Re: An uncountable countable set
- From: "Dik T. Winter" <Dik.Winter@xxxxxx>
- Date: Tue, 11 Jul 2006 12:09:08 GMT
In article <1152604358.062964.188890@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
Dik T. Winter schrieb:
> >
> > If that were true, there must be some n such that 10^(-n) equals 0.
>
> Indeed, if you consider 1.000... - 0.999... = 0
That is true because of the *definition* of that notation.
If we assume 0.999... to be a well-defined (i.e. really existing)
expression, then we need no "special definition" because (9+1) *
0.999... = 9 + 0.999... The only definition required (besides + and *)
is how to express real numbers in the decimal system.
So you have first to define addition and multiplication from infinite
strings. How do you do that?
Without a
special definition that notation has no meaning. (There do not exists
things like infinite sums in arithmetic.) So the definitions:
1.000... =(def) 1 + sum{i = 1..oo} 0/10^i
and
0.999... =(def) 0 + sum{i = 1..oo} 9/10^i
where sum{i = 1..oo} is defined as lim{n -> oo} sum{i = 1..n}.
Note, again, the limit implied in the notation. And, at no place
do we leave the finite real.
Archimedes already knew the infinite sum. It exists whether you think
it is necessary to define a limit or not. And at no place it leaves the
finite real.
In mathematics it is necessary to define things.
> Indeed, if you consider 1.000... - 0.999... = 0 and if you claim that
> all positions are enumerated by natural numbers, then there must be
> natural number n with 10^(-n) = 0.
Why? Can you *prove* that? Because that would mean that there is a
natural number that is not finite, but there is no such number.
I see no proof below, only vague remarks:
If all places of representations like 1.000... and 0.999... would exist
and could be indexed by natural numbers,
Well, we have the set of natural numbers, the first number is represented
as d0 = 1, d_i = 0 for i > 0, the second as d0 = 0, d_i = 9 for i > 0.
then there were only those n
by which these numbers differed: 10(-n) > 0.
As the representations differ in *every* digit position.
As they do not differ, we
would have 10(-n) = 0.
This is incomprehensible nonsense. The numbers they represent are the
same, it is the representations that are different. So how I defined
the number represented by such a representation above.
But you are correct, as there is no infinite
natural number, there is no such place. And that is precisely the
reason, why there are less than aleph_0 places available in 0.999... as
well as in K = 0.111... .
This is again incomprehensible.
Hence K = 0.111... does not exist.
As I defined that notation above as meaning:
sum{i = 1..oo} 1/10^i = lim{n -> oo} sum{i = 1..n} 1/10^i =
= lim{n -> oo} 10^(-1) * (10^(-n) - 1) / (10^(-1) - 1) =
= lim{n -> oo} (1 - 10^(-n))/9 = 1/9 - lim{n -> oo} 10^(-n) / 9 = 1/9
you are arguing that 1/9 does not exist. (When you want to attack
something in methematics it is better to use the definitions mathematics
supplies.)
> > There is an n such that for all p An[p] = K[p]
No, there is not such a list number An which can index every p. But
every p which can be indexed, can be indexed by one single list number
Ap which also can and does index all p' < p.
What logical reasoning do you use to come from the first to the second? I
asked already twice before, and have not yet seen an answer.
The reason lies in the fact, that in a linear sequence like this
sequence of list numbers
Sorry, this ia again not logical reasoning. Again, I ask what logical
steps you take to get from
For all p there is an n such that An[p] = K[p]
to
There is an n such that for all p An[p] = K[p],
I would think you should be able to answer that simple question.
0.1
0.11
0.111
...
it is never necessary to use two elements for the same purpose.
If you think the sentence "all positions of K = 0.111...are indexed by
list numbers" is not equivalent to the sentence "K
is in the list", then you seem to imply that more than one list number
is required to index the digits of K.
Yes, each digit position of K can be indexed by a number in the list. And
it is not equivalent to "K is in the list". The implication you mention
I do not understand. Of course to index the digits of K you need each and
every list number. For starters the first digit is indexed by the first
list number (list numbers considered as natural numbers), the second digit
is indexed by the second list number.
You do not need the first two list numbers 0.1 and 0.11, because the
third alone is sufficient: 0.111 does index the first three digits p =
1 to 3 of 0.111...
I am a bit at a loss here what you mean with "indexing".
That can be carried on. p = 1 to 4 in 0.111... can
be indexed by 0.1111. The digits p = 1 to n can be indexed by list
number n = 0.111...1 with n 1's. If, as you seem to imagine, it may
happen, that two list numbers are required to index some p, then one of
the two is smaller than the other. So the other will suffice for the
task of the smaller. This proves: If a p is finite and, hence, can be
indexed by finite list number, then all digits < p can be indexed by
the same list number. Therefore, all digits which can be indexed can be
indeed by one finite list number alone.
Pray prove the "therefore", because it does not follow. What is valid
for the finite is not necessarily valid for the infinite.
--
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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