I think.
>
> > > A proof that the rationals are uncountable would demonstrate a
> > > contradiction
> > > in mathematics, just the same as if you proved that 1+1=3. All of
> > > mathematics
> > > would be affected.
>
> > Not at all. It has been attempted during the last 100 years or so to
> > base mathematics on set theory. But in fact mathematics is completely
> > independent of set theory. It existed before and will remain after. In
> > mathematics there is at most one infinity. Everything else is
> > matheology.
>
> It is WM's own views of mathematics that are theological instead of
> being merely logical.
>
> What parts of mathematics does WM claim cannot be embedded in set theory?
But doesn`t the VM:s argument go other way around? He consider weaker
theorems than set theory, so they don`t imply set theory - namely
because they are weaker. WM should study subsystems of second order
arithmetic, in those theories a lot of analysis can be done, but they
are not set theories.
WM
wrote:
> On 2008-01-27, in sci.logic, lwal...@lausd.net wrote:
>
> > We know that WM wishes to prove card(N)=card(R),
> > which is false in ZFC. There are two ways to
> > have card(N)=card(R). One way is to deny that
> > certain standard reals actually exist -- and that
> > is what WM usually does with his pi*googleplex and
> > related arguments.
>
> Could you elaborate on this "way"? Constructive reals are
> constructively uncountable, for example,
The question is not about "constructive reals" or "real reals" and
whether they are constructively uncountable or really uncountable. The
question is whether or not there is any possibility to distinguish
uncountable many elements - absolutely. In the first place it is not
even the question of well-ordering, but the fact that only such
elements which are distinct form one another can inflate a set. And
the answer to this question is an absolute no. There is absolutely not
the slightest chance.
We have without any ado:
1) There are only countably many distinct elements.
2) Cantor's diagonal argument is interpreted to produce evidence for
uncountably many distinct elements.
3) Conclusion: The argument or its interpretation or both are wrong.
> and it seems difficult to
> envisage any conception of mathematics on which the reals are
> bijectable with the naturals without equivocating, e.g. taking 'real'
> to mean 'recursive real' in context of classical reasoning.
This part of "classical" reasoning is wrong. We see that the concept
of bijection taken from the finite case cannot meaningfully be
transferred to the infinite case.
Regards, WM
WM
>
> > That's not the point. An ordering of countably many different elements
> > would already be impossible in practice. What I object to is the
> > alleged possibility to well-order an uncountable set of elements which
> > cannot be distinguished in principle because there are only countably
> > many labels.
>
> These worries of yours have nothing to do with set theory as usually
> understood.
But they have to do with the art of mathematics.
> They will be effective in dissuading a poor soul from
> engaging in the fantasy game of set theory only in case of those who,
> for some reason, share your ideas about labels and what not; these
> concepts do not enter into any set theoretic arguments in ordinary
> mathematics.
They are blocked in the same way as reality is blocked by some insane
minds?
Regards, WM
WM
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Jan., 22:00, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> > wrote:
> > > On 2008-01-27, in sci.logic, WM wrote:
>
> > > > How can you believe that your explanations, illustrations,
> > > > explications, practice (?) and such like could convince any sober mind
> > > > when you are unable to explain how you would try to order
> > > > indistinguishable elements but claim that you are able to do?
>
> > > As I have no idea what "ordering indistinguishable elements" means
> > > I've no idea. As to Zermelo's well-ordering theorem, there's no
> > > pretence that any actual human being could in any concrete sense carry
> > > out any ordering.
>
> > That's not the point. An ordering of countably many different elements
> > would already be impossible in practice. What I object to is the
> > alleged possibility to well-order an uncountable set of elements which
> > cannot be distinguished in principle because there are only countably
> > many labels.
>
> Lots of people have trouble with requiring every set to be well
> orderable. So there is no such requirement. The axiom of choice is
> optional.
>
the fact that only such elements which are distinct form one another
can inflate a set. And the answer to this question is an absolute no.
We have without any ado:
1) There are only countably many distinct elements.
2) Cantor's diagonal argument is interpreted to produce evidence for
uncountably many distinct elements.
3) Conclusion: The argument or its interpretation or both are wrong.
>
>
> > So you and your ilk say: for every subset of the uncountable set I can
> > find an element which is different from all the other elements. I call
> > this the first element of that subset.
>
> According to what ordering principle?
>
> > But you know that in most cases
> > it is impossible to distinguish this element from most of the other
> > elements. I do not accept that.
>
> The issue is not whether a single comparison can distinguish one real
> simultaneously from all other reals, but whether it is capable of
> distinguishing one real from another real.
Wrong. Read Zermelo's proof. Or try to visualize an unordered set the
members of which could only be compared one by one.
Regards, WM
WM
Kronecker could not yet know about the limited framework. There is an
upper limit of information available. Hence there are *indices* which
cannot be determined, let alone the digits at these places. Whether a
number exists depends on the information required to talk about it as
an individual. I have a nice little model to demonstrate that:
In this model we can represent every natural number from 0 to 9999999
by choosing seven symbols from the type writer's set of digits. By
means of abbreviations we can represent
123(9*) which is to represent our natural number 12333333333 with
nine
3's, or
1(999*) which is to represent our natural number consisting of 999
1's.
But we cannot express every number between them. So we cannot express
any number between
123(9*) and 124(9*), i.e., between 12333333333 and 12444444444 by
this
kind of abbreviation.
Introducing the symbol "..." (to be understood as a single place
symbol), we can even represent infinite strings like 0.1212... or
1,2,3,... . It is clear, however, that there are no infinite sets of
numbers existing in our model or in any other
Regards, WM
Don Stockbauer
> numbers existing in our model or in any other
>
> Regards, WM
No, there aren't , because in reality, as opposed to Cantor's
nightmare of infinities (over which he killed himself), there are just
two in everyday, real, cybernetic life:
1. The potential
2. The actual
- Tex
Daryl McCullough
>
>On 27 Jan., 16:08, herbzet <herb...@gmail.com> wrote:
>> herbzet wrote:
>> > So, what would be called for here is a listing of ... well,
>> > whatever, and an application of the diagonal argument that
>> > fails to produce an item not already on the list.
>>
>> Hard to see how this might be done, since a "diagonal argument"
>> is the production of an object which, by construction, is
>> different from each item already on a given list.
>
>This was Cantor's original opinion which however turned out wrong. A
>simple example is well known:
>
>Take the list of binaries
>
>1.000...
>0.1000...
>0.11000...
>0.111000...
>...
>
>When substituting 0 by 1 the diagonal 0.111... produces the first
>entry.
Yes, the double representation of reals by binary sequences is
sometimes a pain. But it's not an obstacle to Cantor's
diagonalization procedure. You just have to do the diagonalization
in a way that guarantees that the result will *not* be one of the
reals with multiple representations. Here's one approach to doing
that:
You have your original list of reals to diagonalize out of:
r_0
r_1
r_2
r_3
r_4
.
.
.
Now, make a new list, where between every two reals, alternate
between .000000... or .111...:
r_0
0.0000....
r_1
0.1111...
r_2
0.000...
r_3
0.111...
etc.
Then the diagonal number will be guaranteed not
to end with all 0s or all 1s. And it will be
guaranteed to differ from each r_n in at least
one position. Together, those facts imply that
the diagonal is unequal (as reals) to every r_n.
--
Daryl McCullough
Ithaca, NY
WM
I know these approaches. Cantor's original argument however is
wrong:"da wir sonst vor dem Widerspruch stehen würden, daß ein Ding E0
sowohl Element von M, wie auch nicht Element von M wäre."
And if you recognize that there are countably more multiple
representations like
LIM{n-->oo}SUM{1 to n-1} 7*10-n = LIM{n-->oo}SUM{1 to n} 7*10-n =
LIM{n-->oo}(10-n + SUM{1 to n} 7*10-n ) = ...
then you see that the diagonal argument fails for infinite lists.
Regards, WM
MoeBlee
> Dear members of sci.math and sci.logic,
>
> if there is an infinite sequence (list) of all rational numbers in [0,
> 1] and an antidiagonal number of this list (according to G. Cantor's
> diagonal argument) and this antidiagonal number is a rational number
> in [0, 1]
>
> would destroy this fact the diagonal argument of G. Cantor which
> proves that the set of reals is more infinite than the set of rational
> numbers?
(1) The diagonal argument regarding the reals proves that the set of
reals is uncountable (I suppose that's what you mean by "more
infinite").
(2) If you show that there is an enumeration of the set of rationals
in [0 1] but that the range of that enumeration is not the set of
rationals in [0 1], then you've proven a contradiction in Z set
theory, and if you used only axioms used in the diagonal argument
regarding the reals, then you'll have shown that diagonal argument is
pretty much irrelevent since it uses an inconsistent set of premises.
But be careful - just showing that there is an enumeration whose range
is a denumerable subset of the set of rationals in [0 1] and that
there is another rational in [0 1] not in the range of that
enumeration does NOT prove any contradiction in Z set theory. Rather,
to prove such a contradiction, you must prove that the range of the
enumeration is the set of ALL rationals in [0 1] but that there is a
rational in [0 1] that is not in the range of the enumeration.
(3) You've been posting about set theory for years; Why don't you know
the answers to such basic questions already?
MoeBlee
MoeBlee
> Typically, most adherents of a standard set
> theory, such as ZFC, believe that standard set theory is
> the only theory worth studying, and label all nonstandard
> set theories as "rubbish."
Typically? Name one person.
MoeBlee
MoeBlee
> A proof which shows that the rationals are
> uncountable would distroy the diagonal argument - not the proof that
> the rationals are countable, I think.
A proof from what AXIOMS and using what LOGIC?
If you prove (using first order logic) from the axioms of Z set theory
that the rationals are uncountable then you prove Z set theory to be
inconsistent.
> The diagonal argument of G. Cantor "measures with two measures": The
> real numbers in Cantor's lists are actual inifinite in the sense of
> being completely defined by given an infinite sequence of digits. The
> real numbers symbolise their limits.
> The list itself isn't actual infinite since it lacks of its limit.
>
> E.g. the infinite list
>
> 0.011111...
> 0.101111...
> 0.110111...
> 0.111011...
> 0.111101...
> ...
>
> doesn't contain its limit 0.111111111.... which is the antidiagonal of
> the list. But it is thought that any number of the list _is_ the limit
> of the infinite sum which the number respresents.
>
> This small gap between diffenrent understandings of infinity leads to
> the senseless view of different infinites, I think.
You're wrong. If you'd learn first order predicate logic and then
follow the proofs in first order predicate logic of the theorems from
the axioms, up to and including the theorem that the set of real
numbers is uncountable, then you'd understand why you are wrong as now
you use your fuzzy undefined notions in place of the exact, rigorous
derivations.
MoeBlee
MoeBlee
> So you and your ilk say: for every subset of the uncountable set I can
> find an element which is different from all the other elements. I call
> this the first element of that subset. But you know that in most cases
> it is impossible to distinguish this element from most of the other
> elements. I do not accept that.
I like to prove theorems of mathematics in ZFC. But I don't say what
you just said. So I don't know what "ilk" you have in mind.
> > It's apparent you don't like the abstractions and notions introduced
> > in set theory.
>
> I don't like explicit contradictions as that one explained above.
There is no known proof of a contradiction in ZFC.
MoeBlee
William Hughes
> n --> oo does not mean
> that n gets a non-natural number but that the assertion is correct for
> every natural number however large it may be.
Correct and very important. n can be very large
but n is always a natural number. Note "lim n-> inf"
is not a natural number (indeed, it may not
be any sort of number).
>
> You agree that irrational numbers are in Cantor's list, at least at
> the diagonal. For irrational numbers you accept the limit
> lim{n --> oo} sum(1 to n) a_n*10^-n.
Yes, but note that this is a value, not a digit position.
> Changing one diagonal digit by (+
> or -) 1 at position n
at position n, that is at a natural number
> yields a different number. But changing all
> diagonal digits includes the limit lim{n --> oo} (+ or -) 10^-n.
Piffle. Changing all the diagonal digits means "for each
natural number change the diagonal digit" However, lim{n--> oo}
is not a natural number. There is no digit "at the limit".
This would have to be the last digit. Guess what?
> Otherwise you have not changed every digit of an irrational (or
> periodical rational) number but only an initial segment of digits.
>
Piffle.
The initial segment which contains all the digits
contains all the digits (DUH!). If you change every digit in this
initial seqment you change all the digits (DUH!). So you can change
all the digits by changing "only an initial segment of digits".
- William Hughes
WM
Do you agree that lim{n-->oo} 10^-n = lim{n-->oo} 2*10^-n and that
both these numbers could be diagonal numbers of special Cantor lists?
If not, why could pi or 0.111... be diagonal numbers of Cantor lists?
Regards, WM
Randy Poe
Both of these numbers are 0, and 0 could be a diagonal
number of a list if none of the entries on the list is
0.
- Randy
Virgil
<404fd116-ac40-4326...@i7g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
That is not exactly what it means in mathematics, as it need not refer
to an assertion at all. Lim_{ n --> oo } Sum_{ 1 <= k <= n} 9/10^k
does not assert anything.
>
> You agree that irrational numbers are in Cantor's list, at least at
> the diagonal.
Not necessarily. If the list is n --> pi*(n + 1), then it contains no
rationals at all.
[snip]
>
> 1.000... can never be 0.999...9 for a finite index of the last 9. io
> why are there problems?
That presumes that the sequence of 9's has a last 9.
When that is not the case then the value of equals 1.
I.e. if Sum_{n in N} 9/10^(n+1) is to have any numerical meaning, it
must mean 1.
>
> Regards, WM
Dave Seaman
Actually, 0 and pi could not be the diagonal number if the rule for
specifying the diagonal digits avoids using 0 or 9, which is the usual
way of avoiding the dual-representation problem. Also, the usual rule
for specifying the diagonal number always picks a number between 0 and 1,
and therefore pi is again excluded.
It is not necessary that every real number be a possible diagonal number.
All that matters is that there is a diagonal number for every list, and
the diagonal number is not a member of the list.
--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
Virgil
<4987f132-7c45-4393...@i12g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Jan., 23:06, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> wrote:
> > On 2008-01-27, in sci.logic, lwal...@lausd.net wrote:
> >
> > > We know that WM wishes to prove card(N)=card(R),
> > > which is false in ZFC. There are two ways to
> > > have card(N)=card(R). One way is to deny that
> > > certain standard reals actually exist -- and that
> > > is what WM usually does with his pi*googleplex and
> > > related arguments.
> >
> > Could you elaborate on this "way"? Constructive reals are
> > constructively uncountable, for example,
>
>
> The question is not about "constructive reals" or "real reals" and
> whether they are constructively uncountable or really uncountable. The
> question is whether or not there is any possibility to distinguish
> uncountable many elements - absolutely. In the first place it is not
> even the question of well-ordering, but the fact that only such
> elements which are distinct form one another can inflate a set. And
> the answer to this question is an absolute no. There is absolutely not
> the slightest chance.
>
> We have without any ado:
> 1) There are only countably many distinct elements.
There are only countably many nameable elements, but that only limits
our access to them, it does not limit their existence.
WM insists that existence requires accessibility, but nothing in
mathematics justifies that assumption.
> 2) Cantor's diagonal argument is interpreted to produce evidence for
> uncountably many distinct elements.
> 3) Conclusion: The argument or its interpretation or both are wrong.
WM's interpretation is what is wrong.
>
> > and it seems difficult to
> > envisage any conception of mathematics on which the reals are
> > bijectable with the naturals without equivocating, e.g. taking 'real'
> > to mean 'recursive real' in context of classical reasoning.
>
> This part of "classical" reasoning is wrong. We see that the concept
> of bijection taken from the finite case cannot meaningfully be
> transferred to the infinite case.
it is remarkable the number of things which are not there that WM
manages to see anyway, while managing not to see things which are there.
>
> Regards, WM
Virgil
<0080bb11-57f8-4001...@v4g2000hsf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
AS reality is blocked from you mind, do you mean?
>
> Regards, WM
Virgil
<90979260-ca31-488f...@e23g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> There is an upper limit of information available. Hence there are
> *indices* which cannot be determined, let alone the digits at these
> places. Whether a number exists depends on the information required
> to talk about it as an individual.
According to WM's philosophy, a number which cannot be named can not
exist, but mathematics is not so nice. Mathematics allows unnamed and
even unnamable numbers.
Virgil
<13470a57-872f-40b1...@q77g2000hsh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> We have without any ado:
> 1) There are only countably many distinct elements.
In WM's world, there are only finitely many distinct elements, since
those which are inaccessible to him do not exist. In mathematicians
worlds, what can be imagined in theory can exist in that theory however
inaccessible in practice.
Virgil
<90979260-ca31-488f...@e23g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> Introducing the symbol "..." (to be understood as a single place
> symbol), we can even represent infinite strings like 0.1212... or
> 1,2,3,... . It is clear, however, that there are no infinite sets of
> numbers existing in our model or in any other
Speak only for yourself, WM.
There ARE infinite sets of numbers in many of our models.
That you do not like it is your problem, not ours.
William Hughes
You are trying to change the subject.
The question is:
Can you change all the digits of a real number by changing
each digit in an initial segment?
- William Hughes
Virgil
<3830bb77-09cc-4392...@i3g2000hsf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Cantor's original "diagonal" argument was based on binary sequences,
functions from N to , say, {m, w}, or any other two element set, for
which it is trivial by didagonal argument to show that no list can
contain all such sequences/functions.
And that argument does not fail for infinite lists.
Albrecht
MoeBlee schrieb:
It's basic for you, not for me.
In my opinion the diagonal argument is logical inconsistent. But if
so, how is to prove this fact? Since the diagonal argument leads to
consequences which, if accepted, makes the proof valid in return. The
very interesting system of transfinite numbers is a kind of
selffulfilling system, in my opinion.
Thank you for your direct answer to my question. Thanks to the others
too.
Best regards
Albrecht S. Storz
Virgil
<aa22c4aa-2639-4780...@e6g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> Do you agree that lim{n-->oo} 10^-n = lim{n-->oo} 2*10^-n and that
> both these numbers could be diagonal numbers of special Cantor lists?
There is no such thing as "both such numbers" when one has only one
number.
> If not, why could pi or 0.111... be diagonal numbers of Cantor lists?
Diagonal numbers for decimal Cantor lists do not contain digits 0 or 9,
so as to avoid being one representation of a dually represented number,
so pi could not be one, though 1/9 could be.
WM
Correct.
> and 0 could be a diagonal
> number of a list
Correct.
> if none of the entries on the list is
> 0.
That condition is not necessary, as you can see from the fact that 0
has many representations as an infinite series. Perhaps you can see it
easier from the following representations of 0:
0 = -1 + SUM{n = 1 to oo} 1/2^n
0 = -1/2 + SUM {n = 1 to oo} 1/3^n
(Instead of powers of 10 a number can also be represented by other
powers.)
Regards, WM
WM
> > You agree that irrational numbers are in Cantor's list, at least at
> > the diagonal.
>
> Not necessarily. If the list is n --> pi*(n + 1), then it contains no
> rationals at all.
But possibly. (I showed frequently enough that it is not necessary.)
>
>
> > 1.000... can never be 0.999...9 for a finite index of the last 9. io
> > why are there problems?
>
> That presumes that the sequence of 9's has a last 9.
> When that is not the case then the value of equals 1.
Correct. This shows that for infinite sequences differences in digits
cease to cause differences in values.
> I.e. if Sum_{n in N} 9/10^(n+1) is to have any numerical meaning, it
> must mean 1.
yes,
LIM{n-->oo}SUM{k = 1 to n} 9*10^-k = 1
as well as
LIM{n-->oo}SUM{k = 1 to n-1} 9*10^-k = 1.
whereas the difference of these expressions is
LIM{n-->oo} 9*10^-n = 0
Regards, WM