Re: abundance of irrationals!)

Randy Poe 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).

> > 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?

Ask for the k-th digit of any real number of (0,1). You can find it in
the list, for the first time in line k in number l at position m = i.
>
> 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.

Beyond which point? Consider the list as a whole. Is there any natural
number m such that any b_m cannot be found? Why should you require to
have it in one of the earlier lines? Don't forget: There all m e N.
Your induction proof is valid only for finite subsets of N. Otherwise
you must accept that by Card({2,4,6,...,2n}) < 2n the finiteness of N
has been proven. Further your critique would apply to Cantor's
enumeration of he rationals too. There is always a stage at which
denominators > m+1 are missing yet. Why do you consider the next one
without problems in that proof?

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

> 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.

The limit *is* 0.999... . Do you think 0.999... =/= 1 ???
>
> 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.

Of course theer is not one partial sum. Otherwise there musr be a
number larger than any other. That is nonsense. But *all* partial sums
are in my list, by construction.
>
> > 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.

The third index enumerates the digits of the l-th number in the k-th
line. But it does not make a difference whether N x N or Nx N x N is
the cardinality of my list. It is countable.

> 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.

There is *one* list with a rather easy structure.
In line k = 3 for example there is for l = 1 the number 0.001 which for
m = 1 has the bit 0, for m = 3 it has bit 1.
>
> > 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|.

And that shows that R is countable by this proof whereas it is
uncountable by another. Therefore countability is an ill-defined
concept.

Regards, WM

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