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

On Wed, 06 Feb 2008 09:05:09 +0100, Han de Bruijn
<Han.deBruijn@xxxxxxxxxxxxxx> wrote:


As others have said, the notation S(0) is not defined here or elsewhere,
so the proof you give is just silliness.

But let's try to define it. [...]

Let x, y be any sets and we define the set S(x) by:

y is in S(x) iff there is some n such that for all m > n: y is in S_m(x)

I already posted an approach similar to this one. (*sigh*)

In short

S(x) = lim S_n(x)
n

with some appropriate definition of "lim". (Halmos would call the one
proposed by Jesse F. Hughes "lim sup".)


Here, S_n(x) = s o s o ... o s(x), where there are exactly n compositions.

Right.


We may _define_ s^n _recursively_ for any n in N, using the following
defining functional equations:

s^0 = id (on N),
s^(n+1) = s o s^n (n e N).

Then we would get

s^n(x) = (s o ... o s)(x) = x + n
`----.----´
n-times
for any x e N.



Thus, S(x) we can understand as a kind of a limit,
specifically, a limit as defined above.

Exactly.

But now I have to digress:

With this definition at hand, the following is indeed a theorem.

Theorem: [...]

No, that's just nonsense. (Multiple nonsense, actually.)


Proof:

There can't be a proof for this nonsense. We only can prove _theorems_.
But no theorem has been stated here.


What do you think?


Theorem:

S(0) = N.

Proof: (Left as an exercise to the reader.)


F.

--

E-mail: info<at>simple-line<dot>de
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Relevant Pages

  • Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
    ... fed this fish to the penguins: ... In the lucid notation by G. Frege: ... S= lim S_n ...
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  • Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
    ... In the lucid notation by G. Frege: ... Well, actually, I used a slightly other notation. ... lim S_n= N. ... in this case the notion of /limit/ for sequences of sets. ...
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