Re: abundance of irrationals!)


mueck...@xxxxxxxxxxxxxxxxx wrote:
> Virgil wrote:
>
>
> > >
> > > > > My list:
> > > > > 0.1;
> > > > > 0.01; 0.11;
> > > > > 0.001; 0.101; 0.011; 0.111;
> > > > > ...
> > > > > contains every positive real number of (0,1).
>
>
> > But not the limit, which does not occur as a member of the list.
>
> OK. Drop the assertion that all real numbers are in my list.
>
> Do you agree that all digit-positions of all real numbers of (0,1)
are
> in my list?

I don't know what this means. What are the "digit-positions"
of real number?

Every number on your list has an m-th digit, b_m, for every
natural number m. But for every number on your list, all
of the b_m are equal to zero beyond some point.

> Remember: All digits of a limit of a series are already
> contained in its partial sums. Do you agree?

Let's see if I can make sense of that....

OK, I think you mean that you can pick m, and there always
exists a partial sum S(n) which is identical to the limit
S = lim(n->oo) S(n) in the first m digits. This would
seem to be a consequence of the fact that you can find
S(n) that is within 2^-m of S (if we are looking at
binary expansions).

Two comments on this assertion:

1. It's not true. Consider the increasing sequence 0.9,
0.99, 0.999, .... whose limit is 1. No member of this
series agrees with the limit in even a single decimal
place.

2. Even for sequences in which it is true that you will
eventually match the first m places for every m, for
instance the binary sequence 0.1, 0.101, 0.10101,...
(which converges to 1/3), it is not true that there
exists any partial sum which agrees in ALL digits.

> All these digit positions
> klm (with k: line-number, l: number of number in that line, m: number
> of digit of that number) are enumerated, thus they are countable.

The total number of digits which occurs in ONE countable list,
which is the set of valyes kl in your notation, is countable.
It is the cardinality of N x N.

Your addition of the index "m" is a red herring. You have
only one digit at each position. There's no third index.
I suspect you're trying to use it to enumerate all
digits in all possible expansions. But you're counting
incorrectly. If you let m run from 0 to 1, then you have
only allowed for 2 possible lists. I hope you realize
that more than 2 lists are possible.

> Do you agree that the cardinal number of all digit positions of all
> real numbers is not less than the cardinal number of all real
numbers?

I agree that the number you describe, 2^N, is not less than
|R|. But there are far fewer than 2^N combinations of klm.
Your index set is at most the cardinality of N x N x {0,1},
which is |N|.

- Randy

.

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