Date: Mar 10, 2013 5:02 AM Author: mueckenh@rz.fh-augsburg.de Subject: Re: Matheology § 222 Back to the roots On 9 Mrz., 23:53, William Hughes <wpihug...@gmail.com> wrote:

> On Mar 9, 5:57 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

>

> > What do you understand by the not changeable function (d)?

>

> If x is a potentially infinite sequence of 0's and 1's

> then we say (x) is associated to x, if (x) is an algorithm

> which given a natural number produces a finite string

> of 0's and 1's, such that for every natural number n,

> (x) produces the nth FIS of x.

Ok. So it is clear that a finite number has to be given and a finite

string is produced.

>

> We will say x is coFIS to (y) iff

>

> i. We have (x) associated to x and

> (y) associated to y

>

> ii. For every n, (x) and (y) produce the same

> finite string.

"Every given n" is tantamount to "there is a last given n".

This maximum is the same for line l_max and max FIS of d.

>

> Some comments.

>

> We can never have more than a limited number of the strings

> produced by (x) existing at any time, since we can never

> have more that a limited number of natural numbers.

Correct.

>

> We cannot show that x is coFIS to (y) by using (x)

> and (y) to produce "all possible strings" and

> comparing them, since "all possible natural numbers"

> does not exist. However, we may be able to use induction

> to show "For every n, (x) and (y) produce the same finite

> string"

Induction is fine, but also restricted to the (variable) maximum.

>

> To show that x is coFIS to (y), it is not enough

> to show that every existing FIS of x, is equal to an

> existing FIS of y.

More cannot be shown.

>

> d, the diagonal of the list L, is a potentially infinite

> sequence of 0's and 1's with associated (d):

> for any natural number n produce

> a sequence of n ones.

This sequence is identical to a line l_max of the list L, by

construction of d_max.

>

> l_k, the k_th line of L is a potentially infinite

> sequence of 0's and 1' with associated (l_k):

> for any natural number n produce a sequence

> of k 1's followed by a sequence of n-k 0's

>

> A line of L, l_k is findable iff the index

> k is findable.

>

> Do you agree with the statement

>

> There does not exist

> (in the sense of not findable)

> a natural number m such that

> the mth line of L is coFIS to

> (d)

No. The construction of (d) implies that for every d_k there is a line

l_k = d_1, ..., d_k, hence for every d_max there is a line l_max.

For reference, the list L is given by

1

1,2

1,2,3

...

Regards, WM