# Thread: nested interval theorem with unbounded interval

1. ## nested interval theorem with unbounded interval

I'm trying to show that the Nested Interval Theorem does NOT hold if the "first" interval in the sequence is unbounded.

So, $\displaystyle \big\{I_n\big\}_{n=1} ^\infty$ where $\displaystyle I_{n} = [a_{n}, b_{n}], a,b \in R$

with$\displaystyle I_{1} = [a_{1}, \infty)$, for example, implies that $\displaystyle \bigcap_{n}^{\infty} A_{n} = \varnothing$

The only thing that makes sense to me that I can think of is that $\displaystyle a_{1}, a_{2}, a_{3} ... a_{n}$ is an increasing sequence with no upper bound so eventually there is a possibility that for some large N $\displaystyle [a_{N}, b_{N}]$ becomes an empty set. And the intersection of any set with the empty set is the empty set.

Any suggestion or tips? Thanks

2. ## Re: nested interval theorem with unbounded interval

Originally Posted by director
I'm trying to show that the Nested Interval Theorem does NOT hold if the "first" interval in the sequence is unbounded. So, $\displaystyle \big\{I_n\big\}_{n=1} ^\infty$ where $\displaystyle I_{n} = [a_{n}, b_{n}], a,b \in R$
with$\displaystyle I_{1} = [a_{1}, \infty)$, for example, implies that $\displaystyle \bigcap_{n}^{\infty} A_{n} = \varnothing$
I think that you need to tell us what you mean by " Nested Interval Theorem ", what is its statement?

The examples you gave are confusing. Look at this:
if $\displaystyle n\in\mathbb{Z}^+$ define $\displaystyle A_n=[n,\infty)$ then $\displaystyle \bigcap\limits_n {{A_n}} = \emptyset$.

3. ## Re: nested interval theorem with unbounded interval

Sorry for the confusion.

Originally Posted by Plato
if $\displaystyle n\in\mathbb{Z}^+$ define $\displaystyle A_n=[n,\infty)$ then $\displaystyle \bigcap\limits_n {{A_n}} = \emptyset$.
Plato, I found your answer to a problem similar to the one quoted above and it makes sense to me. I see why the intersection is empty. Thanks

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