The following is the definition of compactness. It is a very useful notion which can be used to
prove existence theorems.
Definition B.2.1A set, K ⊆ ℝ^{p}, is said to be sequentially compact if whenever
{an}
⊆ K is a sequence, there exists a subsequence,
{an }
k
such that this subsequenceconverges to a point of K. Let C be a set of open sets in ℝ^{p}. It is called an open coverfor K if ∪C⊇ K. Then K is called compact if every such open cover has the propertythat there are finitely many sets of C, U_{1},
⋅⋅⋅
,U_{m}such that K ⊆∪_{i=1}^{m}U_{i}.
Lemma B.2.2If 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
(x,rx)
such that B
(x,rx)
∩B
(k,rx)
= ∅. Since K is compact,
finitely many of these balls B
(xi,rxi)
i = 1,
⋅⋅⋅
,m must cover K. But then one could consider
r = min
{rxi,i = 1⋅⋅⋅m }
> 0 and B
(k,r)
would contain points of K which are
not covered by
{B(xi,rxi)}
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.2.3Let I_{k} =
[ ]
ak,bk
and supposethat for all k = 1,2,
⋅⋅⋅
,
Ik ⊇ Ik+1.
Then there exists a point, c ∈ ℝ which is an element of every I_{k}. If
kli→m∞ bk − ak = 0
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 ℝ^{p}.
Definition B.2.4Thediameter of a set S, is defined as
diam (S) ≡ sup {||x− y|| : x,y ∈ S }.
Thus diam
(S)
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.2.5Let I_{k} = ∏_{i=1}^{p}
[ak,bk]
i i
≡
{x ∈ ℝp : xi ∈ [ak,bk]}
i i
and suppose that forall k = 1,2,
⋅⋅⋅
,
Ik ⊇ Ik+1.
Then there exists a point c ∈ ℝ^{p}which is an element of every I_{k}. If
lkim→∞ diam (Ik) = 0,
then the point c is unique.
Proof:For each i = 1,
⋅⋅⋅
,p,
[ ]
aki,bki
⊇
[ ]
aki+1,bki+1
and so, by the nested interval
lemma, there exists a point c_{i}∈
[ ]
aki,bki
for all k. Then letting c ≡
(c1,⋅⋅⋅,cp)
it follows
c ∈ I_{k} for all k. If the condition on the diameters holds, then the lengths of the intervals
lim_{k→∞}
[ ]
aki,bki
= 0 and so by the same lemma, each c_{i} is unique. Hence c is unique.
■
Theorem B.2.6Let
∏p
I = [ak,bk] ⊆ ℝp
k=1
Then I is sequentially compact and compact.
Proof:First consider compactness. Let I_{0} 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 I_{0}. Now we describe a
nested sequence of such products recursively as follows. I_{n} cannot be covered with finitely
many sets of C (like I_{0}) and if I_{n} = ∏_{k=1}^{p}
[ank,bnk]
, let c_{k}^{n} be
n n
ak+2bk
and consider all
products which are of the form ∏_{k=1}^{p}
[αk,βk]
where α_{k}≤ β_{k} and one of α_{k},β_{k} is c_{k}^{n}. Thus
there are 2^{p} of these products
{Jk}
_{k=1}^{2n
} and diam
(Jk)
= 2^{−1}diam
(In)
. If each can be
covered by finitely many sets of C then so can I_{n} contrary to assumption. Hence
one of these cannot be covered with finitely many sets of C. Call it I_{n+1}. Then
iterating the relation on the diameter, it follows that diam
(In+1)
≤ 2^{−n}diam
(I0)
and so the diameters of these nested products converges to 0. By the above nested
interval lemma, Lemma B.2.5, it follows that there exists a unique c in ∩_{n}I_{n}. Then c
is contained in some U ∈C and so, since the diameters of the I_{n} converge to 0,
for large enough n, it follows that I_{n}⊆ U also, contrary to assumption. Thus I is
compact.
Next let
{xn}
be a sequence in I. Using the same nested sequence bisection
method described above, we can choose a nested sequence
{Ik}
such that nested
lim_{k→∞}diam
(Ik)
= 0 and I_{k} contains x_{k} for infinitely many indices k. Letting c = ∩_{k=1}^{∞}I_{k}
one can choose an increasing sequence
{nk}
such that x_{nk}∈ I_{k}. Then it follows that
lim_{k→∞}x_{nk} = c. ■
A useful corollary of this theorem is the following.
Corollary B.2.7Let
{xk}
_{k=1}^{∞}be a bounded sequence. Then it has a convergentsubsequence.
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 ℝ^{p}, the two versions of compactness are equivalent.
Lemma B.2.8Let K≠∅ be sequentially compact in ℝ^{p}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
B
(k,δ)
is contained in some set of O for any k ∈ K. If not, there would be a sequence
{kn}
of points of K such that B
(kn,1∕n)
is not contained in any single set of of O. By sequential
compactness, there is a subsequence, still denoted as
{kn}
which converges to k ∈ K. Now
B
(k,δ)
⊆ O ∈O for some open set O. Consider n large enough that 1∕n < δ∕5 and also
k_{n}∈ B
(k,δ∕5)
. Then
B (k ,1∕n) ⊆ B (k ,δ∕5) ⊆ B (k,2δ∕5) ⊆ B (k,δ) ⊆ O
n n
contrary to the construction of the k_{n}. This shows the claim.
Now pick k_{1}∈ K. If B
(k1,δ)
covers K, stop. Otherwise pick k_{2}∈ K ∖ B
(k1,δ)
. If
B
(k1,δ)
,B
(k2,δ)
covers K, stop. Otherwise pick k_{3} 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
and K would fail to be sequentially compact. Therefore, there is some n such that
{B(ki,δ)}
_{i=1}^{n} is an open cover. However, from the choice of δ,B
(ki,δ)
⊆ O_{i}∈O and so
{O1,⋅⋅⋅,On }
is an open cover. ■
Conversely, we have the following.
Lemma B.2.9If K≠∅ is compact set in ℝ^{p}, then it is sequentially compact.
Proof:Let
{k }
n
be a sequence in K and it is desired to show it has a convergent
subsequence. Suppose it does not. For k ∈ K there must be a ball B
(k,δ )
k
which contains k_{k}
for only finitely many k since otherwise, one could extract a convergent subsequence by
Corollary B.2.7 which would be in K since K is closed (Lemma B.2.2). Since K is compact,
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 = ∏_{k=1}^{p}
[a ,b ]
k k
for some choice of intervals
[a ,b ]
k k
.
Theorem B.2.10A set K ⊆ ℝ^{p}is compact if and only if it is sequentiallycompact 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 x_{n}∈ K such that
∥xn∥
≥ n.
This sequence can have no subsequence which converges. Therefore, K cannot be
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
bounded.
Now suppose K is closed and bounded. Let I ⊇ K where I = ∏_{k=1}^{p}
[ak,bk]
. Now let
{kn}
⊆ K. 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
theorem. ■
Corollary B.2.11Suppose K_{i}is a compact subset of ℝ. Then K ≡∏_{i=1}^{p}K_{i}is acompact subset of ℝ^{p}.
Proof:This is easiest to see in terms of sequential compactness. Let
{xn}
_{n=1}^{∞} be a
sequence in K. Say x_{n} =
( 1 2 p )
xn xn ⋅⋅⋅ xn
. By sequential compactness of
each K_{i}, it follows that taking p subsequences, one can obtain a subsequence, still
denoted by
{xn }
such that for each i ≤ p,lim_{n→∞}x_{n}^{i} = x^{i}∈ K_{i}. Then x_{n}→ x ∈ K.
■
Since ℂ^{p} is just ℝ^{2p}, closed and bounded sets are compact in ℂ^{p} also as a special case of
the above.