nested interval theorem
There are two consequences to nesting of intervals: for :
first of all, we have the inequality for , which means that the sequence is nondecreasing;
Therefore, the limit of the sequence exists, and is just the supremum, say (see proof here (http://planetmath.org/NondecreasingSequenceWithUpperBound)). Similarly the sequence is nonincreasing and bounded from below by all , where , and hence has an infimum .
Now, as the supremum of , for all . But because is the infimum of , . Therefore, the interval is non-empty (containing at least one of ). Since , every interval contains the interval , so their intersection also contains , hence is non-empty.
If is a point outside of , say , then there is some , such that (by the definition of the supremum ), and hence . This shows that the intersection actually coincides with .
Now, since , we have that . So . This means that the intersection of the nested intervals contains a single point . ∎
Remark. This result is called the nested interval theorem. It is a restatement of the finite intersection property for the compact set . The result may also be proven by elementary methods: namely, any number lying in between the supremum of all the and the infimum of all the will be in all the nested intervals.
|Title||nested interval theorem|
|Date of creation||2013-03-22 17:27:12|
|Last modified on||2013-03-22 17:27:12|
|Last modified by||pahio (2872)|