The following is the definition of compactness. It is a very useful notion which can be used to
prove existence theorems.
Definition B.2.1 A set, K ⊆ ℝp, is said to be sequentially compact if whenever
⊆ K is a sequence, there exists a subsequence,
such that this subsequence
converges to a point of K. Let C be a set of open sets in ℝp. It is called an open cover
for K if ∪C⊇ K. Then K is called compact if every such open cover has the property
that there are finitely many sets of C, U1,
,Um such that K ⊆∪i=1mUi.
Lemma B.2.2 If K is a nonempty compact set, then it must be closed.
Proof: If it is not closed, then it has a limit point k not in the set. Thus, K is not a finite
set. For each x ∈ K, there is B
. Since K
finitely many of these balls B
must cover K
. But then one could consider
0 and B
would contain points of
not covered by
because it is a limit point. This is a contradiction.
Recall the nested interval lemma, 4.5.1. If the lengths of the intervals in this lemma
converge to 0, then there is only one point in all of them.
Lemma B.2.3 Let Ik =
and suppose that for all k
Then there exists a point, c ∈ ℝ which is an element of every Ik. If
then there is exactly one point in all of these intervals.
Proof: If the lengths converge to 0, then there can be no more than one point in all the
intervals since otherwise, eventually you would have two points in an interval which has
length smaller than the distance between the two points. ■
This generalizes right away to the following version in ℝp.
Definition B.2.4 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 B.2.5 Let Ik = ∏
and suppose that for
Then there exists a point c ∈ ℝp which is an element of every Ik. If
then the point c is unique.
Proof: For each i = 1,
and so, by the nested interval
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
= 0 and so by the same lemma, each
is unique. Hence c
Theorem B.2.6 Let
Then I is sequentially compact and compact.
Proof: First consider compactness. Let I0 be the given set and suppose it is not compact.
Then there is a set of open sets C which admits no finite sub-cover of I0. Now we describe a
nested sequence of such products recursively as follows. In cannot be covered with finitely
many sets of C (like I0) and if In = ∏
and consider all
products which are of the form
αk ≤ βk
and one of αk,βk
there are 2p
of these products
If each can be
covered by finitely many sets of C
then so can In
contrary to assumption. Hence
one of these cannot be covered with finitely many sets of C
. Call it In+1
iterating the relation on the diameter, it follows that diam
and so the diameters of these nested products converges to 0. By the above nested
interval lemma, Lemma
, it follows that there exists a unique c
. Then c
is contained in some U ∈C
and so, since the diameters of the In
converge to 0,
for large enough n,
it follows that In ⊆ U
also, contrary to assumption. Thus I
be a sequence in
. Using the same nested sequence bisection
method described above, we can choose a nested sequence
such that nested
= 0 and
for infinitely many indices k
. Letting c
one can choose an increasing sequence
xnk ∈ Ik.
Then it follows that
A useful corollary of this theorem is the following.
Corollary B.2.7 Let
k=1∞ be a bounded sequence. Then it has a convergent
Proof: The given sequence is contained in some I as in Theorem B.2.6 which was shown
to be a sequentially compact set. Hence the given sequence has a convergent subsequence.
In ℝp, the two versions of compactness are equivalent.
Lemma B.2.8 Let K≠∅ be sequentially compact in ℝp and let O be an open cover.
Then K is compact.
Proof: First, I claim there is a number δ > 0, called a Lebesgue number such that
is contained in some set of
for any k ∈ K
. If not, there would be a sequence
of points of
such that B
is not contained in any single set of of
. By sequential
compactness, there is a subsequence, still denoted as
which converges to
k ∈ K
⊆ O ∈O
for some open set O
. Consider n
large enough that 1∕n < δ∕
5 and also
kn ∈ B
contrary to the construction of the kn. This shows the claim.
Now pick k1 ∈ K. If B
stop. Otherwise pick k2 ∈ K ∖ B
stop. Otherwise pick k3
not covered. Continue this way
obtaining a sequence of points any pair further apart than δ
. Therefore, this process must
stop since otherwise, there would be no subsequence which could converge to a point of K
would fail to be sequentially compact. Therefore, there is some n
is an open cover. However, from the choice of δ,B
⊆ Oi ∈O
is an open cover.
Conversely, we have the following.
Lemma B.2.9 If K≠∅ is compact set in ℝp, then it is sequentially compact.
be a sequence in
and it is desired to show it has a convergent
subsequence. Suppose it does not. For k ∈ K
there must be a ball B
for only finitely many k
since otherwise, one could extract a convergent subsequence by
which would be in K
is closed (Lemma B.2.2
). Since K
there are finitely many of these balls which cover K
. But now, there are only finitely many
of the indices accounted for, a contradiction. Hence K
is sequentially compact.
This proves most of the following theorem. First, a set is bounded if it is contained in
I = ∏
for some choice of intervals
Theorem B.2.10 A set K ⊆ ℝp is compact if and only if it is sequentially
compact if and only if it is closed and bounded.
Proof: The first equivalence was established in the lemmas. Suppose then that K is
compact. Why is it bounded? If not, then there exist xn ∈ K such that
This sequence can have no subsequence which converges. Therefore, K
sequentially compact. Hence it is not compact either. Thus K
is bounded. It was shown
earlier that K
is closed, Lemma B.2.2
. Thus if K
is compact, then it is closed and
Now suppose K is closed and bounded. Let I ⊇ K where I = ∏
. By Theorem B.2.6
it has a subsequence converging to k ∈ I
. But K
is closed and
so k ∈ K
. Thus K
is sequentially compact and hence is compact by the first part of this
Corollary B.2.11 Suppose Ki is a compact subset of ℝ. Then K ≡∏
i=1pKi is a
compact subset of ℝp.
Proof: This is easiest to see in terms of sequential compactness. Let
sequence in K
. Say xn
By sequential compactness of
it follows that taking p
subsequences, one can obtain a subsequence, still
such that for each
i ≤ p,
= xi ∈ Ki
. Then xn → x ∈ K
Since ℂp is just ℝ2p, closed and bounded sets are compact in ℂp also as a special case of