The following is the definition of compactness. It is a very useful notion which can be used to prove
existence theorems.
Definition 2.2.1A set, K ⊆ ℝ^{p}, is said to be sequentiallycompact if whenever
{an}
⊆ K is asequence, there exists a subsequence,
{ank}
such that this subsequence converges 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 calledcompactif every such open cover has the property that there are finitely many sets of C, U_{1},
⋅⋅⋅
,U_{m}such that K ⊆∪_{i=1}^{m}U_{i}.
Lemma 2.2.2If 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, 1.14.7. If the lengths of the intervals in this lemma converge to 0, then
there is only one point in all of them.
Lemma 2.2.3Let 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
k k
kli→m∞b − a = 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 2.2.4Thediameter 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 2.2.5Let I_{k} = ∏_{i=1}^{p}
[ k k]
ai,bi
≡
{ p [ k k]}
x ∈ ℝ : xi ∈ ai,bi
and suppose that for allk = 1,2,
⋅⋅⋅
,
I ⊇ I .
k k+1
Then there exists a point c ∈ ℝ^{p}which is an element of every I_{k}. If
lim diam (I ) = 0,
k→∞ k
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 2.2.6Let
∏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}
n n
[ak,bk]
, let c_{k}^{n} be
ank+bnk
2
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 2.2.5, 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 2.2.7Let
{xk }
_{k=1}^{∞}be a bounded sequence. Then it has a convergent subsequence.
Proof: The given sequence is contained in some I as in Theorem 2.2.6 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 2.2.8Let K≠∅ be sequentially compact in ℝ^{p}. Then K is compact.
Proof: Let O be an open cover. 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. Taking a subsequence, we can assume
|kn − k|
< 1∕n. 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 (kn,1∕n) ⊆ B (kn,δ∕5) ⊆ B (k,2δ∕5) ⊆ B(k,δ) ⊆ O
contrary to the construction of the k_{n}. This shows the existence of δ the Lebesgue number.
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 2.2.9If K≠∅ is a compact set in ℝ^{p}, then it is sequentially compact.
Proof:Let
{kn}
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,δk)
which contains k_{k} for only finitely many
indices k since otherwise, one could extract a convergent subsequence by Corollary 2.2.7 which would be in
K since K is closed (Lemma 2.2.2). 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. ■
None of this in these two lemmas depends on the context of ℝ^{n}. See Problem 11 below. However, if you
want to get compact being the same as closed and bounded, then you really do need to be in a finite
dimensional setting. See Problem 10.
This proves most of the following theorem. First, a set is bounded if it is contained in I = ∏_{k=1}^{p}
[ak,bk]
for
some choice of intervals
[ak,bk]
.
Theorem 2.2.10A set K ⊆ ℝ^{p}is compact if and only if it is sequentially compact if and only ifit 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 2.2.2. 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 2.2.6 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 2.2.11Suppose K_{i}is a compact subset of ℝ. Then K ≡∏_{i=1}^{p}K_{i}is a compact subsetof 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} =
( )
x1n x2n ⋅⋅⋅ xpn
. 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.