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Math Help - Bounded Sequence

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    Junior Member Kanwar245's Avatar
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    Bounded Sequence

    Suppose (an) is bounded, and that every convergent subsequence of (an) has limit L. Prove that \lim_{n\rightarrow\infty}a_{n} = L
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    Re: Bounded Sequence

    Quote Originally Posted by Kanwar245 View Post
    Suppose (an) is bounded, and that every convergent subsequence of (an) has limit L. Prove that \lim_{n\rightarrow\infty}a_{n} = L
    Well any sequence is a subsequence of itself.
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    Junior Member Kanwar245's Avatar
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    Re: Bounded Sequence

    Okay so from Bolzano-Wiererstrass since a_n is bounded, there exists atleast one convergent subsequence. Let that subsequence be a_n itself and we're done?
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    Re: Bounded Sequence

    Quote Originally Posted by Kanwar245 View Post
    Okay so from Bolzano-Wiererstrass since a_n is bounded, there exists atleast one convergent subsequence. Let that subsequence be a_n itself and we're done?
    Suppose the sequence does not converge to L.

    What does that mean?

    If it does converge to L then if \varepsilon  > 0 and (L - \varepsilon ,L + \varepsilon ) contains almost of the terms. So what is the negation of that?
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    Re: Bounded Sequence

    Quote Originally Posted by Kanwar245 View Post
    Okay so from Bolzano-Wiererstrass since a_n is bounded, there exists atleast one convergent subsequence. Let that subsequence be a_n itself and we're done?
    No, you can say that there exist at least one convergent subsequence but you [b]cannot[b] assert that it happens to be [itex]\{a_n\}[/itself]. You still need to prove that the sequence is itself convergent. Do, rather, what Plato suggested.
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    Junior Member Kanwar245's Avatar
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    Re: Bounded Sequence

    Quote Originally Posted by HallsofIvy View Post
    No, you can say that there exist at least one convergent subsequence but you [b]cannot[b] assert that it happens to be [itex]\{a_n\}[/itself]. You still need to prove that the sequence is itself convergent. Do, rather, what Plato suggested.
    okay
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