Re: abundance of irrationals!)

In article <1115569851.909098.102670@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

> 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?

The mth number in WM's list is zero past the mth place.
In fact, for n < 2^m, the nth number is zerp past the mth place.

Consider the list as a whole.
It is the properties of the individual members of the list that are at
issue.

> Is there any natural
> number m such that any b_m cannot be found?

Irrelevant!

> Why should you require to
> have it in one of the earlier lines?

Nothing is required of any earlier lines, the only requirement is of the
line one is considering at any one moment, whether it has a last
non-zero digit or not. And it always does.

>Don't forget: There all m e N.

So?

> Your induction proof is valid only for finite subsets of N.

It is valid for each and every member of WM's list, so that EVERY member
of that list has a last non-zero digit.


> Otherwise
> you must accept that by Card({2,4,6,...,2n}) < 2n the finiteness of N
> has been proven.

I have seen no such proof, merely more hand waving.

> Further your critique would apply to Cantor's
> enumeration of he rationals too.

AS that ennumeration does not depend on decimal representations, it is
an entirely different issue.

But it is easy to construct injections of the rationals INTO the
naturals, which proves that Card(Q) <= Card(N), and even easier to
inject the naturals into the ratinals proving that Card(N) <= Card(Q).

> There is always a stage at which
> denominators > m+1 are missing yet.

Not the way I do it.

> 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.

More hand waving.
> >
> > > 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.

But {0,1,2,3,4,5,6,7,8,9}^N (the set of all functins from
{0,1,2,3,4,5,6,7,8,9} to N, is not
>
> > 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.

So?
> >
> > > 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.

WM clearly does not understand the difference berween N x N and 2^N.
The short notation for N x N is N^2, not 2^N.

For non-empty sets A and B, A^B is defined to mean the set of all
functions from B to A.
>
> Regards, WM
.

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