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Thread: 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 $\displaystyle n \in \mathbb{N} $ we have a non-empty closed and bounded set $\displaystyle A_n \subset \mathbb{C} $ and

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

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

    Solution:

    Let $\displaystyle x_n \in A_n $. Then there exists a bounded sequence $\displaystyle \{ x_n \} $ with a convergent subsequence $\displaystyle \{x_{k_n} \} $ whose limit is $\displaystyle x \in A_n $. I need to show that $\displaystyle x \in A_m \forall m \in \mathbb{N} $. I can see that $\displaystyle \forall m \le n, x \in A_m $ since $\displaystyle A_1 \supseteq A_2 \supseteq ... \supseteq A_n \ni x $. But how do I show that if $\displaystyle x \in A_n $ then $\displaystyle x \in A_{n+1} $? Any help would be appreciated.
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  2. #2
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    Let $\displaystyle (x_n)$ be a sequence such that $\displaystyle x_n\in A_n$ for all $\displaystyle n$ then since $\displaystyle A_n \subset A_1$ for all $\displaystyle n$ and $\displaystyle A_1,A_n$ are closed and bounded there is a subsequence $\displaystyle x_{n_k}$ with $\displaystyle n_k<n_{k+1}$ such that $\displaystyle x_{n_k} \rightarrow x$. And by construction, and since the sequence is nested, $\displaystyle x \in A_n$ for all $\displaystyle n$.
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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 $\displaystyle ...\subset A_{n+1} \subset A_n \subset ... \subset A_1$ 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 $\displaystyle \{x_n\}_{n=1}^{\infty} $ be a sequence such that $\displaystyle x_n \in A_n \forall n \in \mathbb{N} $ (i.e. $\displaystyle x_1 \in A_1, x_2 \in A_2 $ ... )

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

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

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