Real Analysis True or false, justify your answer?

1) A subsequence of a bounded sequence is bounded.

2) A subsequence of a monotone sequence is monotone.

3) A subsequence of a convergent sequence is convergent

4) A sequence converges if it has a convergent subsequence.

1 Answer

  • 8 years ago
    Favorite Answer

    1) True, fairly trivially. All the values of the subsequence are taken from the sequence, which is bounded.

    As a formal proof, suppose a(n) is bounded, i.e. for all n in N, there exists an M such that |a(n)| <= M. If a(n(m)) is a subsequence of a(n), i.e. n(m) is a strictly monotone increasing sequence from N, then we have n(m) in N, thus |a(n(m))| < M.

    2) Also true. Taking a subsequence means a selection of values from your sequence taken *in order*. In the order given, the sequence points get uniformly larger or uniformly smaller, and taking a selection in order will not change this.

    Formally, if a(n) is a monotone increasing sequence, then a(m) >= a(n) when m > n. Suppose a(n(m)) is a subsequence, i.e. n(m) is a strictly monotone increasing sequence from N. Then, if i > j, then n(i) > n(j), and so a(n(i)) >= a(n(j)). Thus, a(n(m)) is monotone increasing. Decreasing is similar enough that you can do it yourself if you want.

    3) This is true as well. A sequence that converges to some L will be eventually closer to L than any given bound. If we take a subsequence, all that can happen is we miss some terms, and the sequence converges even faster.

    Formally, if a(n) ---> L, then for any e > 0, there exists some N such that n > N ===> |a(n) - L| < e. Fix some e and the corresponding N. Suppose n(m) is a strictly monotone increasing sequence from N. It is a property (provable by induction) of such sequences that, for all m in N, a(m) >= m. Therefore:

    m >= N ===> n(m) >= N ===> |a(n(m)) - L| < e

    So, the subsequence converges by the same choice of N (and hence at the same rate or faster).

    4) False! We just need a counterexample. The classic example:

    0, 1, 0, 1, 0, 1, 0, 1, ...

    It doesn't converge because it's not Cauchy, or because it has two subsequences that converge to different limits, or whatever other property you've proven about convergent sequences that it doesn't have. But, taking the even or odd indexed terms both yield constant subsequences, which are convergent.

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