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Math Help - Cauchy sequence

  1. #1
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    Cauchy sequence

    Prove that a sequence converges iff it is a Cauchy sequence.

    So the sequence  a_n \to L . For  \varepsilon > 0 there is an  N \in \bold{N} such that  j \geq N \implies |a_{n}-L| < \varepsilon/2 . And so  m,n \geq N \implies |a_{m}-a_{n}| = |a_{m}-L +L-a_{n}| \leq |a_{m}-L|+|L-a_{n}| < \varepsilon (by triangle inequality).

    Now suppose  a_n is a Cauchy sequence. Choose  \varepsilon = 1/2 . Then there is an  N \in \bold{N} such that  n,m \geq N \implies |a_{n}-a_{m}| < 1/2 . So  a_N - 1/2 < a_n < a_N+1/2 .

    Is this the correct way of proving this?
    Last edited by particlejohn; July 10th 2008 at 02:23 PM.
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    Prove that a sequence converges iff it is a Cauchy sequence.
    So the sequence  a_n \to L . For  \varepsilon > 0 there is an  N \in \bold{N} such that  j \geq N \implies |a_{n}-L| < \varepsilon/2 . And so  m,n \geq N \implies |a_{m}-a_{n}| = |a_{m}-L +L-a_{n}| \leq |a_{m}-L|+|L-a_{n}| < \varepsilon (by triangle inequality).
    That part is certainly correct and well done. What are you doing with the converse?

    If  a_n is a Cauchy sequence, then you need to find a number to which it converges.
    Again I donít know what theorems you have.

    Here is one way to proceed.
    1. Every sequence contains a monotone subsequence.
    2. Every Cauchy sequence is bounded.
    3. Every bounded monotone sequence converges.
    4. Prove the limit of that subsequence is the limit of the whole sequence.
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  3. #3
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    Quote Originally Posted by particlejohn View Post
    Now suppose  a_n is a Cauchy sequence. Choose  \varepsilon = 1/2 . Then there is an  N \in \bold{N} such that  n,m \geq N \implies |a_{n}-a_{m}| < 1/2 . So  a_N - 1/2 < a_n < a_N+1/2 .
    You proved the forward direction correctly. (In fact in any metric space a convergent sequence is Cauchy. However, not necessarily the other way around. When the converse hold we say the metric space is complete).

    The way I like to prove this fact is by introducing the notion of limsup's and liminf's. Let \{a_n\} be a sequence. If \{a_n\} is not bounded from above then we define \limsup a_n = \infty. Otherwise, we contruct the following sequence: s_1 = \sup\{ s_k|k\geq 1\} = \sup\{ s_1,s_2,...\}, s_2 = \sup\{ s_k|k\geq 2\} = \sup\{ s_2,s_3,...\}, in general s_n = \sup\{ s_k | k\geq n\}. Of couse these sequences make sense because we assumed the sequence was bounded from above. Let us define s_n's to be the superior sequence. Note that s_1\geq s_2\geq s_3 \geq ... . Thus, \{ s_n\} is a non-increasing sequence and it must have a limit (or -\infty). Now we define \limsup a_n = \lim s_n this is the limsup of a_n. The beauty of this is that the sequence a_n does not need to have a limit, not even \pm \infty divergence, but yet the limit superior always exists. Now we can define limit inferior in an analogus fashion. If \{a_n\} is not bounded from below we define \liminf a_n = -\infty. Otherwise construct i_n = \inf\{ s_k | k\geq n\} just like above. Now confirm that i_1 \leq i_2 \leq i_3 \leq ... . This means \{ i_n\} the inferior sequence is a non-decreasing sequence. Thus, it must have a limit (or +\infty). We define \liminf a_n = \lim i_n. Finally if \{a_n\} is bounded (meaning both from above and below) then s_n has a lower bound (i.e. i_1) which automatically means \limsup a_n is a real number. Likewise if \{a_n\} is bounded then i_n has an upper bound (i.e. s_1) which automatically means \liminf a_n is a real number.
    There are a number of nice properties that are not hard to prove, let L = \mathbb{R}\cup \{\pm \infty\}:
    • if \lim a_n = L then \liminf a_n = \limsup a_n = L
    • if \lim a_n = L then \liminf a_n \leq L \leq \limsup a_n
    • \liminf a_n \leq \limsup a_n
    • if \liminf a_n = \limsup a_n = L then \lim a_n = L


    We are now able to elegantly prove that Cauchy sequences are convergent. The first fact we will need is that is \{a_n\} is a Cauchy sequence then \{a_n\} is a bounded sequence. I leave that for you to prove, it happens to be easy. Since a_n is bounded it means \liminf a_n and \limsup a_n are real numbers. To show \{a_n\} is convergent is sufficies to use property # 4 from that list. Let \epsilon > 0. Since \{a_n\} is Cauchy it means there is N\in \mathbb{N} such that |a_n - a_m | < \epsilon for all n,m\geq N. Thus, a_m - \epsilon < a_n < a_m + \epsilon for all n,m\geq N. Thus, a_m + \epsilon is an upper bound for a_n, so s_N = \sup\{ s_k | k\geq N\} \leq a_m + \epsilon. But this show s_N - \epsilon < a_m i.e. s_N - \epsilon is a lower bound for a_m. Thus, s_N - \epsilon \leq \inf \{ a_k | k\geq N \} = i_N. This means \limsup a_n \leq s_N \leq i_N + \epsilon \leq \liminf a_n + \epsilon. This statement is true for any \epsilon > 0 so \limsup a_n \leq \liminf a_n. But by property #3 we have \liminf a_n \leq \limsup a_n. Thus, \liminf a_n = \limsup a_n. This prove that \{a_n\} is convergent.

    The reason why limsup's and liminf's are helpful in proving convergence is that they are always convergent if the sequence is bounded. And so dealing with them is always possible.
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    Quote Originally Posted by Plato View Post
    That part is certainly correct and well done. What are you doing with the converse?

    If  a_n is a Cauchy sequence, then you need to find a number to which it converges.
    Again I don’t know what theorems you have.

    Here is one way to proceed.
    1. Every sequence contains a monotone subsequence.
    2. Every Cauchy sequence is bounded.
    3. Every bounded monotone sequence converges.
    4. Prove the limit of that subsequence is the limit of the whole sequence.

    Now suppose  a_n is a Cauchy sequence. Choose  \varepsilon = 1/2 . Then there is an  N \in \bold{N} such that  n,m \geq N \implies |a_{n}-a_{m}| < 1/2 . So  a_N - 1/2 < a_n < a_N+1/2 for all  n \geq N .

    So  \{a_{n}: n \geq N \} is bounded. Also,  \{a_{n}: n < N \} is also bounded. The union,  \{a_{n}: n \geq N \} \cup  \{a_{n}: n < N \} = \{a_{n}: n \in \bold{N}\} is thus bounded. So it has a convergent subsequence  c_n \to L . For  \varepsilon _1 > 0 we choose  N \in \bold{N} such that  n,m \geq N \implies |a_{n}-a_{m}| < \varepsilon_{1}/2 . So  c_{k} = a_{m_{k}} such that  m_{k} \geq N and  |c_{m_{k}}-L| < \varepsilon_{1}/2 . For  n \geq N , we have  |a_{n}-L| = |a_{n}-c_{k}+c_{k}-L| \leq |a_{n}-c_{k}| + |c_{k}-L| < |a_{n}-a_{m_{k}}| + \varepsilon_{1}/2 < \varepsilon_1 . So  a_n \to L .  \square
    Last edited by particlejohn; July 10th 2008 at 02:46 PM.
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  5. #5
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    That's good. It works.
    You say what the b_n in the proof comes from.
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