Sequence of intervals

kef · #1 $Old$ November 3rd, 2009, 11:22 AM

Hi all,

I'm having a spot of bother on the following question; I simply do not know where to begin. Any help would be massively appreciated.

Cheers,
Kef

Mauritzvdworm · #2 $Old$ November 3rd, 2009, 12:18 PM

assume there are two such points such that $p,q\in J_n$ for all n then show that these two points will be the same. it is not difficult since $x_n\rightarrow 0 \text{ as } n\rightarrow\infty$

you can also show that there has to be at least one point and by the previously mentioned part there can then be only one

kef · #3 $Old$ November 3rd, 2009, 01:12 PM

I see, so I should go for the proof by contradiction method? Could you start me off on how to show that the points are the same? I'm having a lot of difficulty with this one.

Plato · #4 $Old$ November 3rd, 2009, 01:23 PM

Quote:

Originally Posted by kef $View Post$

I see, so I should go for the proof by contradiction method? Could you start me off on how to show that the points are the same? I'm having a lot of difficulty with this one.

By all means you can use contradiction.
But, first of all how do you know the intersection is not empty?
Do you know that monotone bounded sequences converge?
Then how could there be two points in the intersection?

kef · #5 $Old$ November 3rd, 2009, 02:32 PM

Sorry, I don't quite follow...

Plato · #6 $Old$ November 3rd, 2009, 02:39 PM

Quote:

Originally Posted by kef $View Post$

Sorry, I don't quite follow...

Can you prove that $(b_n)$ is a decreasing sequence?
Can you prove that $(a_n)$ is a increasing sequence?
Can you prove that $\left( {\forall n} \right)\left( {\forall m} \right)\left[ {a_n < b_m } \right]$ ?
Does that mean that both sequences converge? WHY?

kef · #7 $Old$ November 4th, 2009, 04:32 AM

Thanks a lot for the help guys, it's done now. Took me a while to understand what the question was actually asking.

Cheers.