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Thread: Sequences

  1. #1
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    Sequences

    For $\displaystyle n \in {\mathbb {N}}$, define $\displaystyle I_n$ = $\displaystyle [{a_n, b_n}]$ to be a sequence of closed and bounded intervals such that $\displaystyle I_1\supseteq {I_2} \supseteq {I_3} ...$

    Prove that there is a real number $\displaystyle x$ such that $\displaystyle x \in I_n$ for all $\displaystyle n$ i.e.

    $\displaystyle {^\infty}$
    $\displaystyle \bigcap {I_n}{\neq}{\oslash}$
    $\displaystyle _{i=1}$
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by yusukered07 View Post
    For $\displaystyle n \in {\mathbb {N}}$, define $\displaystyle I_n$ = $\displaystyle [{a_n, b_n}]$ to be a sequence of closed and bounded intervals such that $\displaystyle I_1\supseteq {I_2} \supseteq {I_3} ...$

    Prove that there is a real number $\displaystyle x$ such that $\displaystyle x \in I_n$ for all $\displaystyle n$ i.e.

    $\displaystyle {^\infty}$
    $\displaystyle \bigcap {I_n}{\neq}{\oslash}$
    $\displaystyle _{i=1}$
    This is a sequence of closed subsets of $\displaystyle [a_1,b_1]$ which have the FIP, apply $\displaystyle [a_1,b_1]$'s compactness.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by yusukered07 View Post
    For $\displaystyle n \in {\mathbb {N}}$, define $\displaystyle I_n$ = $\displaystyle [{a_n, b_n}]$ to be a sequence of closed and bounded intervals such that $\displaystyle I_1\supseteq {I_2} \supseteq {I_3} ...$

    Prove that there is a real number $\displaystyle x$ such that $\displaystyle x \in I_n$ for all $\displaystyle n$ i.e.

    $\displaystyle {^\infty}$
    $\displaystyle \bigcap {I_n}{\neq}{\oslash}$
    $\displaystyle _{i=1}$
    Since $\displaystyle \{a_{n}\}_{n\in\mathbb{N}}$ is nondecreasing and $\displaystyle \{b_{n}\}_{n\in\mathbb{N}}$ is increasing,
    we have $\displaystyle a:=\lim\nolimits_{n\to\infty}a_{n}$ and $\displaystyle b:=\lim\nolimits_{n\to\infty}b_{n}$.
    Then $\displaystyle (a+b)/2\in[a,b]=\cap_{n\in\mathbb{N}}[a_{n},b_{n}]$.
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