8.10.2 The Nested Interval Lemma
First, here is the one dimensional nested interval lemma.
Lemma 8.10.2 Let Ik =
be closed intervals, ak ≤ bk, such that Ik ⊇ Ik+1 for all k. Then
there exists a point c which is contained in all these intervals. If
, then there
is exactly one such point.
Proof: Note that the
are an increasing sequence and that
is a decreasing sequence. Now
note that if
m < n
while if m > n,
It follows that am ≤ bn for any pair m,n. Therefore, each bn is an upper bound for all the am and so if
c ≡ sup
, then for each
, it follows that c ≤ bn
and so for all, an ≤ c ≤ bn
which shows that c
is in all
of these intervals.
If the condition on the lengths of the intervals holds, then if c,c′ are in all the intervals,
then if they are not equal, then eventually, for large enough k, they cannot both be contained
bk − ak <
. This would be a contradiction. Hence
Definition 8.10.3 The diameter of a set S, is defined as
is just a careful description of what you would think of as the diameter. It measures how
stretched out the set is.
Here is a multidimensional version of the nested interval lemma.
Lemma 8.10.4 Let Ik = ∏
and suppose that for all
Then there exists a point c ∈ ℝp which is an element of every Ik. If limk→∞diam
, then the point
c is unique.
Proof: For each i = 1,
and so, by Lemma
, there exists a point
. Then letting c ≡
c ∈ Ik
for all k
. If the condition on the
diameters holds, then the lengths of the intervals limk→∞
= 0 and so by the same lemma, each
unique. Hence c
is unique. ■
I will sometimes refer to the above Cartesian product of closed intervals as an interval to emphasize the
analogy with one dimensions, and sometimes as a box.