The following is the definition of compactness. It is a very useful notion which can be used to prove
Definition 2.2.1 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,
such that K ⊆∪i=1mUi.
Lemma 2.2.2 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
is compact, finitely many of these
must cover K
. But then one could consider r
would contain points of
which are not covered by
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 2.2.3 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 2.2.4 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 2.2.5 Let Ik = ∏
and suppose that for all
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
. 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. ■
Theorem 2.2.6 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
. Thus there are
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
. Then 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
. 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 lim
= 0 and
infinitely many indices k
. Letting c
one can choose an increasing sequence
xnk ∈ Ik.
Then it follows that limk→∞xnk
A useful corollary of this theorem is the following.
Corollary 2.2.7 Let
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 ℝp, the two versions of compactness are equivalent.
Lemma 2.2.8 Let K≠∅ be sequentially compact in ℝp. 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
is contained in some set of
for any k ∈ K
. If not, there would be a sequence
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
. Taking a subsequence, we can assume
. Now B
⊆ O ∈O
for some open set O
. Consider n
large enough that 1∕n < δ∕
also kn ∈ B
contrary to the construction of the kn. This shows the existence of δ the Lebesgue number.
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
is an open cover. However, from the choice of δ,B
⊆ Oi ∈O
is an open cover.
Conversely, we have the following.
Lemma 2.2.9 If K≠∅ is a 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
since otherwise, one could extract a convergent subsequence by Corollary 2.2.7
which would be in
is closed (Lemma 2.2.2
). Since K
is compact, there are finitely many of these balls which cover
. But now, there are only finitely many of the indices accounted for, a contradiction. Hence K
sequentially compact. ■
None of this in these two lemmas depends on the context of ℝn. 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 = ∏
some choice of intervals
Theorem 2.2.10 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
cannot be sequentially compact. Hence it is not compact either. Thus K
bounded. It was shown earlier that K
is closed, Lemma 2.2.2
. Thus if K
is compact, then it is closed and
Now suppose K is closed and bounded. Let I ⊇ K where I = ∏
it has a subsequence converging to k ∈ I
. But K
is closed and so k ∈ K
. Thus K
sequentially compact and hence is compact by the first part of this theorem. ■
Corollary 2.2.11 Suppose Ki is a compact subset of ℝ. Then K ≡∏
i=1pKi is a compact subset
Proof: This is easiest to see in terms of sequential compactness. Let
be a sequence in
. Say xn
By sequential compactness of each Ki,
it follows that
subsequences, one can obtain a subsequence, still denoted by
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 the