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Attempts to Refute Cantor's Uncountability Proof?
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Hatto von Aquitanien
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Joined: 19 Nov 2005
Posts: 410

PostPosted: Fri Jul 21, 2006 5:51 am    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

Dik T. Winter wrote:

Quote:
In article <BrqdneznR_qBjSPZnZ2dnUVZ_u-dnZ2d@speakeasy.net> Hatto von
Aquitanien <abbot@AugiaDives.hre> writes:
Dik T. Winter wrote:

In article <Hd-dnfIlLZvPwiDZnZ2dnUVZ_v-dnZ2d@speakeasy.net> Hatto von
Aquitanien <abbot@

"..." means "and so forth". That is, a pattern is exhibited, and then
repetition is indicated by the ellipsis.

Yes, that is common usage, but not mathematics.

I'm not willing to accept that claim. The ellipsis is used in every math,
chemistry, physics and computer science book I have read in the past 3
decades, and it is used to mean that an exhibited pattern is implied to be
continued either to some finite terminating value, or ad infinitum.

Quote:
I have already indicated that
I
am aware of a connotation which will treat such repetition as performed
"at once" rather than in temporal order.

When you look closely at mathematics you will find that there is no
instance where you can state that the repetition is performed "at
once". In most instances there is no repetition *at all*.

I believe you are not understanding the abstraction I intend by the use of
the term "repetition". I do not merely mean some finite sequence is
repeated. I mean that some rule (algorithm) is repeatedly followed
(iterated) to obtain a subsequent term.

Quote:
But I believe it is wise to
question
the implications of such an interpretation. This is why I have
stressed
the obvious prima facia meaning of "and so forth". That is some
process of repetition, that is to say iteration sequential in time.

Have a look at repeating decimals: 0.999... . The strict definition here
is:
0.999... = lim{n -> oo} sum{k = 1 .. n} 9.10^(-k)
where lim is defined using the epsilon argument. That is, there is *no*
repetition at all.

There is no repetition in the repeating decimal? No repetition in the
evaluation of 9*10^(-k) by substituting the subsequent values 1, 2, ..., oo
for k? You are using repetition to express these ideas. You might claim
this use of repetition is merely heuristic, but in this case I cannot see
how you could possibly sustain the assertion that

0.999... = lim{n -> oo} sum{k = 1 .. n} 9*10^(-k)

does not involve some form of repetition or iteration. In words, that
statement will typically be expressed as 'zero point nine, nine, nine
continued infinitely, equals the limit as n tends to infinity of the sum
from k equals 1 to k equals n of 9*10^(-k).'

Quote:
My understanding of what Cantor presented is that I could move
diagonally downward and to the right along a diagonal on a conceptual
gird of
elements. For each element I encounter, I place its opposite in a new
row to be added to the grid when I happen to construct a sequence not
yet in the grid.

You implicitly imply some sequencing the definition. The definition does
not have sequencing implied. The diagonal is defined as:
element n of D is the opposite of element n of listmember A[n]
so, no sequencing.

What is meant by "einfach unendliche Reihe von Elementen der
Mannigfaltigkeit M"? The translator renders it as: "simply infinite series
of elements of the manifold M". I am using the word 'sequence' where the
translator has used the word 'series'. My justification is as follows:

<quote
author="Konrad Knopp"
book="Infinite Sequences and Series">
Definition: The natural numbers 1,2,3,... ordered according to magnitude
form a sequence, the sequence of natural numbers. If with every one of
these numbers nu there is associated in any manner a single definite
(complex) number z_nu, then these numbers z_0, z_1, ... z_nu,... form an
infinite sequence of numbers, briefly, as sequence.
</quote>

Quote:
Certainly, we can claim that this all happens "at once", and
the
concept of traversing the diagonal is merely heuristic. But how much
meaning such a claim actually has is not clear to me.

The claim has meaning here because there are no dependencies between the
different elements of the diagonal.

I see two plausible objections to this. 1) There does appear to be some
global interdependence dictating the value assigned to each and every
element of the grid. 2) In order for me to contemplate this infinite
ordered set of infinite sequences, I have to contemplate a finite portion
of this structure, and imagine that this conception is extended ad
infinitum.

<quote
url='http://uk.geocities.com/frege@btinternet.com/cantor/diagarg.htm'>

Hier sind die au,v in bestimmter Weise m oder w. Es werde nun eine Reihe
b_1, b_2, ... b_v,..., so definiert, dass bv auch nur gleich m oder w und
von a_v,v verschieden sei.

Ist also a_v,v = m, so ist b_v = w. [My correction, STH]

Betrachten wir alsdann das Element

E_0 = (b_1, b_2, b_3, ...)

von M, so sieht man ohne weiteres, dass die Gleichung

E_0 = E_u

für keinen positiven ganzzahligen Wert von u erfüllt sein kann, da sonst für
das betreffende u und für alle ganzzahligen Werte von v.

</quote>

Notice that Cantor presents his demonstration by means of an implied
sequential visit to every a_v,v upon which a new element b_v is added to
E_0.

Quote:
This comes out a bit garbled. There is no statement that the n-th element
of D will be different from more than one n-th element of the members of
the list. What *is* the claim is that the n-th element of D will be
different from at least one n-th element of the members of the list (we
can name the n-th member here). And also that each member of the list
will be different from D (namely in the n-th element).

What I mean to say is that there will be some n such that

(a_1, a_2, a_3, ..., a_n-1)=(b_1, b_2, b_3, ..., b_n-1) => a_n!=b_n

Quote:
Now, when we convert the argument to an argument with decimal digits (and
take care of dual representations in the proper way), we find that D is
a sequence of digits in the range [0,9]. This is a Cauchy sequence, and
so (because the reals are complete), it denotes a real number. (It is
irrelevant precisely *which* number.) And as it is different from all
real numbers on the list, it is a new real number that was not listed.
This does not work with a list of rational numbers. In that case you
have also that D is a real number, but in order to show that the list
of rational numbers is incomplete you have to show that D is a rational
number. And you can not do that.

I am not absolutely confident of my suggestion above. I have had a headache
for over a week, and have not been able (or willing) to focus on this
matter. I will continue to mull it over. If I conclude that it is in
error, I will offer my acknowledgment of that. /Assuming/ that it is
correct, then I must conclude that I could indeed claim that this
new 'number' E_0 is 'rational'.

Quote:
On the other hand, I think that his proof that the reals (or any
complete, densely ordered set of numbers) is not countable is even
completely valid at that time.

But what does "densely ordered set of numbers" mean? The set of
rational numbers appears to me to qualify for such a definition.

Yes, the rationals are also densely ordered. But see also the word
"complete" above.

By implication you are saying the rationals are incomplete. What argument
do you have for that conclusion? If this is due to the "empirical"
observation that there are irrational formulae such as pi and sqrt(2), then
I will suggest that we are adding structure to our model of the "continuum"
which may not be meaningfully expressible in terms of decimal point
notation. If we draw a fundamental distinction between a terminating
decimal point representation, and an infinite decimal point representation
(by calling the latter a limit), then how meaningful is Cantor's argument
wich appears to appeal to the model of decimal point representations as
infinite sums?

I'll have to meditate on that last matter a bit.

Quote:
Is the base 10 form of the proof incorrect?

Depends. The proof is correct when you assume the proper axioms.

It seems to me that the proof is just as valid for the rational numbers
as it is for some other set.

No, see above.

Likewise.
--
Nil conscire sibi
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mike3
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Joined: 27 May 2005
Posts: 52

PostPosted: Thu Jul 20, 2006 8:16 am    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

imaginatorium@despammed.com wrote:
Quote:
mike4ty4@yahoo.com wrote:

snip

In case you're curious, another (informal) argument, I don't know how
good, could
also be made for the real numbers' uncountability. Consider the
interval [-1, 1]. With
just the endpoints, we have 2. Add the middle, 0, and we have 3. We can
always
find a smaller number to stick in, and thus our total keeps going up.
We can never
"fill the gap" with naturals, because no matter how many points we
denumerate,
we can always make more in the same interval, so it evades all our
attempts to try
and assign a natural to each element. Since [-1, 1] is countless, and
it can be
mapped to all the real numbers via the one-to-one function f(x) =
tan(pix/2),
we conclude it's impossible to count the reals. If there's any way to
give the above
argument rigor, I'd be curious to hear about it.

I doubt it. Your argument above also appears to apply to the rationals
in [-1, 1]. There are an unlimited number of rationals, and any attempt
to count them (in the meaning of normal language, in which "counting"
is a terminating process) will fail. You will never get to the end,
even though you get to every individual rational eventually. (Last
sentence may induce QD in the feeble-brained.)


The thing that I'm trying to get across is that no matter how many
reals
you denumerate, there will always be more that "slip through your
grasp",
so to speak, because you can always find something "in the gaps", so
you "run out of naturals" to tag them all.

Quote:
Brian Chandler
http://imaginatorium.org
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Dik T. Winter
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Joined: 25 Mar 2005
Posts: 1359

PostPosted: Wed Jul 19, 2006 1:34 pm    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

In article <BrqdneznR_qBjSPZnZ2dnUVZ_u-dnZ2d@speakeasy.net> Hatto von Aquitanien <abbot@AugiaDives.hre> writes:
Quote:
Dik T. Winter wrote:

In article <Hd-dnfIlLZvPwiDZnZ2dnUVZ_v-dnZ2d@speakeasy.net> Hatto von
Aquitanien <abbot@AugiaDives.hre> writes:

It's not so much that I'm thinking about notation.

But you are. You state that the notations "to" and "..." seem to imply
some
kind of iteration. However those notations have in mathematics no meaning
until there is a definition. And when you look at the definition you will
find that there is no form of iteration at all.

"..." means "and so forth". That is, a pattern is exhibited, and then
repetition is indicated by the ellipsis.

Yes, that is common usage, but not mathematics.

Quote:
I have already indicated that I
am aware of a connotation which will treat such repetition as performed "at
once" rather than in temporal order.

When you look closely at mathematics you will find that there is no
instance where you can state that the repetition is performed "at
once". In most instances there is no repetition *at all*.

Quote:
But I believe it is wise to question
the implications of such an interpretation. This is why I have stressed
the obvious prima facia meaning of "and so forth". That is some process of
repetition, that is to say iteration sequential in time.

Have a look at repeating decimals: 0.999... . The strict definition here is:
0.999... = lim{n -> oo} sum{k = 1 .. n} 9.10^(-k)
where lim is defined using the epsilon argument. That is, there is *no*
repetition at all.

Quote:
My understanding of what Cantor presented is that I could move diagonally
downward and to the right along a diagonal on a conceptual gird of
elements. For each element I encounter, I place its opposite in a new row
to be added to the grid when I happen to construct a sequence not yet in
the grid.

You implicitly imply some sequencing the definition. The definition does
not have sequencing implied. The diagonal is defined as:
element n of D is the opposite of element n of listmember A[n]
so, no sequencing.

Quote:
Certainly, we can claim that this all happens "at once", and the
concept of traversing the diagonal is merely heuristic. But how much
meaning such a claim actually has is not clear to me.

The claim has meaning here because there are no dependencies between the
different elements of the diagonal.

Quote:
One reason for my focusing on the notion of a sequential traversal is
because I have certain misgivings regarding the distinction between
applying this method to the set of rational numbers as opposed to the set
of real numbers. As I understand things, Cantor is asserting that there
will always be some n such that A_nn (the A_n in the new sequence) is
different from all the other A_n in the existing sequences which match the
new sequence to that point.

This comes out a bit garbled. There is no statement that the n-th element
of D will be different from more than one n-th element of the members of
the list. What *is* the claim is that the n-th element of D will be
different from at least one n-th element of the members of the list (we
can name the n-th member here). And also that each member of the list
will be different from D (namely in the n-th element).

Now, when we convert the argument to an argument with decimal digits (and
take care of dual representations in the proper way), we find that D is
a sequence of digits in the range [0,9]. This is a Cauchy sequence, and
so (because the reals are complete), it denotes a real number. (It is
irrelevant precisely *which* number.) And as it is different from all
real numbers on the list, it is a new real number that was not listed.
This does not work with a list of rational numbers. In that case you
have also that D is a real number, but in order to show that the list
of rational numbers is incomplete you have to show that D is a rational
number. And you can not do that.

Quote:
On the other hand, I think that his proof that the reals (or any complete,
densely ordered set of numbers) is not countable is even completely valid
at that time.

But what does "densely ordered set of numbers" mean? The set of rational
numbers appears to me to qualify for such a definition.

Yes, the rationals are also densely ordered. But see also the word
"complete" above.

Quote:
Is the base 10 form of the proof incorrect?

Depends. The proof is correct when you assume the proper axioms.

It seems to me that the proof is just as valid for the rational numbers as
it is for some other set.

No, see above.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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Hatto von Aquitanien
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Joined: 19 Nov 2005
Posts: 410

PostPosted: Wed Jul 19, 2006 10:56 am    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

Dik T. Winter wrote:

Quote:
In article <Hd-dnfIlLZvPwiDZnZ2dnUVZ_v-dnZ2d@speakeasy.net> Hatto von
Aquitanien <abbot@AugiaDives.hre> writes:

It's not so much that I'm thinking about notation.

But you are. You state that the notations "to" and "..." seem to imply
some
kind of iteration. However those notations have in mathematics no meaning
until there is a definition. And when you look at the definition you will
find that there is no form of iteration at all.

"..." means "and so forth". That is, a pattern is exhibited, and then
repetition is indicated by the ellipsis. I have already indicated that I
am aware of a connotation which will treat such repetition as performed "at
once" rather than in temporal order. But I believe it is wise to question
the implications of such an interpretation. This is why I have stressed
the obvious prima facia meaning of "and so forth". That is some process of
repetition, that is to say iteration sequential in time.

Quote:
In the
circumstance,
notation is what is used to present the proof. The proof itself
appears to rely on the notational structure.

Yes, that is what all proofs rely on. If you have not defined what you
are meaning with a particular notation, there is no proof.

I meant more than semantics in this case.


Quote:
Nonetheless, I
believe his argument is basically lexicographical.

Yup, it was about sets of infinite sequences.

If I do address the proof given in decimal point notation, then what I
see is an argument which says: if I run down the diagonal of a decimal
point representation of an ordered set of real numbers far enough
(0.a_11a_22a_33...a_ii...), I will encounter a number not yet in the
list.
these are all finite concepts as far as I can see. At some diagonal
a_nn, I will generate a new number which has not yet been put in 1-1
correspondence with N. Is this not the argument?

No. You again focus on real numbers, but let that pass. If we agree on
sequences of digits where 0.999... is different from 1.000... as a
sequence,
we can continue. (Actually he uses only two "digits", but because they do
not represent real numbers, but only sequences, there is no problem.)

Whether we are talking about simple string comparison or the standard
interpretation of decimal point notation is irrelevant to the point I'm
trying to clarify.

Quote:
The problem is that you state "I will encounter". Given any list of
sequences,
you can *define* a sequence that is not in the list. The ultimate
discussion is not whether you can do so, but whether the definition is a
definition
indeed. First some notation. I will use (like Cantor) the symbols w and
m. And (like Cantor) I will focus on infinite sequences of those two
symbols. Now let it occur that I am given a list A_i, where each A_i is a
sequence of
the symbols w and m, and where i is a natural number. The claim is that
that
list does not contain all possible sequences of w and m. Because define
the sequence K where the i-th element of K is w when the i-th element of
A_i is
m, and the reverse. The claim is that K is different from all A_i, and
that is easy to show. The base problem is that those that oppose it say
that the definition of K is not proper.

My understanding of what Cantor presented is that I could move diagonally
downward and to the right along a diagonal on a conceptual gird of
elements. For each element I encounter, I place its opposite in a new row
to be added to the grid when I happen to construct a sequence not yet in
the grid. Certainly, we can claim that this all happens "at once", and the
concept of traversing the diagonal is merely heuristic. But how much
meaning such a claim actually has is not clear to me.

One reason for my focusing on the notion of a sequential traversal is
because I have certain misgivings regarding the distinction between
applying this method to the set of rational numbers as opposed to the set
of real numbers. As I understand things, Cantor is asserting that there
will always be some n such that A_nn (the A_n in the new sequence) is
different from all the other A_n in the existing sequences which match the
new sequence to that point. That n is a finite number.

[...]

Quote:
On the other hand, I think that his proof that the reals (or any complete,
densely ordered set of numbers) is not countable is even completely valid
at that time.

But what does "densely ordered set of numbers" mean? The set of rational
numbers appears to me to qualify for such a definition.

Quote:
Assuming that you did allow that such did even exist.
Quote Kronecker on his remark about Liouville's (I think) proof that pi
was transcendental: "a remarkable work, but without value, such numbers
do not exist". In those cases you come in the realm of philosophy: what
numbers do exist?

[...]

Quote:
Is the base 10 form of the proof incorrect?

Depends. The proof is correct when you assume the proper axioms.

It seems to me that the proof is just as valid for the rational numbers as
it is for some other set. I can follow the demonstration that purports to
show that the rational numbers are countable. Now, if I were to neglect
that demonstration when considering the diagonal proof by contradiction, it
seems that proof by contradiction would say the rational numbers cannot be
put into 1-1 correspondence with the natural numbers. I believe what I am
saying is there is a contradiction here.

To tell the truth, I'm not even sure how I can conceptually distinguish
between a representation of the rational numbers as infinite sums of
decimal point places, and a similar representation of the real numbers
using the same method. I believe there are irrational numbers because
there are proofs which demonstrate that specific numbers cannot be
expressed as ratios of natural numbers. I can further persuade myself that
it is meaningful to consider these irrational numbers to be "expressible"
as an infinite sum of decimal point places, though clearly, I cannot write
such a number down.

These are concepts which I must add to the notion of infinite decimal place
notation after I have established the basic model.
--
Nil conscire sibi
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Dik T. Winter
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Joined: 25 Mar 2005
Posts: 1359

PostPosted: Wed Jul 19, 2006 1:59 am    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

In article <Hd-dnfIlLZvPwiDZnZ2dnUVZ_v-dnZ2d@speakeasy.net> Hatto von Aquitanien <abbot@AugiaDives.hre> writes:
Quote:
Dik T. Winter wrote:

In article <1153169578.808273.40000@i42g2000cwa.googlegroups.com> "David R
Tribble" <david@tribble.com> writes:
David R Tribble writes:
1/3 = sum{1 to oo} 3 x 10^-1 = 0.333...

Dik T. Winter wrote:
You should stress that sum{1 to oo} is *not* an iterative process.
...
It's the limitations of the notations that appear to the be crux of
Hatto's problem:
1/3 = sum{i=1 to oo} 3x10^-i
1/3 = 3x10^-1 + 3x10^-2 + 3x10^-3 + ...
1/3 = 0.333...

Of course these are not "iterative processes", but the notations
"to" and "..." seem to imply some kind of iteration. It's a question
of comprehension and not reading too much into the notation.

Yes, it is the problem with anybody who thinks too much about notation,
and thinks that that will provide him with all information.

It's not so much that I'm thinking about notation.

But you are. You state that the notations "to" and "..." seem to imply some
kind of iteration. However those notations have in mathematics no meaning
until there is a definition. And when you look at the definition you will
find that there is no form of iteration at all.

Quote:
In the circumstance,
notation is what is used to present the proof. The proof itself appears to
rely on the notational structure.

Yes, that is what all proofs rely on. If you have not defined what you are
meaning with a particular notation, there is no proof.

Quote:
I know that Cantor's argument can be
expressed in ways which do not use decimal (base 10) notation. Cantor's
paper did not even use decimal notation of any base.

Again a myth. Cantor has never used the diagonal argument to show that
the reals had a greater cardinality than the natural numbers. That proof
is quite different. The diagonal argument is used to show that some
infinite sets have greater cardinality than other sets. It was Zermelo
(I think) who did show first how that proof can be changed to a proof
that the reals had greater cardinality than the naturals.

Quote:
Nonetheless, I
believe his argument is basically lexicographical.

Yup, it was about sets of infinite sequences.

Quote:
If I do address the proof given in decimal point notation, then what I see
is an argument which says: if I run down the diagonal of a decimal point
representation of an ordered set of real numbers far enough
(0.a_11a_22a_33...a_ii...), I will encounter a number not yet in the list.
these are all finite concepts as far as I can see. At some diagonal a_nn,
I will generate a new number which has not yet been put in 1-1
correspondence with N. Is this not the argument?

No. You again focus on real numbers, but let that pass. If we agree on
sequences of digits where 0.999... is different from 1.000... as a sequence,
we can continue. (Actually he uses only two "digits", but because they do
not represent real numbers, but only sequences, there is no problem.)

The problem is that you state "I will encounter". Given any list of sequences,
you can *define* a sequence that is not in the list. The ultimate discussion
is not whether you can do so, but whether the definition is a definition
indeed. First some notation. I will use (like Cantor) the symbols w and m.
And (like Cantor) I will focus on infinite sequences of those two symbols.
Now let it occur that I am given a list A_i, where each A_i is a sequence of
the symbols w and m, and where i is a natural number. The claim is that that
list does not contain all possible sequences of w and m. Because define the
sequence K where the i-th element of K is w when the i-th element of A_i is
m, and the reverse. The claim is that K is different from all A_i, and
that is easy to show. The base problem is that those that oppose it say
that the definition of K is not proper.

In my opinion (note, just opinion) that was true in Cantor's time. It is
a proper definition if you allow an additional axiom. There are more
places where Cantor assumes something that is not necessarily true (the
axiom of choice).

On the other hand, I think that his proof that the reals (or any complete,
densely ordered set of numbers) is not countable is even completely valid
at that time. Assuming that you did allow that such did even exist.
Quote Kronecker on his remark about Liouville's (I think) proof that pi
was transcendental: "a remarkable work, but without value, such numbers
do not exist". In those cases you come in the realm of philosophy: what
numbers do exist?

One final remark. When Cantor wrote his papers, the thinking about it
was still at its infancy. Quite a few things were changed from his
original thinking to modern set theory. So, when you quote Cantor you
should also consider context. What did he mean at that time, and how
would it be formulated currently?

Quote:
Is the base 10 form of the proof incorrect?

Depends. The proof is correct when you assume the proper axioms.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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Hatto von Aquitanien
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Joined: 19 Nov 2005
Posts: 410

PostPosted: Tue Jul 18, 2006 11:17 pm    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

Lasse wrote:

Quote:
Yes, the series 0.3, 0.33, 0.333, etc.., does indeed converge to 1/3.
But the (countable) sequence of these values does not actually contain
the limit value 0.333... (1/3).

The limit is called a supremum - IIRC from 10 years ago. But the only
way I can know that 0.333...=1/3 is by induction, which implies a
repeated
process. It's really nice that 1/3 happens to have a simple repeating
pattern, but that is not ture of all such infinite decimal point
representations of numbers.

Nonsense. Induction is a method that can be used to prove a statement
for all natural numbers. What is that statement here?

That for every decimal place n places to the right in the decimal point
representation of 1/3 the digit will be 3.

Quote:
Also, do you have
the conception that induction is the *only* way to prove a statement
for all natural numbers? This is decidedly *not* the case.

Starting with Peano's axioms, I'm not sure how much can be said about the
natural numbers as a whole without appeal, directly or implicitly to
induction.

Quote:
Anyway, for the number x = 0.333... , it is easy to see from the
definition, and basic rules for limits, that 10*x = 3 + x. (In fact,
you can make this the definition of 0.333... . It's not a bad
definition for numbers with periodic decimal extensions, since it
avoids the use of limits altogether.)

Hence we have 9*x = 3, and thus x = 1/3.

Do you see induction anywhere in the above proof?

I don't see how you are getting 0.333... from the basic rules of limits if
not through induction. More to the point, the concept of decimal point
notation is based on the idea of summing of negative powers of 10
multiplied by single digit integers.

Quote:

The form I was given for Cantor's diagonal proof by contradiction used
infinite decimal point notation[*]. It therefore assumes that it is
meaningful to represent the entire ordered set of real numbers in such a
form. I am first told to assume that there is a bijection between the
elements of this set and the natural numbers.

No, you do not have to assume any such thing. Given *any* function from
the naturals into the reals, you can show that there is some real
number which is not in the image of this function. I posted a proof of
this (or rather, the underlying fact that no set is equipollent to its
power set) a while ago. You should try to understand this proof; it is
not that long.

I asked what the axioms are for your set theory. The different
presentations of set theory I am drawing on are not mutually consistent in
the details of their axioms. I'll have to look again to see if you replied
to that request.

Quote:
The rest of your post seems rather confused and is difficult to
understand, but it seems to me that you have a problem grasping the
following: given a list of real numbers (i.e., a function from N to R),
we find an element which is not on this list. Could we not then just
add that element to our list?

The answer is: of course we can. However, by the same argument as
before, this list will still be missing some real number, and so on.
After all, we have proved the theorem not for a *particular* list of
real numbers, but for *all possible* lists!

Consider the following statements:

(i) For every real number x, there is a function f:N -> R whose image
contains x.

I'm not sure exactly what is meant by R, or even N in this case.

--
Nil conscire sibi
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Hatto von Aquitanien
science forum Guru


Joined: 19 Nov 2005
Posts: 410

PostPosted: Tue Jul 18, 2006 10:22 pm    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

Dik T. Winter wrote:

Quote:
In article <1153169578.808273.40000@i42g2000cwa.googlegroups.com> "David R
Tribble" <david@tribble.com> writes:
David R Tribble writes:
1/3 = sum{1 to oo} 3 x 10^-1 = 0.333...

Dik T. Winter wrote:
You should stress that sum{1 to oo} is *not* an iterative process.
...
It's the limitations of the notations that appear to the be crux of
Hatto's problem:
1/3 = sum{i=1 to oo} 3x10^-i
1/3 = 3x10^-1 + 3x10^-2 + 3x10^-3 + ...
1/3 = 0.333...

Of course these are not "iterative processes", but the notations
"to" and "..." seem to imply some kind of iteration. It's a question
of comprehension and not reading too much into the notation.

Yes, it is the problem with anybody who thinks too much about notation,
and thinks that that will provide him with all information.

It's not so much that I'm thinking about notation. In the circumstance,
notation is what is used to present the proof. The proof itself appears to
rely on the notational structure. I know that Cantor's argument can be
expressed in ways which do not use decimal (base 10) notation. Cantor's
paper did not even use decimal notation of any base. Nonetheless, I
believe his argument is basically lexicographical.

If I do address the proof given in decimal point notation, then what I see
is an argument which says: if I run down the diagonal of a decimal point
representation of an ordered set of real numbers far enough
(0.a_11a_22a_33...a_ii...), I will encounter a number not yet in the list.
these are all finite concepts as far as I can see. At some diagonal a_nn,
I will generate a new number which has not yet been put in 1-1
correspondence with N. Is this not the argument?

Is the base 10 form of the proof incorrect?

--
Nil conscire sibi
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Dik T. Winter
science forum Guru


Joined: 25 Mar 2005
Posts: 1359

PostPosted: Tue Jul 18, 2006 1:25 am    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

In article <1153169578.808273.40000@i42g2000cwa.googlegroups.com> "David R Tribble" <david@tribble.com> writes:
Quote:
David R Tribble writes:
1/3 = sum{1 to oo} 3 x 10^-1 = 0.333...

Dik T. Winter wrote:
You should stress that sum{1 to oo} is *not* an iterative process.
....
It's the limitations of the notations that appear to the be crux of
Hatto's problem:
1/3 = sum{i=1 to oo} 3x10^-i
1/3 = 3x10^-1 + 3x10^-2 + 3x10^-3 + ...
1/3 = 0.333...

Of course these are not "iterative processes", but the notations
"to" and "..." seem to imply some kind of iteration. It's a question
of comprehension and not reading too much into the notation.

Yes, it is the problem with anybody who thinks too much about notation, and
thinks that that will provide him with all information.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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David R Tribble
science forum Guru


Joined: 21 Jul 2005
Posts: 1005

PostPosted: Mon Jul 17, 2006 8:52 pm    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

Hatto von Aquitanien wrote:
Quote:
We have a tendency to accept .333... as being as good a specification of a
number as is 3, for example. That is to say "for all intents and
purposes, .333... is as good a specification of a number as is 3." But is
it? Are there some attributes which we should not attribute to .333...
which we attribute to 3 by virtue of its being a number?


David R Tribble writes:
Quote:
1/3 = sum{1 to oo} 3 x 10^-1 = 0.333...


Dik T. Winter wrote:
Quote:
You should stress that sum{1 to oo} is *not* an iterative process.

I thought I did (in the next sentence).

Quote:
As it
stands, it appears that you are taking the sum of infinitely many terms,
and that is not the case. sum{1 to oo} is (in my opinion) abuse of
notation, as it is defined as lim{n -> oo} sum{1 to n}. Which (in turn)
also is not an iterative process (think epsilon and delta).

It's the limitations of the notations that appear to the be crux of
Hatto's problem:
1/3 = sum{i=1 to oo} 3x10^-i
1/3 = 3x10^-1 + 3x10^-2 + 3x10^-3 + ...
1/3 = 0.333...

Of course these are not "iterative processes", but the notations
"to" and "..." seem to imply some kind of iteration. It's a question
of comprehension and not reading too much into the notation.
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Lasse
science forum addict


Joined: 29 Apr 2005
Posts: 71

PostPosted: Mon Jul 17, 2006 12:56 pm    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

Quote:
Yes, the series 0.3, 0.33, 0.333, etc.., does indeed converge to 1/3.
But the (countable) sequence of these values does not actually contain
the limit value 0.333... (1/3).

The limit is called a supremum - IIRC from 10 years ago. But the only way I
can know that 0.333...=1/3 is by induction, which implies a repeated
process. It's really nice that 1/3 happens to have a simple repeating
pattern, but that is not ture of all such infinite decimal point
representations of numbers.

Nonsense. Induction is a method that can be used to prove a statement
for all natural numbers. What is that statement here? Also, do you have
the conception that induction is the *only* way to prove a statement
for all natural numbers? This is decidedly *not* the case.

Anyway, for the number x = 0.333... , it is easy to see from the
definition, and basic rules for limits, that 10*x = 3 + x. (In fact,
you can make this the definition of 0.333... . It's not a bad
definition for numbers with periodic decimal extensions, since it
avoids the use of limits altogether.)

Hence we have 9*x = 3, and thus x = 1/3.

Do you see induction anywhere in the above proof?


Quote:
The form I was given for Cantor's diagonal proof by contradiction used
infinite decimal point notation[*]. It therefore assumes that it is
meaningful to represent the entire ordered set of real numbers in such a
form. I am first told to assume that there is a bijection between the
elements of this set and the natural numbers.

No, you do not have to assume any such thing. Given *any* function from
the naturals into the reals, you can show that there is some real
number which is not in the image of this function. I posted a proof of
this (or rather, the underlying fact that no set is equipollent to its
power set) a while ago. You should try to understand this proof; it is
not that long.

The rest of your post seems rather confused and is difficult to
understand, but it seems to me that you have a problem grasping the
following: given a list of real numbers (i.e., a function from N to R),
we find an element which is not on this list. Could we not then just
add that element to our list?

The answer is: of course we can. However, by the same argument as
before, this list will still be missing some real number, and so on.
After all, we have proved the theorem not for a *particular* list of
real numbers, but for *all possible* lists!

Consider the following statements:

(i) For every real number x, there is a function f:N -> R whose image
contains x.

(ii) For every function f:N -> R, there is some real number x which is
*not* in the image of f.

Both of these statements are true, and there is no contradiction (look
at the order of the quantifiers)!
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Virgil
science forum Guru


Joined: 24 Mar 2005
Posts: 5536

PostPosted: Sun Jul 16, 2006 5:39 pm    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

In article <XfSdnQDiqbiIrifZnZ2dnUVZ_oWdnZ2d@speakeasy.net>,
Hatto von Aquitanien <abbot@AugiaDives.hre> wrote:

Quote:
Virgil wrote:

In article <iJOdnfATh6UuuCXZnZ2dnUVZ_qGdnZ2d@speakeasy.net>,
Hatto von Aquitanien <abbot@AugiaDives.hre> wrote:

Perhaps you can explain to me how one arrives at 1/3 = .333... without
appeal to an iterative process extended ad infinitum.

The process of multiplying the numerator by 10 then finding the integer
quotient and remainder on division by 3 is immediately seen to be
cyclic. Once something is cyclic like this, there is no need to view it
as an iterative process that needs extension ad infinitum.

That's what cyclic means.

Cyclic means that further repetition does not provide any different
result.
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Hatto von Aquitanien
science forum Guru


Joined: 19 Nov 2005
Posts: 410

PostPosted: Sun Jul 16, 2006 12:36 pm    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

Virgil wrote:

Quote:
In article <iJOdnfATh6UuuCXZnZ2dnUVZ_qGdnZ2d@speakeasy.net>,
Hatto von Aquitanien <abbot@AugiaDives.hre> wrote:

Perhaps you can explain to me how one arrives at 1/3 = .333... without
appeal to an iterative process extended ad infinitum.

The process of multiplying the numerator by 10 then finding the integer
quotient and remainder on division by 3 is immediately seen to be
cyclic. Once something is cyclic like this, there is no need to view it
as an iterative process that needs extension ad infinitum.

That's what cyclic means.

--
Nil conscire sibi
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Virgil
science forum Guru


Joined: 24 Mar 2005
Posts: 5536

PostPosted: Sun Jul 16, 2006 8:00 am    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

In article <iJOdnfATh6UuuCXZnZ2dnUVZ_qGdnZ2d@speakeasy.net>,
Hatto von Aquitanien <abbot@AugiaDives.hre> wrote:

Quote:
Nathan wrote:

Hatto von Aquitanien wrote:
Nathan wrote:

You're supposed to learn to understand the meaning of
the limit of a sequence, particularly the idea of the sum of an
infinite series as the limit of the sequence of partial sums.

Not everybody does learn to understand this, which explains the
.999...=1 threads.

Is that what happens to all the smart people in college? They have their
common sense psychologically pounded out of them. They either give lip
service to the orthodox canon, accept it as truth, or transfer to
computer science.

No, there's another option you left out; the one I gave.
They can also learn to understand the concept of a limit.
As I said, this is what is supposed to happen.
This doesn't require that they "accept it as truth", as some sort of
dogma.

There may indeed be those who "give lip service to the canon",
because they really don't grasp it. There certainly are those who
"accept it as truth"; it can be very hard to help certain students
to learn to understand concepts and not just to do mechanical
calculations. Some in that category do transfer to
computer science, where they still tend to have trouble
understanding concepts.

And no, this doesn't necessarily happen in college. I said
"around the second semester of calculus", which for a lot of
"the smart people" means in high school.

Perhaps you can explain to me how one arrives at 1/3 = .333... without
appeal to an iterative process extended ad infinitum.

The process of multiplying the numerator by 10 then finding the integer
quotient and remainder on division by 3 is immediately seen to be
cyclic. Once something is cyclic like this, there is no need to view it
as an iterative process that needs extension ad infinitum.
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Dik T. Winter
science forum Guru


Joined: 25 Mar 2005
Posts: 1359

PostPosted: Sat Jul 15, 2006 11:36 pm    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

In article <1152917292.168488.112220@i42g2000cwa.googlegroups.com> "David R Tribble" <david@tribble.com> writes:
Quote:
Hatto von Aquitanien wrote:
....
We have a tendency to accept .333... as being as good a specification of a
number as is 3, for example. That is to say "for all intents and
purposes, .333... is as good a specification of a number as is 3." But is
it? Are there some attributes which we should not attribute to .333...
which we attribute to 3 by virtue of its being a number?

1/3 = sum{1 to oo} 3 x 10^-1 = 0.333...

You should stress that sum{1 to oo} is *not* an iterative process. As it
stands, it appears that you are taking the sum of infinitely many terms,
and that is not the case. sum{1 to oo} is (in my opinion) abuse of
notation, as it is defined as lim{n -> oo} sum{1 to n}. Which (in turn)
also is not an iterative process (think epsilon and delta).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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Hatto von Aquitanien
science forum Guru


Joined: 19 Nov 2005
Posts: 410

PostPosted: Sat Jul 15, 2006 4:54 am    Post subject: Re: Attempts to Refute Cantor's Uncountability Proof? Reply with quote

Virgil wrote:

Quote:
Given the sequence f(n) = 1/n, and any epsilon greater than zero, one
can show that |1/n - 0| > epsilon only when n < 1/epsilon, so that it
only occurs for finitely many values for n.

What happens for n >= 1/epsilon is irrelevant to the proof.

It may, in this case be trivially dispensed with, but you have implicitly
considered every value of n >= 1/epsilon. Somewhere in your thought
process you have applied a rule that says for n >= 1/epsilon the truth
of |1/n| > epsilon doesn't change, and the statement is false under these
conditions. It may seem so blatantly obvious that you don't see a need to
write out the induction rule, but that really is the only formal way to say
that such values are "irrelevant".
--
Nil conscire sibi
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