E.1 The Nested Interval Lemma
First, here is the one dimensional nested interval lemma.
Lemma E.1.1 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
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 in
bk − ak <
. This would be a contradiction. Hence
Definition E.1.2 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 E.1.3 Let Ik = ∏
and suppose that for all
Then there exists a point c ∈ ℝp which is an element of every Ik. If limk→∞diam
the point c is unique.
Proof: For each i = 1,
and so, by Lemma
, there exists a
point ci ∈
. 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
is unique. Hence c
is unique. ■