Re: Is continuum completely filled up?


"Randy Poe" <poespam-trap@xxxxxxxxx> wrote in message
news:1171305792.356636.212540@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

We don't. We define 0.999... as the limit of the sequence
S_n = sum(k=1,n) 9/10^k

Yes, and value of this sum = 1, at the same time.

The LIMIT of this SEQUENCE is 1.

I agree.

I accept this.
This is difinition of the sum of Cauchy sequence.

Yes. The "infinite sum" is not actually a sum. It
is a limit of a sequence of sums.

OK.

I accept your opinion.

Good. But the fact that you continue means you do
not in fact accept my opinion.

I accept your argument including above so far . I shall show the reason why
I
don't accept your conclusion below.


You know difinition of pi, but probably you cannot compare two infinite
decmals.

Yes, I can compare infinite decimals. For example,
I can ask whether 52163/16604 > pi, 52163/16604 < pi,
or 52163/16604 = pi. Both of these numbers have
infinite decimal expansions. Nevertheless, I can
compare them.

pi = 3.14159 26535 89793 238...
52163/16604 = 3.14159 238737 653...

Now in each case I have stopped at a finite number
of digits, and there are uncountably many numbers
which share those initial digits.

Nevertheless, I have enough information to determine
that 52163/16604 < pi. As I said, if any number is unequal
to pi, then it will differ at a finite digit. In this case,
it differs at the 7-th digit after the decimal point.

What is the limit of a sequence of sums? Does the limit belong
to a sequence, if this sequence is decided with all its infinite number
of finite size terms? That is, does pi belong to a sequence of its
increasing finite parts?
3
3.1
3.14
3.141
................
pi : Digits of this is all in finite turn.

But when we are given a decimal
3.14159... (When we assume that such thing exists),

Starting from basic axioms, we can prove that such a sequence
has a well-defined limit for all possible choices of digits.
Therefore we don't need to assume it exists, we prove
it exists.

Always question comes to axioms.

What question is that?

ZFC starts from infinite set, and proves the existence of infinite set. It
is the reason why you don't need to assume it exists. In your all possible
choices, negative one is not included.

can we know this to be
the same number as pai?

There are uncountably-many numbers which could be
represented by 3.14159... Exactly one member of that
uncountable set is pi.

You cannot but use AC to perform this.

What did I "perform"?

I shall state in other way. You can easily find pi among its resemblances.
But in neighborhood of pi on a line, there are many numbers which we
cannot find, until infinitely many counting have been done.

This premise the axiom of infinity
and
axiom of power set. If you want to use these axioms by no means, it is
sure
that my premise is incomvenient.

I do not know what you are trying to say.

I would restrain myself to explain more not to add more confusion. It's
trifle
in the whole context. but it is important to define the existence of
infinite
set at first.

By checking |pai - this number|?

"This number" is actually a pattern which is matched
by uncountably many numbers.

Certainly.

The circumstences
is the same also in case of 0.999....

No it's not, because the set of sequences meant by
that pattern has only one member, not uncountably
many members.

I don't think so.

The convention meaning of 0.999... is the limit of one
and only one particular sequence.

You think that 0.999... stands for more than one possible
sequence?

If you know all digits of pi, you might say that pi is the limit of one
and only one particular sequence.
0.999... has also possibility of other choice of digits in every units.
It means that 0.999... is surrounded by these numbers. It is the
same for two cases that they are clearly defined.
Difference is that pi cannot be calculated by formura of a sum of
infinite series. In this sense I accept your assertion.
As for difinition of the sum 0.999..., lim(n =>oo)9/10^n is problm.
I cannot think other value except 1 as the limit, but it is question
whether it is correctry 1 or not.

The sum of Cauchy sequense is made by difinition. And a
number "1" cannot be distinguished from other numbers in the limit of
its
unlimitedly small neighborhood.

What?

Please ignore. Only my expression is poor. I used similler expression
in another places.

We can pick up it individually, when we deal
with countable numbers.

What?

I shall explane when I should recover my spilits.

That member has a value. We can use the definition
of that value to find |1 - 0.999...| and prove that
it is equal to 0.

It is easy to say that 9 continue infinitely,

Yes, by convention, 0.999... means that all digits
are 9s.

But no such convention restricts the meaning of the
dots when we write 3.14159...

At first 0.999... came in mind of humans. And its meanig was thought
later.

What?

...........................................


but when we are given this number
actually, how do we distinguish this from other number?

By defining the meaning of the notation.

If you mean something different by 0.999... for
instance "a real number which starts with 0.999
but has arbitrary digits after that", then again
you're back to an uncountable set.

When we produce new number, we may do it by adding digits. If it is
infinite
decimal, it takes infinite procedure. When we are given a infinite
decimal,
we
need infinite procedure agane to verify its value.

We do not need to "verify its value" to reason about its
properties.

This is a point where opinions may be devided.

We do not need to define all of its digits to define
the number. We define it by its properties.

Yes I agree.

Can we use the way
which you have offered? And division of 1 by 3 never end.

Division of 1/3 ends after one step, with the rational
value 1/3.

In one sense, but I mean its calculation. We cannot know this result
wheter to be 1/3 or not.

Yes, we can know that 1 divided by 3 is 1/3 without
calculating all of its digits.

In fractional expression, Yes.

Thank you for long answer, and I am sorry. I am puzzling how to answer you
for similler question which seems not to come to agreement. My answer of
this time might be hard to understand, because I have wrote alredy similler
expression many times. I understand your opinion. It is one of logic. I
don't
know how do you think my and other similler opinions.
You may already know well deadly faults of my idea. But I 'll show my idea
more concretely, as you may find faults of it better.

Regards
Ozaki Toshiaki


.

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