Compactness
Intuitively compactness can be thought of as a notion of “quasi-finiteness”.
Definition 1 Let X be a topological space.
S An open cover of X is a collection {Uα } of open sets such that X = US
α . A subcover is a subcollection
{Uβ } ⊆ {Uα }, which is also a cover, ie. X = Uβ .
Definition 2 A topological space is said to be compact if every open cover has
a finite subcover.
Example 3 Any finite topological space is trivially compact. There are only
finitely many open sets and thus any open cover is necessarily finite.
Example 4 Rn is not compact. For we can take
S∞the following open cover of
n
Rn by open balls {B(m, 0)}∞
m=1 . Clearly R =
m=1 B(m, 0). If there were a
finite subcover, then we would have Rn = B(m, 0) for some m, which is clearly
not the case.
1 ∞
, 1 m=2
Example 5 The open interval (0, 1) is not compact. The open cover m
has no finite subcover.
Before we can exhibit a nontrivial example of a compact topological space,
we need to prove some preliminary results concerning properties of topological
spaces. First let us look at a slight reformulation of the previous definitions.
Definition 6 Let X be a topological space and let A be a subset.
S We say that
a collection {Uα } of open sets in X is a cover of A in
X
if
A
⊆
Uα and that
S
a subcollection {Uβ } ⊆ {Uα } is a subcover if A ⊆ Uβ . We say that A is a
compact subset of X if any open cover of A in X has a finite subcover.
It is clear that A is a compact subset of X if and only if A is compact topological
space with respect to the subspace topology induced by the topology on X. For
{Uα } is a open cover of A in X if and only if {Uα ∩ A} is a open cover of A
regarded as a topological space in its own right. Thus the notion of a compact
set A is absolute, not relative like the notion of open set, ie. it does not depend
on which topological space of which it is a subspace.
Proposition 7 If X is a compact topological space and A is a closed subset,
then A is also compact.
Proof. Let {Uα } be an open cover of A in X. Then {Uα } ∪ {X\A} is an open
cover of X. By compactness, this has a finite subcover. If we delete X\A from
this finite subcover (if it is inside the subcover), we obtain a finite subcover of
{Uα } which covers A.
Proposition 8 Let (X, d) be metric space regarded as a topological space. If A
is a compact subset of X, then it is closed.
1
We want to show X\A is open, ie. that anyn y ∈ X\A isoan interior
point. To see this consider the collection of open balls B d(a,y)
. This
2 ,a
Proof.
a∈A
is an open cover of A in X, since
A=
[
a∈A
{a} ⊆
[
B
a∈A
d(a, y)
,a .
2
n
n om
om
. Let r = min ( d(a2i ,y)
.
By compactness there is a finite subcover B d(a2i ,y) , ai
i=1
i=1
Then
m
m
[
[
d(ai , y)
d(ai , y)
d(ai , y)
A∩B(r, y) ⊆
B
, ai ∩B(r, y) ⊆
, ai ∩B
,y = ∅
B
2
2
2
i=1
i=1
Thus B(r, y) ⊆ X\A and y is an interior point of X\A. This concludes the
proof.
Remark 9 The previous result does not hold for arbitrary topological spaces.
For if X is a finite set containing at least two points and X is endowed with
the indiscrete topology T = {∅, X}, then any proper nonempty subset A is
trivially compact but not closed. For Proposition 8 to hold the topological space
has to satisfy the Hausdorff condition: any two distinct points have disjoint
neighborhoods. We leave it as an exercise to redo the above proof using this
condition.
We will also need the following generalization of the nested intervals property
of the real numbers.
Theorem 10 (Nested Boxes Theorem) Let
B1 ⊇ B2 ⊇ B3 ⊇ . . .
be a nested sequence of closed boxes in Rn , ie.
B m = [am1 , bm1 ] × [am2 , bm2 ] × · · · × [amn , bmn ].
m
Suppose that limm→∞ (bmi − ami ) = 0 for each i = 1, 2, . . . , m. Then ∩∞
m=1 B
is a single point.
Proof. For each i = 1, 2, . . . , m the intervals {[ami , bmi ]}∞
m=1 satisfy the nested
intervals condition. Hence ∩∞
m=1 [ami , bmi ] is a single point. Consequently
m
∞
∞
∩∞
= (∩∞
m=1 B
m=1 [am1 , bm1 ]) × (∩m=1 [am2 , bm2 ]) × · · · × (∩m=1 [amn , bmn ])
is a single point.
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Theorem 11 Let K(r, 0) ⊂ Rn be a closed box of radius r > 0 (in the box
metric) centered at 0, ie.
K(r, 0) = [−r, r] × [−r, r] × · · · × [−r, r] = {x ∈ Rn | |x|∞ ≤ r}.
Then K(r, 0) is compact.
Proof. Suppose that {Uα } is open cover of K(r, 0) in Rn which has no finite
subcover. We use the method of iterated bisection to derive a contradiction.
Denote K(r, 0) by K0 for short. In the first step, we subdivide K(r, 0) into 2n
boxes of radius 2r :
[ r r K0 = K(r, 0) = K
, e ,
2 2
e
where e ranges over all n-tuples (±1, ±1, . . . , ±1). For n = 2 we have the following picture of this subdivision:
6
6
•
r
−r r
2 , 2
•
r r
2, 2
•
?
•
−r −r
2 , 2
2r
•
r −r
2, 2
?
Now it can not be the case that finitely many Uα cover each K 2r , 2r e . For
then we could combine the 2n finite subcovers of each K 2r , 2r e into a finite
subcover of K0 . Thus there must at least one K 2r , 2r e which is not covered
by finitely many Uα . Pick one such and call it K1 for short.
Now iterate this process. Subdivide K1 into 2n boxes of radius 4r and pick one
which is not covered by finitely many Uα , etc. Thus we obtain a decreasing
sequence of boxes
K0 ⊃ K1 ⊃ K2 ⊃ . . .
with the following properties:
(a) Km has radius
r
2m
in the box metric.
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(b) Km is not covered by finitely many Uα .
We can now apply the nested boxes theorem: ∩∞
m=0 Km = {a} for some point a.
Now a ∈ Uα0 for some α0 . Since Uα0 is open, B∞ (, a) ⊆ Uα0 for some > 0.
Now pick m large enough so that 2rm < 2 . Then Km ⊆ B∞ (, a) ⊆ Uα0 . For if
b denotes the center of the box Km and x ∈ Km we have
|x − a|∞ ≤ |x − b|∞ + |b − a|∞ ≤
r
r
+ m < + =
m
2
2
2 2
Thus Km ⊆ Uα0 contradicting (b). This contradiction establishes the result.
Remark 12 This argument fails for infinite dimensional metric spaces. For
instance if we take
∞
R∞ = {{xn }∞
n=1 | {{xn }n=1 is a bounded sequence of real numbers} ,
(ie. we consider a sequence to be a ∞-tuple of real numbers) with the infinitedimensional analog of the box metric:
∞
∞
d∞ ({xn }∞
n=1 , {yn }n=1 ) = sup{|xn − yn |}n=1 ,
then
K(r, 0) = {{xn }∞
n=1 | |xn | ≤ 1 for all n}
fails to be compact. The above argument breaks down at the first step. For
the decomposition of K(r, 0) into boxes of radius 2r is an infinite decomposition.
We then can’t conclude that one of the boxes can’t be covered by finitely many
Uα .
However if we take a different kind of box in (R∞ , d∞ ), namely
{{xn }∞
n=1 | |xn | ≤ an for all n} ,
where {an } is a fixed sequence of positive real numbers which converges to 0,
then the above argument still works, and we can conclude that such boxes are
compact. These kinds of boxes are called Hilbert cubes.
An immediate consequence of the preceding result is the following.
Theorem 13 (Heine-Borel) Let Rn be endowed with a metric induced by a
norm. Then a subset A ⊆ Rn is compact if and only if it is closed and bounded.
Proof. We may as well use the box metric. If A is bounded, then A ⊆ K(r, 0)
for r large enough. Since K(r, 0) is compact by Theorem 11, and A is a closed
subset, A is compact by Proposition 7.
Conversely if A is a compact subset of Rn , then A is closed by Proposition 8.
Also A is bounded, since otherwise {B∞ (m, 0)}∞
m=1 would be an open cover of
A in Rn with no finite subcover.
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Remark 14 The Heine-Borel theorem does not work for all metrics on Rn
which are equivalent to the standard ones. If A is closed with respect to such a
metric, then it is closed in any of the standard metrics. However if A is bounded
with respect to such a metric, it need not be bounded with respect to one of
the standard metrics. This follows from the Exercise 15 below. However if d1
and d2 are comparable metrics on X, then a A ⊆ X is bounded in (X, d1 ) if
and only if it is bounded in (X, d2 ).
Exercise 15 If (X, d) is a metric space, then
d0 (x, y) =
d(x, y)
1 + d(x, y)
is an equivalent metric such that d0 (x, y) < 1 for all x, y ∈ X. Hints:
•
x
1+x+y
+
y
1+x+y
≤
x
1+x
+
y
1+y
for all nonnegative x, y.
• If {xn }∞
n=1 is a sequence of nonnegative real numbers, then
lim xn = 0
n→∞
⇐⇒
lim
n→∞
xn
=0
1 + xn
Proposition 16 Let f : X → Y be a continuous surjection between topological
spaces (ie. f is onto). If X is compact, so is Y .
Proof. Let {Vα } be an open cover of Y . Then f −1 (Vα ) is an open cover of
m
X. Thus there is a finite subcover f −1 (Vαi ) i=1 . It then follows that
m
[
Vαi =
i=1
m
[
f f
−1
(Vαi ) = f
i=1
m
[
!
f
−1
(Vαi )
= f (X) = Y.
i=1
m
Thus {Vαi }i=1 is a finite subcover of Y .
Corollary 17 (Extreme Value Theorem) Let A be a closed and bounded nonempty
subset of Rn (with respect to a metric induced by a norm). Let f : A → R be a
continuous function. Then f has an absolute maximum and minimum.
Proof.
By Theorem 13, A is compact. By Proposition 16, f (A) is compact. Again by Theorem 13, f (A) is closed and bounded. Since it is bounded,
sup f (A) and inf f (A) both exist. Since f (A) is closed, sup f (A) ∈ f (A) and
inf f (A) ∈ f (A). Hence sup f (A) is an absolute maximum of f and inf f (A) is
an absolute minimum of f .
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Remark 18 The extreme value theorem, when A is a closed finite interval, is of
fundamental importance in single variable calculus. Not only does it guarantee
the existence of maxima and minima, but it also underpins the proof of Rolle’s
theorem which in turn is used to prove the mean value theorem. That in turn
is the major ingredient in the proof of the fundamental theorem of calculus
Theorem 19 (Bolzano-Weierstrass) If (X, d) is a compact metric space, then
any sequence {xn } in X has a convergent subsequence.
Proof. We use proof by contradiction. Suppose that {xn } has no convergent
subsequence. Then for any z ∈ X, there is an open ball B(rz , z) of z such
that xn ∈ B(rz , z) for only finitely many n (possibly for no n). For otherwise
1
, z contains xn for infinitely many n. Pick
for m = 1, 2, 3, . . . , each B m
xn1 ∈ B (1, z), then xn2 ∈ B 21 , z with n2 > n1 , etc. Then {xnm } is a
subsequence converging to z.
Now {B(rz , z)}z∈X is an open
Sm cover of X. Hence there is finite subcover
{B(rzi , zi )}m
.
Then
x
∈
n
i=1
i=1 B(rzi , zi ) = X for only finitely many n. This
is a contradiction which establishes the theorem.
Definition 20 Let (X, d) be a metric space and let {xn } be a sequence in X.
We say that {xn } is a Cauchy sequence, if for any > 0 we can find an index
M such that d(xn , xm ) < for all m, n ≥ M (ie. the sequence bunches up).
We say that (X, d) is a complete metric space if every Cauchy sequence in X
converges.
Proposition 21 If (X, d) is a compact metric space, then it is complete.
Proof. Let {xn } be a Cauchy sequence. By Theorem 19, there is a convergent
subsequence {xni }, which converges to some z ∈ X. This means that for any
> 0 there is an I such that d(xni , z) < 2 for all i ≥ I. Now pick M such that
d(xn , xm ) < 2 for all m, n ≥ M . Pick j so that j ≥ I and nj ≥ M . Then for
n ≥ nj we have
d(xn , z) ≤ d(xm , xnj ) + d(xnj , z) <
+ =
2 2
Therefore limn→∞ xn = z.
Proposition 22 Let Rn be endowed with a metric d induced by a norm. Then
Rn is a complete metric space.
Proof. Let {xn } be a Cauchy sequence. Pick M so that d(xn , xm ) < 1 for
all m, n ≥ M . Let R be the largest of d(0, xi ) for i = 1, 2, . . . , M . Then
d(0, xn ) < R + 1, for all n. This is obviously true if n ≤ M . If n > M , then
d(0, xn ) ≤ d(0, xM ) + d(xM , xn ) < R + 1.
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Thus the Cauchy sequence is contained in a closed ball centered at 0 of radius
R + 1. By Theorem 13, this closed ball is compact. Hence by Proposition 21,
the Cauchy sequence converges. Thus (Rn , d) is complete.
Remark 23 As is the case with boundedness, completeness is not a topological
property. There are equivalent metrics d on Rn with (Rn , d) not being complete.
For example the metric d(x, y) = | arctan(x) − arctan(y)| is such a metric on R,
with {n}∞
n=1 being a Cauchy sequence in (R, d) which does not converge. [This
is left as an exercise.] Also as is the case with boundedness, if d1 and d2 are
comparable metrics (ie. strongly equivalent), then (X, d1 ) is complete if and
only if (X, d2 ) is complete.
Another important consequence of compactness for metric spaces is uniform
continuity.
Definition 24 Let f : (X, d1 ) −→ (Y, d2 ) be a function between metric spaces.
We say that f is uniformly continuous if for any > 0, there is a δ > 0 such
that for all x, z ∈ X
d1 (x, z) < δ
=⇒
d2 (f (x), f (z)) < .
Remark 25 Clearly uniform continuity implies continuity. In the definition of
continuity at z ∈ X is held fixed. In the definition of uniform continuity both x
and z are allowed to vary. It is also worth noting that uniform continuity is not
a topological notiion – it can not be reformulated in terms of open sets. This
is because the conditions d1 (x, z) < δ and d2 (f (x), f (z)) < specify subsets
of X × X and Y × Y respectively, rather than subsets of X and Y . There
is a generalization of the notion of metric space where the notion of uniform
continuity can be formulated. This is the notion of uniformity on a set X,
which specifies a collection of subsets of X × X satisfying certain properties.
However this notion is a much less useful generalization of metric space than is
the notion of topological space.
Theorem 26 Suppose f : (X, d1 ) −→ (Y, d2 ) is a continuous function between
metric spaces and X is compact. Then f is uniformly continuous.
Proof. We prove this by contradiction. Suppose that f is not uniformly continuous. Then for some > 0, there is no corresponding δ > 0. Taking
δ = n1 , this means that we can find xn and zn such that d1 (xn , zn ) < n1
∞
but d2 (f (x), f (z)) ≥ . These choices define sequences {xn }∞
n=1 , {zn }n=1 in
X. Since X is compact, by the Bolzano-Weierstrass theorem, we can find a
convergent subsequence {xnj }∞
j=1 . Let limj→∞ xnj = a.
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Next we note that limj→∞ znj = a. This follows from the squeeze theorem and
the triangle inequality:
0 ≤ d1 (a, znj ) ≤ d1 (a, xnj ) + d1 (xnj , znj ) ≤ d1 (a, xnj ) +
so
1
lim d1 (a, xnj ) +
=0
j→∞
j
=⇒
1
1
≤ d1 (a, xnj ) +
nj
j
lim d1 (a, znj ) = 0.
j→∞
Thus limj→∞ znj = a.
By continuity of f we must have limj→∞ f (xnj ) = f (a) and limj→∞ f (znj ) =
f (a). This leads
a large enough index j so that
to a contradiction. Choose
d2 f (xnj ), f (a) < 2 and d2 f (znj ), f (a) < 2 . Then we obtain
≤ d2 f (xnj ), znj ≤ d2 f (xnj ), f (a) + d2 f (znj ), f (a) < + = .
2 2
Thus < . This contradiction completes the proof.
Theorem 27 (Ultimate Heine-Borel Theorem) Suppose that (X, d) is a metric
space with the following two properties
(1) (X, d) is complete, ie. every Cauchy sequence converges.
(2) Any open ball of any radius R > 0 can be covered by finitely many open
balls of radius R2 .
Then a subset A ⊆ X is compact if and only if A is closed and bounded. Conversely if (X, d) is a metric space such A ⊆ X is compact if and only if A is
closed and bounded, then (X, d) satisfies (1) and (2).
Remark 28 Condition (2) is a very weak form of finite-dimensionality. See for
instance, Wikipedia articles on Hausdorff/Minkowski dimension.
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