Help with nested interval proof

• Nov 5th 2010, 08:06 AM
zebra2147
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.
• Nov 5th 2010, 08:26 AM
Plato
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 $\displaystyle \left( N \right)\left[ {x_0 - a_N < \frac{\delta }{2}\;\& \,b_N - x_0 < \frac{\delta }{2}} \right]$
• Nov 5th 2010, 08:27 AM
Drexel28
Quote:

Originally Posted by zebra2147
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?
• Nov 6th 2010, 12:50 PM
zebra2147
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 $\displaystyle (x_0-\delta,x_0+\delta)$.
• Nov 6th 2010, 01:04 PM
Drexel28
Quote:

Originally Posted by zebra2147
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.