Hi,

I have been strugling with a proof which I can't seem to understand.
It concerns theorem 28.1 of Ostrowski's "Solutions of equations and systems of equations"

Theorem 28.1
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Assume a bounded sequence S of points E_t in Rn for
which E(t+1) Et -> 0. Then the derived set S' of S is a continuum, if
Et does not converge in the usual sense.

2. Remark. We remind the reader that a continuum is defined
as a closed set of points which cannot be decomposed into the sum
of two closed sets of points without common points.

3. Proof of Theorem 28.1.
S' is obviously closed. Suppose we
have S' = C1 + C2 where C1 and C2 are both closed and have no
points in common. Then there exists a positive p such that the distance of any point of C1 from every point of C2 does not exceed p.

By hypothesis we have for a certain n0
|E(t+1) - Et| <+ p/3 for all t >= n0 (28.1)

Take a point P from C1 There exist then arbitrarily large indices
m > n0 such that |Em, P| < p/3. As the points Ev with v > m have
a cluster point in C2, there exist indices k > m such that |Ek, C2\ <= 2p/3.
Assume that k is the smallest such index. Then we have certainly
|E(k-1), C2| > 2p/3, and therefore by 28.1 |Ek, C2| > p/3.
We see that
p/3 < |Ek, C2| <= 2p/3, k > m. (28.2)

4. There exists therefore an infinite sequence of indices kl,k2,...
for which (28.2) holds. A cluster point c of this sequence belongs
to S and satisfies the relation

p/3 <= |c, C2| <= 2p/3

It therefore does not belong to C2. It must then lie in C1 while its distance from C2 is less than p. With this contradiction our theorem is proved.
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The first confusing part was
"any point of C1 from every point of C2 does not exceed p", but I think that they actually mean |C1, C2| >= p because of the last paragraph of the proof and the fact that this makes me understand the proof until the part about there being an infinite sequence of indices kl,k2,....
However I cannot understand why there must be an infinite amount of indices k1,k2,..., could anyone explain that to me ?

Thanks in advance for your help.

Sincerely,

Ben