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Math Help - Closed, non-empty and bounded sets.

  1. #1
    Senior Member slevvio's Avatar
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    Closed, non-empty and bounded sets.

    Suppose that for each  n \in \mathbb{N} we have a non-empty closed and bounded set  A_n \subset \mathbb{C} and

     A_1 \supseteq A_2 \supseteq ... \supseteq A_n \supseteq A_{n+1} \supseteq ...

    Prove that \bigcap_{n=1}^{\infty} A_n is non empty. [Hint: use Bolzano-Weierstrass]

    Solution:

    Let  x_n \in A_n . Then there exists a bounded sequence  \{ x_n \} with a convergent subsequence  \{x_{k_n} \} whose limit is  x \in A_n . I need to show that  x \in A_m \forall m \in \mathbb{N} . I can see that  \forall m \le n, x \in A_m since  A_1 \supseteq A_2 \supseteq ... \supseteq A_n \ni x . But how do I show that if  x \in A_n then  x \in A_{n+1} ? Any help would be appreciated.
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  2. #2
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    Let (x_n) be a sequence such that x_n\in A_n for all n then since A_n \subset A_1 for all n and A_1,A_n are closed and bounded there is a subsequence x_{n_k} with n_k<n_{k+1} such that x_{n_k} \rightarrow x. And by construction, and since the sequence is nested, x \in A_n for all n.
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  3. #3
    Senior Member slevvio's Avatar
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    sorry could you explain nested is that a precise mathematical definition?
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  4. #4
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    The fact that ...\subset A_{n+1} \subset A_n \subset ... \subset A_1 is expressed saying that it's a sequence of nested sets.
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  5. #5
    Senior Member slevvio's Avatar
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    ok i will have a look a this once i have had some sleep. thanks for the help.
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  6. #6
    Senior Member slevvio's Avatar
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    I have thought about this, does the way I have written this make sense and is it a valid proof?

    Let  \{x_n\}_{n=1}^{\infty} be a sequence such that  x_n \in A_n \forall n \in \mathbb{N} (i.e.  x_1 \in A_1, x_2 \in A_2 ... )

     \exists k_1 < k_2 < ... such that  \{ x_{k_n} \}_{n=1}^{\infty} converges to limit  x \in A_1 since  A_1 closed.

    Since the sets are nested the sequence  \{x_n \}_{n=m}^{\infty} is in  A_m with subsequence  \{x_{k_n} \}_{n=m}^{\infty} which converges to x as  \{x_{k_n} \}_{n=m}^{\infty} is a subsequence of a convergent sequence (the first m terms have simply gone).

    Hence  x \in A_m \forall m \in \mathbb{N} and so  x \in \bigcap_{n=1}^{\infty} A_n .
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