Re: 1-1/2+1/3-1/4+1/5-1/6+1/7

In article
<bda3cb47-3266-4f6d-9f5c-ce34029c8f03@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Han.deBruijn@xxxxxxxxxxxxxx wrote:

On 8 feb, 00:57, G. Frege <nomail@invalid> wrote:
On Thu, 07 Feb 2008 22:49:56 +0100, G. Frege <nomail@invalid> wrote:

Ooops... Of course

   s[A] := {s(x) : x e A}   (A c N)

   "the image of A under s"
                   ~~~     ~~~~

Am I allowed to hook in here, with one of my small but certain steps ?

Let's repeat the definition of S(). I hope this is the latest version
and that it has been found correct by you and Jesse:

y is in S(x) iff there is some n e N such that for all m e N ,
with m > n , y is in S_m(x) .
Here, S_m(x) = s o s o ... o s(x), where there are exactly
m compositions. (m e N)

I have a minor problem with this definition. It seems to me that n is
an unused quantity and that it could be left out. The definition then
would read as follows:

y is in S(x) iff , for all m e N , y is in S_m(x) .
Here, S_m(x) = s o s o ... o s(x), where there are exactly
m compositions.

Is this correct? And if not, why not?

If s is the successor function, and S_m is m-fold composition of s with
itself, as indicated, what does "y is in S_m(x)" mean?

Unless we are using something very like the the von Neumann model, with
0 = {} and s(x) = {x,{x}}, "in" makes no sense.

And if we ARE using that model, how does S_m(x) differ from x + m?

And if it does not differ, how does y in S(x) differ in your definition
from "for all m in N, y < m + x"?

And if it does not differ, how does S(x) differ from x?
.

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