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Math Help - Bounded Sequence having a convergent subsequence

  1. #1
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    Bounded Sequence having a convergent subsequence

    Ok, so i did this proof and I'm unsure about it?

    I wnat to prove that {a_n} has a convergent subsequence.

    Proof:
    Since {a_n} is a bounded sequence of real numbers, it must have at least one accumulation point. Let A be that accum pt. By definition, A is an accum pt of {a_n} iff each neighborhood contains points of {a_n} which are not A. Let \varepsilon>0 be arbitrary. So, we can let a_n _k elemnet of (A- \varepsilon,A+ \varepsilon). As \varepsilon--->0,(A- \varepsilon,A+ \varepsilon) accumulates at A. Therefore, A is an accum pt of a_n _k.

    I'm afraid I wasn't specific enough and should have included the case for finitely many distinct terms.
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    Ok, so i did this proof and I'm unsure about it?

    I wnat to prove that {a_n} has a convergent subsequence.

    Proof:
    Since {a_n} is a bounded sequence of real numbers, it must have at least one accumulation point. Let A be that accum pt. By definition, A is an accum pt of {a_n} iff each neighborhood contains points of {a_n} which are not A. Let \varepsilon>0 be arbitrary. So, we can let a_n _k elemnet of (A- \varepsilon,A+ \varepsilon). As \varepsilon--->0,(A- \varepsilon,A+ \varepsilon) accumulates at A. Therefore, A is an accum pt of a_n _k.

    I'm afraid I wasn't specific enough and should have included the case for finitely many distinct terms.
    If you can find a monotomic subsequence then that would complete the proof. See this and post #9.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    If you can find a monotomic subsequence then that would complete the proof. See this and post #9.
    .
    So this proves that the sequences accumulates at A? therefore, it converges?
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  4. #4
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    Quote Originally Posted by kathrynmath View Post
    By definition, A is an accum pt of {a_n} iff each neighborhood contains points of {a_n} which are not A.
    The above is the key to this proof.
    There is a point of the sequence a_{n_1 }  \in \left( {A - 1,A + 1} \right)
    Then \lambda _2  = \min \left\{ {\frac{1}{2},\left| {A - a_{n_1 } } \right|} \right\}\; \Rightarrow \;\left( {\exists a_{n_2 }  \in \left( {A - \lambda _2 ,A + \lambda _2 } \right)} \right)
    Now if k \geqslant 3\;\;\& \;\lambda _k  = \min \left\{ {\frac{1}{k},\left| {A - a_{n_{k - 1} } } \right|} \right\}\; \Rightarrow \;\left( {\exists a_{n_k }  \in \left( {A - \lambda _k ,A + \lambda _k } \right)} \right)

    Now clearly you have a sequence of distinct terms and \left( {a_{n_k } } \right) \to A<br />
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