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Thread: Help with nested interval proof

  1. #1
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    Help with nested interval proof

    The proof I'm working on states:

    Let $\displaystyle [a_{n},b_{n}]$ be nested intervals such that $\displaystyle b_{n}-a_{n}$ is null. Let $\displaystyle x_{0}$ be the real number satisfying $\displaystyle \bigcap_{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$. Let $\displaystyle \delta >0$. Prove there is an $\displaystyle N$, such that $\displaystyle [a_{N},b_{N}]\subseteq (x_{0}-\delta ,x_{0}+\delta )$.

    Here is the direction that I think I need to go but I'm not sure...
    Since $\displaystyle \bigcap_{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$, and $\displaystyle x_{0}-\delta <x_{0}<x_{0}+\delta$ then there must be an interval $\displaystyle [a_{N},b_{N}]$ that is also contained in $\displaystyle (x_{0}-\delta ,x_{0}+\delta )$. But all that might be completely wrong.
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  2. #2
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    Here some useful observations.
    $\displaystyle \left( {\forall n} \right)\left[ {a_n \leqslant x_0 \leqslant b_n } \right]$.

    $\displaystyle \left( {a_n } \right) \to x_0 \;\& \;\left( {b_n } \right) \to x_0$

    If $\displaystyle \delta>0$ find an $\displaystyle \left( N \right)\left[ {x_0 - a_N < \frac{\delta }{2}\;\& \,b_N - x_0 < \frac{\delta }{2}} \right]$
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zebra2147 View Post
    The proof I'm working on states:

    Let $\displaystyle [a_{n},b_{n}]$ be nested intervals such that $\displaystyle b_{n}-a_{n}$ is null. Let $\displaystyle x_{0}$ be the real number satisfying $\displaystyle \bigcap_{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$. Let $\displaystyle \delta >0$. Prove there is an $\displaystyle N$, such that $\displaystyle [a_{N},b_{N}]\subseteq (x_{0}-\delta ,x_{0}+\delta )$.

    Here is the direction that I think I need to go but I'm not sure...
    Since $\displaystyle \bigcap_{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$, and $\displaystyle x_{0}-\delta <x_{0}<x_{0}+\delta$ then there must be an interval $\displaystyle [a_{N},b_{N}]$ that is also contained in $\displaystyle (x_{0}-\delta ,x_{0}+\delta )$. But all that might be completely wrong.
    Suppose not, then $\displaystyle [a_N,b_N]\supseteq (x_{0}-\delta,x_0+\delta)$ for all $\displaystyle N\in\mathbb{N}$ so that $\displaystyle \displaystyle (x_0-\delta,x_0+\delta)\subseteq\bigcap_{n\in\mathbb{N} }[a_n,b_n]$...but what's wrong with that?
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  4. #4
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    Well, we stated that $\displaystyle \bigcap_{n\in\mathbb{N}}[a_n,b_n]$ contained only one element, $\displaystyle x_{0}$, but the way you stated it in the last line of your last post, $\displaystyle \bigcap_{n\in\mathbb{N}}[a_n,b_n]$ contains numerous point is the interval $\displaystyle (x_0-\delta,x_0+\delta)$.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zebra2147 View Post
    Well, we stated that $\displaystyle \bigcap_{n\in\mathbb{N}}[a_n,b_n]$ contained only one element, $\displaystyle x_{0}$, but the way you stated it in the last line of your last post, $\displaystyle \bigcap_{n\in\mathbb{N}}[a_n,b_n]$ contains numerous point is the interval $\displaystyle (x_0-\delta,x_0+\delta)$.
    That's right.
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