Re: An uncountable countable set


*** T. Winter schrieb:

In article <J21Gtn.BCI@xxxxxx> "*** T. Winter" <***.Winter@xxxxxx> writes:

Let me expand on this.

> In article <1152282364.618937.306970@xxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:

You start with the list of natural numbers in unary notation:
A0 = 0.
A1 = 0.1
A2 = 0.11
A3 = 0.111
...
calculate the diagonal number K:
K = 0.111...
and claim that K equals An for some value of n.

I stated: either it does or it does not.

Note the following:
> > It is not by pure accident that I chose 0.111... . This number has also
> > a meaning as decimal representation of 1/9. So we can do the
> > calculations.

Indeed. Let's do the calculations, and let each An also have the meaning as
decimal representation, so
An = (1 - 10^(-n))/9
and now you claim that for some n
(1 - 10^(-n))/9
eqals K = 1/9.

If that were true, there must be some n such that 10^(-n) equals 0.

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. Hence, this assumption is obviously
wrong.

> > may not, simply cannot occur. We get from (2)
> > A p E n : p = n
> > and by construction of n = 0.111...1 we obtain that all 1's in K
> > coincide with the 1's of n.
>
> Right. But this does *not* prove that there is an n that is equal to K.
> To reformulate clearly
> For all p there is an n such that An[p] = K[p]

This was not entirely correct:
For all p there is an n0 such that for all n >= n0 An[p] = K[p].
> this does *not* imply that
> There is an n such that for all p An[p] = K[p],
> which is what you are arguing.
>
> Pray explain how you come (in logical steps) from the first statement
> to the second.

Note that when we expand the An's with an infinite (countable, mind) sequence
of digits 0 we can do something different. After the decimal point we change
each 0 to 1 and each 1 to 0. We get the following sequence:
A0 = 0.11111111...
A1 = 0.01111111...
A2 = 0.00111111...
and so
K = 0.000000...
Now we see:
For all p there is an n0 such that for all n >= n0 An[p] = K[p].
and your claim is that from that it follows that
There is an n such that for all p An[p] = K[p],
But as An = 1/(9 . 10^n), this would imply that K (the limit of the sequence)
must be in the list and for some n, An must be 0.

So, are you now arguing that the limit of a sequence is an element of that
sequence? Or what else? And if so, for what natural number (eh, finite...)
is 1/(9 . 10^n) equal to 0?

Either K is in the list (which would contradict analysis in the same
way as my original example), then there is an n with 10^(-n) = 0. Or K
is not in the list. Then there must be a position which cannot be
enumerated by natural numbers (the unary representations of them are
now the zeros behind the decimal point). Then this position is
undefined, because it cannot be found and, hence, does not exist.

Regards, WM

.

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