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 Post subject: theorem 28.1 of Ostrowski's "Solutions of equations andPosted: Thu, 21 Jul 2011 18:30:33 UTC
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Joined: Thu, 21 Jul 2011 18:29:16 UTC
Posts: 3
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
__________________________________________________ ____________________
Assume a bounded sequence S of points E_t in Rn for
which E(t+1) â€” 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.
__________________________________________________ _______________________

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 ?

Sincerely,

Ben

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 Post subject: Posted: Thu, 21 Jul 2011 18:57:56 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
Because the distance between Et and E(t+1) goes to zero.

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 Post subject: Posted: Fri, 22 Jul 2011 09:36:39 UTC
 S.O.S. Newbie

Joined: Thu, 21 Jul 2011 18:29:16 UTC
Posts: 3

I started thinking about what you said.

Because the distance between consecutive terms goes to zero the terms have to alternate between being in a neighbourhood of C1 and C2.
So for each mi you choose (larger than the last chosen k(i-1)), there is always a ki after that in the sequence.
Is this correct ?

Cheers,

Ben

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 Post subject: Re: theorem 28.1 of Ostrowski's "Solutions of equationsPosted: Fri, 22 Jul 2011 11:26:05 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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I'm not sure what your notation and mean. However, the existence of a p such that "the distance of any point of C1 from every point of C2 does not exceed p" (are you sure it is "does not exceed"? Strange because you will later want <=2p/3 which suggests you want the Hausdorff distance between C_1 and C_2 to be at least p instead) requires some justification.

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 Post subject: Posted: Fri, 22 Jul 2011 13:24:26 UTC
 S.O.S. Newbie

Joined: Thu, 21 Jul 2011 18:29:16 UTC
Posts: 3
I think |C1, C2| means the shortest distance between both sets and |c, C2| means the shortest distance between c and C2.

I copied the text "does not exceed" from my book, but this is also the part that I found confusing. But if I replace this by "the distance between any point from C1 and any point of C2 is at least p" (i.e. |C1, C2| >= p), then the proof makes sense to me.

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