Compactness
Contents
3.8. Compactness#
The material in this section is primarily based on [2, 41].
Recall from Definition 1.21 that for a subset
3.8.1. Open Covers#
(Open cover)
A family of open subsets
(Lindelöf)
Every open cover of a subset of
Proof. We call a point
Let
be a subset of .Let
be an open cover of (possibly uncountable).Thus,
.For each
,Choose an index
such that .Pick a rational point
and a rational positive number such that .
Consider the collection
.Since the set of rational points is countable and the set of rational numbers is countable, hence the set of all open balls centered at rational points with rational radii is countable.
Hence,
which is a subset of open balls with rational points as centers and rational radii, is at most countable.Thus,
is an at most countable open cover of .Since each
is a subset of an , hence there exists an at most countable subcover of for .
3.8.2. Compact Sets#
(Compact set)
Let
(Compact metric space)
Let
The set
Consider the family of open intervals:
For every
, there is a natural number such that .Thus,
.Thus,
implying that
is an open cover of .At the same time for every
, we have:since
.Thus,
But there is no finite subcover of
in .If there was a finite subcover, we could pick a maximum
among those intervals.But then
won’t belong to any of those intervals in the finite subcover.Hence
is not compact.
We later show in Theorem 3.76
that every compact set is closed and bounded.
Hence, an easy way to say that
3.8.3. Characterization of Compactness#
We have defined compactness as a property where every open
cover can be reduced to a finite subcover.
The characterization of a property involves identifying
other properties which are equivalent in the sense that
property A
We will see later that compactness of a set
(Lebesgue number)
Let
If every sequence in
Any such number
Proof. Assume the claim is false.
Then, for each
, there exists some such that is not a subset of any .In particular, for each
, there exists some such that holds for each .Consider the sequence
.By the hypothesis, every sequence has a subsequence that converges to a point of
.Let
be the limit of such a subsequence of .Since
, for at least one .Pick some
such that .Choose some
such that .Recall that
is open and is its interior point. So such an can be chosen.
Now, select some
such that and .It follows that
.But this contradicts with the selection of
such that is not contained in any .Thus, a
must exist satisfying the condition that for each , for at least one .
(Existence of finite cover of open balls)
Let
This lemma simply claims that we can construct a finite open cover
of open balls for
Proof. Assume the claim to be false. Choose
Pick some
.Choose some
. We can choose since doesn’t have a finite open ball cover.Assuming that
have been chosen inductively, choose from the set .By design,
holds for every .Then, no subsequence of
can converge.This contradicts with the hypothesis that every sequence has a subsequence that converges in
.Hence, the claim must be true.
(Characterization of compactness)
Let
is compact.Every infinite subset of
has an accumulation point in .Every sequence in
has a subsequence which converges to a point of .
Proof. (1)
Let
be an infinite subset of .Assume, by way of contradiction, that
has no accumulation point in .Then for every
, there exists a deleted neighborhood of that is disjoint with .That is,
.Then,
can either be empty or it can at most contain .Thus,
.Consider the open cover
.It is easy to see that
.Since
is compact (by hypothesis), there exists a finite set such that .But then
Thus,
must be a finite set containing at most elements.We arrive at a contradiction as
is infinite.Hence,
has an accumulation point in .
(2)
We assume that every infinite subset of
has an accumulation point in .Let
be an arbitrary sequence of .If
is constant, then every subsequence is constant and convergent.If
takes a finite number of distinct values, then there is at least one value which must occur infinite times. Thus, at least one subsequence is a constant sequence and hence convergent.We are left with the case where
contains infinite distinct values.Then, we can choose a subsequence
of which consists of all distinct values; i.e., whenever .Let
.Assuming
have been selected, choose such that for .This is possible since
has infinite distinct values and so far only distinct values have been chosen.Now, let
.It is clear that
is a subsequence of consisting of all distinct values.
Consider the set
(not the sequence but the set).By our hypothesis (2), since
is an infinite subset of , hence it must have an accumulation point in .Let
be an accumulation point of .Assume that
for each . If for some , then drop the first elements of .This ensures that
for every .We will now select a subsequence of
that converges to .Towards this, choose
such that .Now, inductively, assuming that
have been chosen, choose such thatSince
, hence .Since
is an accumulation point of , hence for every , there exists a point such that .Thus, it is possible to pick a suitable
such that .Also, by design,
must be greater than as .Thus,
is a subsequence of .Hence,
is a subsequence of too.Since
, it follows that .Thus,
has a convergent subsequence which converges to a point of .
(3)
(Bolzano-Weierstrass property)
A set
A compact set has the Bolzano-Weierstrass property.
3.8.4. Closedness and Boundedness#
(Compact sets are closed and bounded)
A compact set is closed and bounded.
Proof. Assume
Choose an open cover for
as .Since
is compact, hence there exist finite set of points such that .Let
.For any
, choose such that and .Then, by triangle inequality, we have:
Thus,
is finite and is bounded.
We now show that if
Let
.Due to Theorem 3.31, there exists a sequence
of with .Due to Theorem 3.75,
has a subsequence that converges to a point of (since is compact by hypothesis).As per Theorem 3.35, subsequences of a convergent sequence converge to the same limit.
Thus,
must be in .Thus,
. Thus, . Thus, is closed.
Although every compact set is closed and bounded, the converse need not be true. See Example 3.29 for an example of closed and bounded set (in discrete space) which is not compact. In fact, discrete space is a complete metric space. Yet, it has closed and bounded sets which are not compact.
In the specific case of Euclidean spaces, all closed and bounded sets are compact too. See Heine-Borel theorem below.
3.8.5. Nested Sequence of Compact Sets#
(Nonempty nested intersection property)
Let
Then their intersection is nonempty; i.e.,
Proof. We prove this result by way of contradiction.
Assume for contradiction that
.Let
for every .Since
for every hence for every .Then
is a collection of open sets.Since
, henceHence
is an open cover for every as .In particular,
is an open cover for .Since
is compact, hence there exists a finite subcover of indexed by a finite set of natural numbers .Let
and the finite subcover be .Since
is a finite set, it has a maximum element.Let
.Then
is also a finite subcover of since .But then
since for every .Hence
.Then
.But
, hence a contradiction.Hence, the intersection of the nested compact sets must be nonempty.
3.8.6. Continuity#
(Continuous images of compact sets are compact)
Let
Proof. We prove this by showing that any open cover of
Let
be an open cover for .Then
.Since
is continuous, hence is an open subset of for every .Since
is compact, there exist indices such that .Then
Thus,
is compact.
3.8.7. Lipschitz Continuity#
Let
for every
Proof. We proceed as follows.
Since
is locally Lipschitz continuous on hence is continuous by Theorem 3.57.Thus, by Theorem 3.78,
is compact.Hence,
is closed and bounded.Thus, there exists
such that for any , .For contradiction, assume that
is not Lipschitz on .Then, there is no
such thatThen, there exist two sequences
and of such thatBut
.Thus,
Since
is compact, hence has a convergent subsequence.Let
be a convergent subsequence with .By compactness of
, .Then,
cannot be not locally Lipschitz continuous at .This contradicts our hypothesis.
Thus,
must be Lipschitz continuous on .
3.8.8. Homeomorphism#
(Homeomorphism preserves compactness)
Let
Proof. Let
Let
3.8.9. Compact Spaces#
Every closed subset of a compact space is compact.
Proof. Let
Let
be an open cover of . We have, . .Then,
.Since all
are subsets of , hence we can write it as: .Since
is closed, hence is open.Thus,
is an open cover of .But
is compact. Hence, there exist finite indices such that .But then
and imply that: .Thus,
is compact.
(Continuous maps are closed)
Let
Proof. Let
Due to Theorem 3.81,
is compact.Since
is continuous, hence, due to Theorem 3.78, is compact.As per Theorem 3.76, since
is compact, hence is closed.Thus,
maps every closed set to a closed set.Thus,
is a closed mapping.
Now, assume that
Thus,
exists. Let .Then, due to bijection property
holds for every subset of .Thus,
is a closed subset of whenever is a closed subset of .Thus, as per Theorem 3.42
, is continuous.Thus, both
and are continuous.Thus,
is a homeomorphism.
(Compact domain + Continuity = Uniform continuity)
Let
Proof. We proceed as follows.
Let
.Since
is continuous on , hence for every , there exists such that holds whenever .The collection of open balls
covers ; i.e., .Since
is compact, there exists a set of finite number of points such that .Now, let
.Since
is the minimum of a finite number of positive numbers, hence .Now, pick any
that satisfy .There exists an integer
such that (due to the finite open cover).Therefore,
.Now, by triangle inequality:
holds true.
Thus,
, since .Thus,
Thus,
is uniformly continuous.
3.8.10. Euclidean Spaces#
Recall that
By
The compact subsets of a Euclidean space are precisely those sets which are closed and bounded.
(Heine-Borel theorem)
A subset of a Euclidean space is compact if and only if it is closed and bounded.
Compare this to Heine-Borel theorem for the real line.
We established there that for a closed and bounded subset of
We start here with the definition that a compact set is one for which
any open cover can be reduced to a finite subcover.
We then show in this theorem that in the special case of
Euclidean spaces
Proof. We have shown in Theorem 3.76 that every compact set is closed and bounded.
For the converse, we assume
Since
is bounded, we can pick such that for all .Fix an element
.Let
be some arbitrary point of . Thenholds for every
.Thus, the set of real numbers consisting of the
-th coordinates of the elements of is a bounded set.Choose an arbitrary sequence
of .Recall from Bolzano Weierstrass theorem for real numbers that every bounded sequence of real numbers has a convergent subsequence.
Note that if we form the sequence of real numbers from any particular coordinate of
, (say first coordinates or second coordinates) then the sequence is bounded by .Every such sequence of real numbers (from a fixed coordinate of
) will have a convergent subsequence.Thus, there is a subsequence
of whose first coordinates form a sequence in that converges in .Now, we choose
as a subsequence of so that the corresponding sequence of second coordinates of converge in .Proceeding in this manner, after
steps, we have a subsequence of with the property that for each , the sequence of its -th coordinates forms a convergent subsequence in .Since each of the coordinates of
converges in , hence, converges in .But since
is closed, hence converges to a point of .Thus, every sequence of
has a convergent subsequence in .Thus, by Theorem 3.75,
is compact.
Note that the property convergence in individual coordinates
implies convergence in
(Attainment of minimum and maximum values)
Let
Proof. Let
Then,
is compact in due to Theorem 3.78.Then,
is closed and bounded due to Heine-Borel theorem.Since
is bounded, hence it has an infimum and supremum.Since
is closed, hence its infimum and supremum lie inside itself.Thus,
attains a maximum and minimum value in .
Every sequence in a closed and bounded set in
Proof. Let
By Theorem 3.84,
is compact.By Theorem 3.75, every sequence in
has a subsequence which converges to a point of .
Every bounded sequence in
Proof. Let
3.8.11. Totally Bounded Metric Spaces#
(Totally bounded space)
A metric space
(Compactness implies total boundedness)
A compact metric space is totally bounded.
Proof. Let
Let
.Consider the family of open balls
.Then
.Since
is compact, there is a finite subcover of open balls.Thus, for every
, there is a union of finite open balls.Thus,
is totally bounded.
We showed earlier in Example 3.24
that
However
Let
.Pick any
. Thus, .Consider the points
. . .Thus,
Thus,
with the caveat that the first and last balls are restricted within the set
.Thus, we have a finite union of open balls.
Thus,
is totally bounded.
(Completeness and Compactness)
A metric space is compact if and only if it is complete and totally bounded.
Proof. Let
is totally bounded (Theorem 3.88).Since
is compact, hence every sequence has a convergent subsequence that converges to a point in (Theorem 3.75 (3)).Thus, if
is a Cauchy subsequence of , then it has a subsequence with limit leading to .Thus, every Cauchy sequence of
converges in .Thus,
is complete.
For the converse, assume that
Let
be an infinite subset of .Since
is totally bounded, there exists a finite subset such thatWe can extend the open balls with closed balls without problem:
Thus:
Since
is infinite, there is some such that is an infinite set.If
were finite for each , then would be finite as a finite union of finite sets.
Since
is totally bounded, we can again find a finite subset such thatSince
is infinite, there is some such that is an infinite set.Proceeding inductively, if
have been chosen, we can choose such that the setis infinite.
Define
for each
.Then for each
: is nonempty and closed. is infinite. . . .
Due to Theorem 3.59, there exists
, such that for each .Now, if
, thenThus, for every
, we can pick such that for every , .Thus,
is an accumulation point of .Thus, every infinite subset of
has an accumulation point in .Thus,
is compact.
3.8.12. Equivalent Metrics#
(Metric equivalence and compactness)
Let
Then, a set
In other words, the compact sets in the two metric spaces are identical.
Proof. Assume
Let
be an open cover of in ; i.e. are open in .Since
and are equivalent, hence are open in too.Thus,
is an open cover for in too.Since
is compact in , hence, there exist finite indices such that .But then,
are open in too.Thus,
is a finite open subcover of in .Thus, every open cover of
in can be reduced to a finite subcover.Thus,
is compact in .
A similar reasoning establishes that if