Re: An uncountable countable set


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.

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.

> 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.

If all places of representations like 1.000... and 0.999... would exist
and could be indexed by natural numbers, then there were only those n
by which these numbers differed: 10(-n) > 0. As they do not differ, we
would have 10(-n) = 0. 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... . Hence K = 0.111... does not exist.

> > 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

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. But that is wrong, as you can see
from the following proof:

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... 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.

To put it in other words: If two list numbers are capable of indexing
the first p digits of 0.111..., then always one of them is sufficient.
If a list number An is capable of indexing some digit of 0.111... ,
then no other list number is required to index smaller indices. And if
another list number Am with m > n is capable of indexing even more
digits of 0.111..., then An is no longer required.

If you believe that it could be of any use to employ more than one
list number An for indexing any number, then you should be able to give
an example where more than one An are required. Of course it must be a
finite example, because there are, by definition, only finite list
numbers An = natural numbers.

Regards, WM

.

Relevant Pages

  • Re: An uncountable countable set
    ... That is true because of the *definition* of that notation. ... As the representations differ in *every* digit position. ... is required to index the digits of K. ... each digit position of K can be indexed by a number in the list. ...
    (sci.math)
  • Re: demonic numbers !
    ... >> But an interesting experiment might be a multiple choice question you ... This is nonsensical except to some computer programmers. ... ...and they learn that scientific notation is used as a terse notation ... for numbers whose significant digits lie far from the decimal point. ...
    (comp.lang.lisp)
  • Re: Recurring decimal - international question
    ... How would you indicate more than one repeating ... digit with the dot notation? ... one each over the first & last digits ... notation, but I'm interested to hear that it exists, because I re-invented ...
    (sci.math)
  • Re: Newbye quetion: Why a double can store more number than a long ?
    ... floating point numbers and why they have a greater range then eg. long. ... with 5 digits only ... Say you use 2 digits for the exponent ... sientific notation, are you?). ...
    (comp.lang.cpp)
  • Re: Newbye quetion: Why a double can store more number than a long ?
    ... floating point numbers and why they have a greater range then eg. long. ... with 5 digits only ... Say you use 2 digits for the exponent ... sientific notation, are you?). ...
    (comp.unix.programmer)