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I am looking for classes of sequence, that converge iff they contain a converging sub-sequence.

  • The basic example of such sequences are monotone sequences of real numbers.
  • A more interesting examples comes from metric fixed point theory:
    Let $B$ be a Banach space and $f\colon B \to B$ be a continuous mapping that is non-expansive (i.e. $\lVert f(x) - f(y)\rVert \le \lVert x -y\rVert$).
    Define $x_{n+1} := \frac{1}{2} x_n + \frac{1}{2} f(x_n)$ for any startingpoint $x_0\in B$. This is the so called Krasnoselski iteration.
    One can show that any accumulation point $\tilde{x}$ of $(x_n)$ is a fixed point of $f$. Since $f$ is non-expansive, it follows that
    $\lVert x_{n+1}-\tilde{x}\rVert = \frac{1}{2}\lVert (x_{n}-\tilde{x}) + (f(x_{n}) - f(\tilde{x}))\rVert\le \lVert x_n -\tilde{x}\rVert$.
    Hence $(x_n)$ converges iff it contains a converging sub-sequence.

    This is a special case of Ishikawa's fixed point theorem. (The Krasnoselski-Mann iteration - a generalization of the Krasnoselski iteration - also has this property.)

I am interested in this sequence because they provide very nice applications of the Bolzano-Weierstrass principle.

Do you know of any other examples of sequences with this property?
Do you know other proofs that uses this property together with the Bolzano-Weierstrass principle to prove the convergence of a sequence?

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Much too trivial for an answer (perhaps even too trivial for a comment), but of course any Cauchy sequence has this property. (It had better be in a non-complete space, to be at least minimally interesting.) –  Harald Hanche-Olsen May 13 '10 at 16:23
    
This isn't quite what you're looking for, but the question reminds me of an interesting fact: a sequence converges to a point x if and only if for any subsequence, there's a sub-subsequence converging to x. –  Kevin Ventullo May 13 '10 at 16:46
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If you go beyond sequences, and to nets, then the ultranets satisfy this (and their analogue ultrafilters too): if there is a convergent subnet it is itself convergent. –  Henno Brandsma May 13 '10 at 18:31
    
@Henno: you are right - this even seems to hold for any ultra-limit and the sequence converging to it. I was hoping for a more constructive way to find such sequences, but this is definitely an answer to the first question. –  alexod May 14 '10 at 6:23
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1 Answer

up vote 2 down vote accepted

The following version of the mean ergodic theorem is taken from the book of Krengel, "ergodic theorems".

Let T be a bounded linear operator in a Banach space X. The Birkhoff averages are denoted by $A_n = {1\over n} \ \Sigma_{k=0}^{n-1} \ T^k$. Assume that the sequence of operator norms $||A_n||$ is bounded independently of $n$. Then for any x and y in B, the following is equivalent :

-- y is a weak cluster point of the sequence $(A_nx)$,

-- y is the weak limit of the sequence $(A_nx)$,

-- y is the strong limit of the sequence $(A_nx)$.

(note that we talk about cluster points instead of converging subsequences because we didn't assume B separable. Hence the weak topology is not necessarily metrizable.)

This theorem implies e.g. the ergodic theorem for Markov operators on $C(K)$ (sequential compactness follows from Azrela-Ascoli), or the ergodic theorem for power bounded operators defined on reflexive Banach spaces (sequential compactness follows from Eberlein-Smulian).

There is a whole set of theorems in ergodic theory along these lines. Let me mention the convergence of the one sided ergodic Hilbert transform, discussed in Cohen and Cuny (see Th 3.2) as another example.

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