Re: An uncountable countable set

In article <1152282364.618937.306970@xxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
*** T. Winter schrieb:

0.111... is not a natural number in unary notation. So it is inherently
different from all elements of the list.

> "To be different" means for all unary representations of n
> An : 0.111... - n =/= 0

How do you propose to define that subtraction when 0.111... is not a
natural number?

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.

I have no idea what you are meaning here.

> not forall n: 0.111... - n =/= 0

But that is a wrong conclusion. Let's call 0.111... (as a sequence of
digits) K. And let's define K[i] is the i'th digit of K and An[i] the
i'th digit of An. The following statement is correct:
There is no i such that for all n K[i] != An[i] (1)

> En : 0.111... - n = 0.

And that is wrong, because that means:
There is an n such that for all i K[i] = An[i]. (2)
pray tell us under what logical reasoning you transform (1) to (2).

And to simplify it, take A0 = 0.10, A1 = 0.01, K = 0.11. (1) is
satisfied, (2) is not satisfied.

Your error is the invention of numbers like 0.01. They are not unary
representations of natural numbers.
By construction of the list we have to compare K = 0.111... only with
numbers

1 0.1
2 0.11
3 0.111
...
n 0.111...1
...

Those are unary representations of 1, 2, 3, ..., n, ... which we may
abbreviate conveniently by n.

There are (by tertium non datur) two possible alternatives:
1) K has a 1 at a position p where no n has a 1.
E p A n : p =/= n
2) K has not a position p where no n has a 1.
~(E p A n : p =/= n)

If (1) is true, then the position p in question cannot be enumerated by
a natural number, hence it is undefined.
If (2) is true, then we observe that for any position where the 1
coincides in p and n also all positions with m < n do coincide.

Indeed.

The
case you mentioned above that position n may coincide but position n-1
may not, simply cannot occur.

Where do I mention such a case (except in my example to show the fallacy
of your reasoning).

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 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.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.

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