# Nested Interval

• Sep 27th 2006, 11:24 AM
MKLyon
Nested Interval
I need to prove the following:
1.) The intersection (infinity on top, n=1 on bottom) of the open interval (0,1/n) = the null set
2.) the intersection (infinity on top, m=1 on bottom) of the interval [m, infinity) = the null set

How do you show these? Why don't they adhere to the Nested Interval Property?
• Sep 27th 2006, 12:01 PM
ThePerfectHacker
Quote:

Originally Posted by MKLyon
I need to prove the following:
1.) The intersection (infinity on top, n=1 on bottom) of the open interval (0,1/n) = the null set

If you had the closed intervals,
[0,1/n]
Then by the nested interval theorem, it would be a single point since lim 1/n---> 0
That point would be zero.
Since, these are open intervals that point (zero) is not included. Thus, the intersection is empty.
• Sep 27th 2006, 12:16 PM
Plato
Here again is another way.
In a previous reply I gave you this lemma.
If e>0 then is a positive integer K such that (1/K)<e.
Looking at the intersection of the collection of sets (0,1/n), if it were not empty then it would contain some e>0. But e is not in set (0,1/K) in the collection, hence a contradiction. The intersection must be empty.
• Sep 28th 2006, 11:05 AM
MKLyon
How would you go about proving the second one:

the intersection (infinity on top, m=1 on bottom) of the interval [m, infinity) = the null set.

Thanks for any help.
• Sep 28th 2006, 11:46 AM
Plato
Well, once again if there were a number in the intersection of the [m,oo) that number would be an upper bound for the positive integers. That is a contradiction.