Results 1 to 6 of 6

Math Help - If lim sup a_n = lim inf a_n = a, then a_n->a

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    If lim sup a_n = lim inf a_n = a, then a_n->a

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Or maybe perhaps someone can provide a different/simpler proof for theorem 1?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post
    Theorem 1: It can easily be seen that if S is the set of all sub sequential limits of \left\{a_n\right\} then \limsup\text{ }a_n=\sup\text{ }S and \liminf\text{ }a_n=\inf\text{ }S. But, since \lim\text{ }a_n exists and equals a we know that every subsequential limit converges to a so that \limsup\text{ }a_n=\sup\text{ }S=\sup\text{ }\{a\}=a=\inf\text{ }\{a\}=\inf\text{ }S=\liminf\text{ }a_n

    Theorem 2: Using the same idea as above we see that since a=\inf\text{ }S\leqslant s\leqslant a=\sup\text{ }S for every s\in S we see that S=\{a\} and so every subsequential limit converges to a. The conclusion follows.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    Smile

    Quote Originally Posted by Drexel28 View Post
    Theorem 1: It can easily be seen that if S is the set of all sub sequential limits of \left\{a_n\right\} then \limsup\text{ }a_n=\sup\text{ }S and \liminf\text{ }a_n=\inf\text{ }S. But, since \lim\text{ }a_n exists and equals a we know that every subsequential limit converges to a so that \limsup\text{ }a_n=\sup\text{ }S=\sup\text{ }\{a\}=a=\inf\text{ }\{a\}=\inf\text{ }S=\liminf\text{ }a_n

    Theorem 2: Using the same idea as above we see that since a=\inf\text{ }S\leqslant s\leqslant a=\sup\text{ }S for every s\in S we see that S=\{a\} and so every subsequential limit converges to a. The conclusion follows.
    Somehow my textbook presents lim sup in a different way. It defines the lim sup as simply the limit of the seqeunce of supremums of the tails, and nothing related to subsequences and subsequential limits. (so I admit there is some gap between your proof and my background)

    Is there an alternative way to prove theorem 1 using more basic definitions/principles? (e.g. epsilon-limit proofs, upper bounds, least upper bounds, etc.)

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post
    Somehow my textbook presents lim sup in a different way. It defines the lim sup as simply the limit of the seqeunce of supremums of the tails, and nothing related to subsequences and subsequential limits. (so I admit there is some gap between your proof and my background)

    Is there an alternative way to prove theorem 1 using more basic definitions/principles? (e.g. epsilon-limit proofs, upper bounds, least upper bounds, etc.)

    Thank you!
    I mean, it's just direct application. If \forall\varepsilon>0 there exists some N\in\mathbb{N} such that N\leqslant n\implies \left|a_n-a\right|<\varepsilon what can we say about \left|\sup_{N\leqslant n}a_n-a\right|?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by Drexel28 View Post
    I mean, it's just direct application. If \forall\varepsilon>0 there exists some N\in\mathbb{N} such that N\leqslant n\implies \left|a_n-a\right|<\varepsilon what can we say about \left|\sup_{N\leqslant n}a_n-a\right|?
    |a_n -a| < ε for all n≥N
    => a-ε < a_n < a+ε for all n≥N
    => in particular,
    a-ε < a_N < a+ε........................(1)

    sup a_n ≥ a_n for all n≥N (by definition of an upper bound)
    n≥N
    => in particular,
    sup a_n ≥ a_N............................(2)
    n≥N

    (1) and (2) show that we will always have
    sup a_n > a-ε
    n≥N

    But how to do it for the "upper" bound a+ε?
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum