26.5
Compact Sets
The property of compactness is very important in topology and analysis. We provide a quick
review geared towards the study of surfaces and for details, we refer the reader to Munkres
[81], Schwartz [89]. In this section, we will need to assume that the topological spaces are
Hausdorff spaces. This is not a luxury, as many of the results are false otherwise.
There are various equivalent ways of defining compactness. For our purposes, the most
convenient way involves the notion of open cover.
Definition 26.20. Given a topological space, E, for any subset, A, of E, an open cover,
(Ui)i∈I, of A is a family of open subsets of E such that A ⊆
U
i∈I
i. An open subcover of an
open cover, (Ui)i∈I, of A is any subfamily, (Uj)j∈J, which is an open cover of A, with J ⊆ I.
An open cover, (Ui)i∈I, of A is finite if I is finite. The topological space, E, is compact if it
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CHAPTER 26. TOPOLOGY
is Hausdorff and for every open cover, (Ui)i∈I, of E, there is a finite open subcover, (Uj)j∈J,
of E. Given any subset, A, of E, we say that A is compact if it is compact with respect to
the subspace topology. We say that A is relatively compact if its closure A is compact.
It is immediately verified that a subset, A, of E is compact in the subspace topology
relative to A iff for every open cover, (Ui)i∈I, of A by open subsets of E, there is a finite
open subcover, (Uj)j∈J, of A. The property that every open cover contains a finite open
subcover is often called the Heine-Borel-Lebesgue property. By considering complements, a
Hausdorff space is compact iff for every family, (Fi)i∈I, of closed sets, if
F
i∈I
i = ∅, then
F
j∈J
j = ∅ for some finite subset, J , of I .
Definition 26.20 requires that a compact space be Hausdorff. There are books in which a
compact space is not necessarily required to be Hausdorff. Following Schwartz, we prefer
calling such a space quasi-compact.
Another equivalent and useful characterization can be given in terms of families having
the finite intersection property. A family, (Fi)i∈I, of sets has the finite intersection property
if
F
j∈J
j = ∅ for every finite subset, J , of I . We have the following proposition:
Proposition 26.21. A topological Hausdorff space, E, is compact iff for every family,
(Fi)i∈I, of closed sets having the finite intersection property, then
F
i∈I
i = ∅.
Proof. If E is compact and (Fi)i∈I is a family of closed sets having the finite intersection
property, then
F
F
i∈I
i cannot be empty, since otherwise we would have
j∈J
j = ∅ for some
finite subset, J, of I, a contradiction. The converse is equally obvious.
Another useful consequence of compactness is as follows. For any family, (Fi)i∈I, of closed
sets such that Fi+1 ⊆ Fi for all i ∈ I, if
F
i∈I
i = ∅, then Fi = ∅ for some i ∈ I . Indeed,
there must be some finite subset, J, of I such that
F
j∈J
j = ∅ and since Fi+1 ⊆ Fi for all
i ∈ I, we must have Fj = ∅ for the smallest Fj in (Fj)j∈J. Using this fact, we note that R
is not compact. Indeed, the family of closed sets, ([n, +∞[ )n≥0, is decreasing and has an
empty intersection.
Given a metric space, if we define a bounded subset to be a subset that can be enclosed
in some closed ball (of finite radius), then any nonbounded subset of a metric space is not
compact. However, a closed interval [a, b] of the real line is compact.
Proposition 26.22. Every closed interval, [a, b], of the real line is compact.
Proof. We proceed by contradiction. Let (Ui)i∈I be any open cover of [a, b] and assume that
there is no finite open subcover. Let c = (a + b)/2. If both [a, c] and [c, b] had some finite
open subcover, so would [a, b], and thus, either [a, c] does not have any finite subcover, or
[c, b] does not have any finite open subcover. Let [a1, b1] be such a bad subinterval. The
same argument applies and we split [a1, b1] into two equal subintervals, one of which must be
bad. Thus, having defined [an, bn] of length (b − a)/2n as an interval having no finite open
26.5. COMPACT SETS
755
subcover, splitting [an, bn] into two equal intervals, we know that at least one of the two has
no finite open subcover and we denote such a bad interval by [an+1, bn+1]. The sequence
(an) is nondecreasing and bounded from above by b, and thus, by a fundamental property
of the real line, it converges to its least upper bound, α. Similarly, the sequence (bn) is
nonincreasing and bounded from below by a and thus, it converges to its greatest lowest
bound, β. Since [an, bn] has length (b − a)/2n, we must have α = β. However, the common
limit α = β of the sequences (an) and (bn) must belong to some open set, Ui, of the open
cover and since Ui is open, it must contain some interval [c, d] containing α. Then, because
α is the common limit of the sequences (an) and (bn), there is some N such that the intervals
[an, bn] are all contained in the interval [c, d] for all n ≥ N, which contradicts the fact that
none of the intervals [an, bn] has a finite open subcover. Thus, [a, b] is indeed compact.
The argument of Proposition 26.22 can be adapted to show that in
m
R , every closed set,
[a1, b1] × · · · × [am, bm], is compact. At every stage, we need to divide into 2m subpieces
instead of 2.
The following two propositions give very important properties of the compact sets, and
they only hold for Hausdorff spaces:
Proposition 26.23. Given a topological Hausdorff space, E, for every compact subset, A,
and every point, b, not in A, there exist disjoint open sets, U and V , such that A ⊆ U and
b ∈ V . As a consequence, every compact subset is closed.
Proof. Since E is Hausdorff, for every a ∈ A, there are some disjoint open sets, Ua and Vb,
containing a and b respectively. Thus, the family, (Ua)a∈A, forms an open cover of A. Since
A is compact there is a finite open subcover, (Uj)j∈J, of A, where J ⊆ A, and then
U
j∈J
j
is an open set containing A disjoint from the open set
V
j∈J
j containing b. This shows that
every point, b, in the complement of A belongs to some open set in this complement and
thus, that the complement is open, i.e., that A is closed.
Actually, the proof of Proposition 26.23 can be used to show the following useful property:
Proposition 26.24. Given a topological Hausdorff space, E, for every pair of compact
disjoint subsets, A and B, there exist disjoint open sets, U and V , such that A ⊆ U and
B ⊆ V .
Proof. We repeat the argument of Proposition 26.23 with B playing the role of b and use
Proposition 26.23 to find disjoint open sets, Ua, containing a ∈ A and, Va, containing B.
The following proposition shows that in a compact topological space, every closed set is
compact:
Proposition 26.25. Given a compact topological space, E, every closed set is compact.
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CHAPTER 26. TOPOLOGY
Proof. Since A is closed, E − A is open and from any open cover, (Ui)i∈I, of A, we can form
an open cover of E by adding E − A to (Ui)i∈I and, since E is compact, a finite subcover,
(Uj)j∈J ∪ {E − A}, of E can be extracted such that (Uj)j∈J is a finite subcover of A.
Remark: Proposition 26.25 also holds for quasi-compact spaces, i.e., the Hausdorff separa-
tion property is not needed.
Putting Proposition 26.24 and Proposition 26.25 together, we note that if X is compact,
then for every pair of disjoint closed, sets A and B, there exist disjoint open sets, U and V ,
such that A ⊆ U and B ⊆ V . We say that X is a normal space.
Proposition 26.26. Given a compact topological space, E, for every a ∈ E, for every
neighborhood, V , of a, there exists a compact neighborhood, U , of a such that U ⊆ V
Proof. Since V is a neighborhood of a, there is some open subset, O, of V containing a. Then
the complement, K = E − O, of O is closed and since E is compact, by Proposition 26.25, K
is compact. Now, if we consider the family of all closed sets of the form, K ∩F , where F is any
closed neighborhood of a, since a /
∈ K, this family has an empty intersection and thus, there
is a finite number of closed neighborhood, F1, . . . , Fn, of a, such that K ∩ F1 ∩ · · · ∩ Fn = ∅.
Then, U = F1 ∩ · · · ∩ Fn is a compact neigborhood of a contained in O ⊆ V .
It can be shown that in a normed vector space of finite dimension, a subset is compact
iff it is closed and bounded. For
n
R , the proof is simple.
In a normed vector space of infinite dimension, there are closed and bounded sets that
are not compact!
More could be said about compactness in metric spaces but we will only need the notion
of Lebesgue number, which will be discussed a little later. Another crucial property of
compactness is that it is preserved under continuity.
Proposition 26.27. Let E be a topological space and let F be a topological Hausdorff space.
For every compact subset, A, of E, for every continuous map, f : E → F , the subspace f(A)
is compact.
Proof. Let (Ui)i∈I be an open cover of f(A). We claim that (f−1(Ui))i∈I is an open cover of
A, which is easily checked. Since A is compact, there is a finite open subcover, (f −1(Uj))j∈J,
of A, and thus, (Uj)j∈J is an open subcover of f(A).
As a corollary of Proposition 26.27, if E is compact, F is Hausdorff, and f : E → F
is continuous and bijective, then f is a homeomorphism. Indeed, it is enough to show
that f −1 is continuous, which is equivalent to showing that f maps closed sets to closed
sets. However, closed sets are compact and Proposition 26.27 shows that compact sets are
mapped to compact sets, which, by Proposition 26.23, are closed.
26.5. COMPACT SETS
757
It can also be shown that if E is a compact nonempty space and f : E → R is a continuous
function, then there are points a, b ∈ E such that f(a) is the minimum of f(E) and f(b)
is the maximum of f (E). Indeed, f (E) is a compact subset of R and thus, a closed and
bounded set which contains its greatest lower bound and its least upper bound.
Another useful notion is that of local compactness. Indeed, manifolds and surfaces are
locally compact.
Definition 26.21. A topological space, E, is locally compact if it is Hausdorff and for every
a ∈ E, there is some compact neighborhood, K, of a.
From Proposition 26.26, every compact space is locally compact but the converse is false.
It can be shown that a normed vector space of finite dimension is locally compact.
Proposition 26.28. Given a locally compact topological space, E, for every a ∈ E, for every
neighborhood, N , of a, there exists a compact neighborhood, U , of a, such that U ⊆ N.
Proof. For any a ∈ E, there is some compact neighborhood, V , of a. By Proposition 26.26,
every neigborhood of a relative to V contains some compact neighborhood U of a relative
to V . But every neighborhood of a relative to V is a neighborhood of a relative to E and
every neighborhood N of a in E yields a neighborhood, V ∩ N, of a in V and thus, for every
neighborhood, N , of a, there exists a compact neighborhood, U , of a such that U ⊆ N.
It is much harder to deal with noncompact surfaces (or manifolds) than it is to deal with
compact surfaces (or manifolds). However, surfaces (and manifolds) are locally compact and
it turns out that there are various ways of embedding a locally compact Hausdorff space into
a compact Hausdorff space. The most economical construction consists in adding just one
point. This construction, known as the Alexandroff compactification, is technically useful,
and we now describe it and sketch the proof that it achieves its goal.
To help the reader’s intuition, let us consider the case of the plane,
2
R . If we view the
plane,
2
3
R , as embedded in 3-space, R , say as the xOy plane of equation z = 0, we can
consider the sphere, Σ, of radius 1 centered on the z-axis at the point (0, 0, 1) and tangent
to the xOy plane at the origin (sphere of equation x2 + y2 + (z − 1)2 = 1). If N denotes
the north pole on the sphere, i.e., the point of coordinates (0, 0, 2), then any line, D, passing
through the north pole and not tangent to the sphere (i.e., not parallel to the xOy plane)
intersects the xOy plane in a unique point, M , and the sphere in a unique point, P , other
than the north pole, N . This, way, we obtain a bijection between the xOy plane and the
punctured sphere Σ, i.e., the sphere with the north pole N deleted. This bijection is called
a stereographic projection. The Alexandroff compactification of the plane puts the north
pole back on the sphere, which amounts to adding a single point at infinity ∞ to the plane.
Intuitively, as we travel away from the origin O towards infinity (in any direction!), we
tend towards an ideal point at infinity ∞. Imagine that we “bend” the plane so that it
gets wrapped around the sphere, according to stereographic projection. A simpler example
takes a line and gets a circle as its compactification. The Alexandroff compactification is a
generalization of these simple constructions.
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CHAPTER 26. TOPOLOGY
Definition 26.22. Let (E, O) be a locally compact space. Let ω be any point not in E,
and let Eω = E ∪ {ω}. Define the family, Oω, as follows:
Oω = O ∪ {(E − K) ∪ {ω} | K compact in E}.
The pair, (Eω, Oω), is called the Alexandroff compactification (or one point compactification)
of (E, O).
The following theorem shows that (Eω, Oω) is indeed a topological space, and that it is
compact.
Theorem 26.29. Let E be a locally compact topological space. The Alexandroff compactifi-
cation, Eω, of E is a compact space such that E is a subspace of Eω and if E is not compact,
then E = Eω.
Proof. The verification that Oω is a family of open sets is not difficult but a bit tedious.
Details can be found in Munkres [81] or Schwartz [89]. Let us show that Eω is compact. For
every open cover, (Ui)i∈I, of Eω, since ω must be covered, there is some Ui of the form
0
Ui = (E − K
0
0) ∪ {ω}
where K0 is compact in E. Consider the family, (Vi)i∈I, defined as follows:
Vi = Ui if Ui ∈ O,
Vi = E − K if Ui = (E − K) ∪ {ω},
where K is compact in E. Then, because each K is compact and thus closed in E (since E
is Hausdorff), E − K is open, and every Vi is an open subset of E. Furthermore, the family,
(Vi)i∈(I−{i0}), is an open cover of K0. Since K0 is compact, there is a finite open subcover,
(Vj)j∈J, of K0, and thus, (Uj)j∈J∪{i0} is a finite open cover of Eω.
Let us show that Eω is Hausdorff. Given any two points, a, b ∈ Eω, if both a, b ∈ E, since
E is Hausdorff and every open set in O is an open set in Oω, there exist disjoint open sets,
U, V (in O), such that a ∈ U and b ∈ V . If b = ω, since E is locally compact, there is some
compact set, K, containing an open set, U , containing a and then, U and V = (E −K)∪{ω}
are disjoint open sets (in Oω) such that a ∈ U and b ∈ V .
The space E is a subspace of Eω because for every open set, U, in Oω, either U ∈ O
and E ∩ U = U is open in E, or U = (E − K) ∪ {ω}, where K is compact in E, and thus,
U ∩ E = E − K, which is open in E, since K is compact in E and thus, closed (since E
is Hausdorff). Finally, if E is not compact, for every compact subset, K, of E, E − K is
nonempty and thus, for every open set, U = (E −K)∪{ω}, containing ω, we have U ∩E = ∅,
which shows that ω ∈ E and thus, that E = Eω.
Finally, in studying surfaces and manifolds, an important property is the existence of a
countable basis for the topology. Indeed, this property guarantees the existence of triangua-
tions of surfaces, a crucial property.
26.5. COMPACT SETS
759
Definition 26.23. A topological space E is called second-countable if there is a countable
basis for its topology, i.e., if there is a countable family, (Ui)i≥0, of open sets such that every
open set of E is a union of open sets Ui.
It is easily seen that
n
R
is second-countable and more generally, that every normed
vector space of finite dimension is second-countable. It can also be shown that if E is a
locally compact space that has a countable basis, then Eω also has a countable basis (and
in fact, is metrizable). We have the following properties.
Proposition 26.30. Given a second-countable topological space E, every open cover (Ui)i∈I,
of E contains some countable subcover.
Proof. Let (On)n≥0 be a countable basis for the topology. Then, all sets On contained in
some Ui can be arranged into a countable subsequence, (Ωm)m≥0, of (On)n≥0 and for every
Ωm, there is some Ui such that Ω
. Furthermore, every U
m
m ⊆ Uim
i is some union of sets Ωj ,
and thus, every a ∈ E belongs to some Ωj, which shows that (Ωm)m≥0 is a countable open
subcover of (Ui)i∈I.
As an immediate corollary of Proposition 26.30, a locally connected second-countable
space has countably many connected components.
In second-countable Hausdorff spaces, compactness can be characterized in terms of ac-
cumulation points (this is also true for metric spaces).
Definition 26.24. Given a topological Hausdorff space, E, given any sequence, (xn), of
points in E, a point, l ∈ E, is an accumulation point (or cluster point) of the sequence (xn)
if every open set, U , containing l contains xn for infinitely many n.
Clearly, if l is a limit of the sequence, (xn), then it is an accumulation point, since every
open set, U , containing a contains all xn except for finitely many n.
Proposition 26.31. A second-countable topological Hausdorff space, E, is compact iff every
sequence, (xn), has some accumulation point.
Proof. Assume that every sequence, (xn), has some accumulation point. Let (Ui)i∈I be some
open cover of E. By Proposition 26.30, there is a countable open subcover, (On)n≥0, for E.
Now, if E is not covered by any finite subcover of (On)n≥0, we can define a sequence, (xm),
by induction as follows:
Let x0 be arbitrary and for every m ≥ 1, let xm be some point in E not in O1 ∪ · · · ∪ Om,
which exists, since O1 ∪ · · · ∪ Om is not an open cover of E. We claim that the sequence,
(xm), does not have any accumulation point. Indeed, for every l ∈ E, since (On)n≥0 is an
open cover of E, there is some Om such that l ∈ Om, and by construction, every xn with
n ≥ m + 1 does not belong to Om, which means that xn ∈ Om for only finitely many n and
l is not an accumulation point.
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CHAPTER 26. TOPOLOGY
Conversely, assume that E is compact, and let (xn) be any sequence. If l ∈ E is not
an accumulation point of the sequence, then there is some open set, Ul, such that l ∈ Ul
and xn ∈ Ul for only finitely many n. Thus, if (xn) does not have any accumulation point,
the family, (Ul)l∈E, is an open cover of E and since E is compact, it has some finite open
subcover, (Ul)l∈J, where J is a finite subset of E. But every Ul with l ∈ J is such that
xn ∈ Ul for only finitely many n, and since J is finite, xn ∈
U
l∈J
l for only finitely many n,
which contradicts the fact that (Ul)l∈J is an open cover of E, and thus contains all the xn.
Thus, (xn) has some accumulation point.
Remark: It should be noted that the proof showing that if E is compact, then every se-
quence has some accumulation point, holds for any arbitrary compact space (the proof does
not use a countable basis for the topology). The converse also holds for metric spaces. We
will prove this converse since it is a major property of metric spaces.
Given a metric space in which every sequence has some accumulation point, we first prove
the existence of a Lebesgue number .
Lemma 26.32. Given a metric space, E, if every sequence, (xn), has an accumulation point,
for every open cover, (Ui)i∈I, of E, there is some δ > 0 (a Lebesgue number for (Ui)i∈I) such
that, for every open ball, B0(a, ), of radius
≤ δ, there is some open subset, Ui, such that
B0(a, ) ⊆ Ui.
Proof. If there was no δ with the above property, then, for every natural number, n, there
would be some open ball, B0(an, 1/n), which is not contained in any open set, Ui, of the
open cover, (Ui)i∈I. However, the sequence, (an), has some accumulation point, a, and since
(Ui)i∈I is an open cover of E, there is some Ui such that a ∈ Ui. Since Ui is open, there is
some open ball of center a and radius
contained in Ui. Now, since a is an accumulation
point of the sequence, (an), every open set containing a contains an for infinitely many n
and thus, there is some n large enough so that
1/n ≤ /2 and an ∈ B0(a, /2),
which implies that
B0(an, 1/n) ⊆ B0(a, ) ⊆ Ui,
a contradiction.
By a previous remark, since the proof of Proposition 26.31 implies that in a compact
topological space, every sequence has some accumulation point, by Lemma 26.32, in a com-
pact metric space, every open cover has a Lebesgue number. This fact can be used to prove
another important property of compact metric spaces, the uniform continuity theorem.
26.5. COMPACT SETS
761
Definition 26.25. Given two metric spaces, (E, dE) and (F, dF ), a function, f : E → F , is
uniformly continuous if for every
> 0, there is some η > 0, such that, for all a, b ∈ E,
if dE(a, b) ≤ η then dF (f(a), f(b)) ≤ .
The uniform continuity theorem can be stated as follows:
Theorem 26.33. Given two metric spaces, (E, dE) and (F, dF ), if E is compact and f : E →
F is a continuous function, then it is uniformly continuous.
Proof. Consider any
> 0 and let (B0(y, /2))y∈F be the open cover of F consisting of open
balls of radius /2. Since f is continuous, the family,
(f −1(B0(y, /2)))y∈F ,
is an open cover of E. Since, E is compact, by Lemma 26.32, there is a Lebesgue number,
δ, such that for every open ball, B0(a, η), of radius η ≤ δ, then B0(a, η) ⊆ f−1(B0(y, /2)),
for some y ∈ F . In particular, for any a, b ∈ E such that dE(a, b) ≤ η = δ/2, we have
a, b ∈ B0(a, δ) and thus, a, b ∈ f−1(B0(y, /2)), which implies that f(a), f(b) ∈ B0(y, /2).
But then, dF (f (a), f(b)) ≤ , as desired.
We now prove another lemma needed to obtain the characterization of compactness in
metric spaces in terms of accumulation points.
Lemma 26.34. Given a metric space, E, if every sequence, (xn), has an accumulation point,
then for every
> 0, there is a finite open cover, B0(a0, ) ∪ · · · ∪ B0(an, ), of E by open
balls of radius .
Proof. Let a0 be any point in E. If B0(a0, ) = E, then the lemma is proved. Otherwise,
assume that a sequence, (a0, a1, . . . , an), has been defined, such that B0(a0, )∪· · ·∪B0(an, )
does not cover E. Then, there is some an+1 not in B0(a0, ) ∪ · · · ∪ B0(an, ) and either
B0(a0, ) ∪ · · · ∪ B0(an+1, ) = E,
in which case the lemma is proved, or we obtain a sequence, (a0, a1, . . . , an+1), such that
B0(a0, ) ∪ · · · ∪ B0(an+1, ) does not cover E. If this process goes on forever, we obtain an
infinite sequence, (an), such that d(am, an) >
for all m = n. Since every sequence in E
has some accumulation point, the sequence, (an), has some accumulation point, a. Then,
for infinitely many n, we must have d(an, a) ≤ /3 and thus, for at least two distinct natural
numbers, p, q, we must have d(ap, a) ≤ /3 and d(aq, a) ≤ /3, which implies d(ap, aq) ≤ 2 /3,
contradicting the fact that d(am, an) >
for all m = n. Thus, there must be some n such
that
B0(a0, ) ∪ · · · ∪ B0(an, ) = E.
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CHAPTER 26. TOPOLOGY
A metric space satisfying the condition of Lemma 26.34 is sometimes called precompact
(or totally bounded ). We now obtain the Weierstrass–Bolzano property.
Theorem 26.35. A metric space, E, is compact iff every sequence, (xn), has an accumula-
tion point.
Proof. We already observ