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CHAPTER 6
Compactness
In this section we shall introduce the notion of compactness for topological spaces.
Compactness of a topological space is among the most fundamental and important
properties in topology and has far reaching consequences both for subspaces and maps
on the compact space. As the name suggests, a compact space can be thought of as
being “small”. In that sense, the function of the compactness property is similar to
that of separability and second countability introduced in chapter 2. It is generally
difficult to tell whether or not a space is compact. However, in the case of metric
topologies we will be able to work out several sufficient and necessary compactness
conditions (see theorems ?? and ??).
6.1. Definition and first examples
Definition 6.1. Let (X, TX ) be a topological space.
(a) An open cover F of X is a collection of open subsets of X whose union is all of
X.
(b) A subcover F 0 of F is a subset of F which is still an open cover of X. A proper
subcover of F is any subcover of F not equal to F.
(c) A refinement F 00 of F is an open cover of X with the property that for every
U ∈ F 00 there exists a V ∈ F such that U ⊂ V .
Definition 6.2. A topological space (X, TX ) of compact if every open cover F of
X has a finite subcover.
Example 6.3. Any topological space (X, TX ) with TX a finite set, is compact. In
particular, if X is a finite set then (X, TX ) is compact for any choice of topology TX .
Example 6.4. The Euclidean line (R, TEu ) is not compact. This can be seen by
considering the open cover
F = {ha − 1, a + 1i | a ∈ Z}
There are no proper subcovers of F at all, let alone finite subcovers, since each integer
a ∈ Z lies in precisely one element of F, namely ha − 1, a + 1i. A similar argument can
be used to that the n-dimensional Euclidean space (Rn , TEu ) is also not compact.
Example 6.5. Consider the finite complement topology Tf c on R and let F be an
open cover. Pick any U ∈ F with U 6= ∅. Then U = R − {x1 , ..., xn } for some points
xi ∈ R, i = 1, ..., n. For each xi let Vi ∈ F be such that xi ∈ Vi (such Vi have to exist
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6. COMPACTNESS
since F is an open cover). But then {U, V1 , ..., Vn } is a finite subcover of F showing
that (R, Tf c ) is a compact space.
Example 6.6. Consider the included point topology Tp on R. The infinite open
cover
F = {{p, x} | x ∈ R − {p} }
has no proper subcovers. Consequently (R, Tp ) is not compact.
Example 6.7. Consider the excluded point topology T p on R and let F be any of
its open covers. Since R is the only non-empty open set that contains p, R must be a
member of every open cover. Given this observation, the cover {R} is a finite subcover
of any open cover. Therefore, (R, T p ) is compact.
We turn to some simple properties of compact spaces.
Proposition 6.8. Compactness is a topological invariant.
Proof. Let f : X → Y be a homeomorphism and assume that X is compact. Let
FY be an open cover of Y and let FX = {f −1 (V ) | V ∈ FY }. Then FX is an open cover
of X and hence, by compactness of X, has a finite subcover FX0 = {f −1 (V1 ), ..., f −1 (Vn )}
(for some V1 , ..., Vn ∈ FY ). But then FY0 = {V1 , ..., Vn } is a finite subcover of FY and
so Y is compact also.
Theorem 6.9. Let X and Y be two topological spaces.
(a) If X is compact and A is a closed subspace of X then A is also compact.
(b) If f : X → Y is a continuous function and X is compact then f (X) is a compact
subspace of Y .
(c) If X is Hausdorff and A ⊂ X is a compact subspace then A is closed in X.
Proof. (a) Let FA = {Vi | i ∈ I} be an open cover of A and let Ui be open subsets
of X such that Vi = A ∩ Ui . Then FX = {X − A, Ui | i ∈ I} is an open cover of X
and thus has a finite subcover FX0 that looks like FX0 = {Ui1 , ..., Uin , X − A} or FX0 =
{Ui1 , ..., Uin } for some choice of indices i1 , ..., in ∈ I. In either case, FA0 = {Vi1 , ..., Vin }
is a finite subcover of FA showing that A is compact.
(b) Let Ff (X) = {Vi | i ∈ I} be an open cover of f (X) and consider the open
cover FX = {f −1 (Vi ) | i ∈ I} of X. By compactness of X, this latter cover has a
finite subcover FX0 = {f −1 (Vi1 ), ..., f −1 (Vin )} for some indices i1 , ..., in ∈ I. But then
Ff0 (X) = {Vi1 , ..., Vin } is a finite subcover of Ff (X) .
(c) Fix a point a ∈ A. The Hausdorff condition guarantees that for every point
x ∈ X − A there exist disjoint neighborhoods Ua,x and Vx,a of a and x respectively. The
collection {Ua,x ∩ A | a ∈ A} is an open cover of A and so by compactness, it has a finite
subcover {Ua1 ,x , ..., Uan ,x } for some points a1 , ..., an ∈ A. Let Vx = ∩ni=1 Vx,ai . Clearly
Vx is a neighborhood of x but moreover, Vx and A are disjoint. For if not, we could
find a point y ∈ Vx ∩ A. Then we would have y ∈ Uaj ,x for some j ∈ {1, ..., n}. But
on the other hand y ∈ Vx ⊂ Vx,aj which is impossible since Uaj ,x ∩ Vx,aj = ∅. Finally,
we note that X − A = ∪x∈X−A Vx showing that X − A is open and therefore that A is
closed.
6.1. DEFINITION AND FIRST EXAMPLES
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Corollary 6.10. Let X be a compact space and Y a Hausdorff space and let
f : X → Y be a continuous bijection. Then f is a homeomorphism.
Proof. We need to show that f −1 : Y → X is continuous. According to part (a)
of theorem 3.9, it suffices to show that f is a closed map. Thus, let A ⊂ X be a closed
subset of X. By part (a) of theorem 6.9, A is a compact subspace of X. By part (b)
of theorem 6.9, f (A) is a compact subspace of Y . Finally, by part (c) of theorem 6.9,
f (A) must be closed.
Theorem 6.11. Let (X, TX ) and (Y, TY ) be two topological space and let X × Y be
given the product topology TX×Y . Then X × Y is compact if and only if each of X and
Y are compact.
Proof. =⇒ Suppose firstly that X and Y are both compact and let W = {Wi | i ∈
I} be an open cover of X × Y . For each Wi ∈ W, find open sets Ui,j ⊂ X and
Vi,j ⊂ Y , j ∈ Ji such that Wi = ∪j∈Ji Ui,j × Vi,j . If we can show that the open cover
f = {Ui,j × Vi,j | j ∈ Ji , i ∈ I} has a finite subcover, then clearly so does W. Thus,
W
without loss of generality, we can (and will) assume that each Wi has the form Ui × Vi ,
i ∈ I, to begin with.
Pick a point y ∈ Y and consider the subspace X × {y} ⊂ X × Y . As we saw in
corollary 5.9, X × {y} is homeomorphic to X and so X × {y} must also be compact
(according to proposition 6.8). Therefore, the induced open cover
Wy = {(Ui × Vi ) ∩ (X × {y}) | i ∈ I}
of X × {y}, must have a finite subcover Wy0 = {Uj × {y} | j ∈ Jy } for some finite subset
of indices Jy ⊂ I. Let Vy = ∩j∈Jy Vj and note that Vy is a neighborhood of y ∈ Y with
the property that Uj × Vy ⊂ Uj × Vj for all j ∈ Jy .
Perform the above procedure for all y ∈ Y to arrive at the open cover {Vy | y ∈ Y }
of Y . By compactness of Y , this has a finite subcover {Vy1 , ..., Vyn } for some points
y1 , ..., yn ∈ Y . We claim that then the set
W 0 = {Uj × Vj | j ∈ Jyi , i = 1, ..., n}
is a finite subcover of W . The finiteness of W 0 is clear, so let’s show that it is a cover.
Given any point (x, y) ∈ X × Y , there exists an index i ∈ {1, ..., n} such that y ∈ Vyi .
Since the sets Uj , j ∈ Jyi cover X, we find that x ∈ Uj for some j ∈ Jyi . But then
(x, y) ∈ Uj × Vyi ⊂ Uj × Vj inW 0 showing that W 0 is indeed an open cover.
⇐= Since the projections maps πX : X × Y → X and πY : X × Y → Y are both
continuous and surjective(part (a) of theorem 5.6), and since the continuous image
of a compact space is compact (part (b) of theorem 6.9), the compactness of X × Y
immediately implies that of both X and Y .
By induction, the result of the preceding theorem extends without difficulty to
arbitrary finite products to show that X = X1 × ... × Xn is compact (with respect to
the product topology) if and only if each of the factors Xi is compact. The case of
infinite products is more subtle. On an infinite product there are two natural choices
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6. COMPACTNESS
of topologies available: the product and the box topology (see definition 5.13). The
compactness of an infinite product X = ×i∈I Xi is still implied by the compactness
of the factors Xi , provided we choose the right topology on X. The reader curious
about which of Tprod or Tbox is the right choice of a topology on X, may skip ahead to
section ?? where we prove Tychonoff’s compactness theorem, one of the foundational
theorems of point set topology.
In this introductory section on compact spaces we saw a few examples and some
of the easier to verify properties of compact space. We will list more compactness
attributes in section 6.3, right after digressing to take a closer look at compactness
among metric spaces in the next section.
6.2. Compactness for metric spaces
This section takes a closer look at the compactness property when the topological
space in questions is a metric space (X, d) equipped with the metric topology Td . For
this class of topological spaces there are several characterizations of compactness, see
theorems 6.17 and 6.21 below. An important corollary of these results yields a very
explicit descriptions of compact subspaces of Euclidean space Rn , see corollary 6.22.
Given a metric space (X, d), recall that Bx (r) denotes the open ball in X with
center x ∈ X and radius r > 0: Bx (r) = {y ∈ X | d(y, x) < r}.
Definition 6.12. Let (X, d) be a metric space.
(a) (X, d) is called totally bounded if forSevery ε > 0 there exist finitely many points
xi ∈ X, i = 1, ..., n such that X = ni=1 Bxi (ε).
(b) (X, d) is said to be bounded if there exists some r > 0 such that d(x, y) < r for
all x, y ∈ X, otherwise we call (X, d) unbounded.
(c) If (X, d) is bounded, we define its diameter diam(X) as
diam(X) = sup{d(x, y) | x, y ∈ X}
while if (X, d) is unbounded we define diam(X) to be infinite.
Using the triangle inequality (relation (3) from the axioms of a metric, see example
2.6), it is easy to show that a a totally bounded metric space is always bounded. The
converse is not generally true, see exercise ??.
In what follows we will always assume, mostly without mention, that a metric space
(X, d) has been made into a topological space by giving it the metric topology Td as
defined in example 2.6.
Lemma 6.13. Let (X, d) be a totally bounded metric space. Then X is separable.
Proof. Let εn = 1/n for n ∈ N. For each such εn , there is a finite set of points
An = {xn1 , ..., xn`n } such that
[
X=
Bx (1/n)
x∈An
Let A = ∪n∈N An and note that A is countable. We claim that A is also dense. To
see this, let U ⊂ X be any non-empty open set and let x ∈ U be any element. Then
6.2. COMPACTNESS FOR METRIC SPACES
85
there exists an ε > 0 such that Bx (ε) ⊂ U . Find an integer n large enough so that
1/n < ε/2. Then An must intersect Bx (ε) for it it didn’t, the point x would not be
contained in ∪x∈An Bx (1/n), a contradiction. We conclude that A ∩ U 6= ∅ and since
U was arbitrary, it follows that Ā = X (according to corollary 2.27).
Remark 6.14. Recall from lemma 2.43 that a metric space is second countable
if and only if it is separable. In particular, a totally bounded metric space must be
second countable.
Lemma 6.15. Let (X, d) be a metric space with the property that every sequence
xk ∈ X has a convergent subsequence. Then (X, d) is totally bounded.
Proof. Suppose that X is not totally bounded. Then there is some ε > 0 so that
X cannot be covered by a finite number of balls of radius ε. Using this, we use this
to construct a sequence xk ∈ X as follows: let x1 be arbitrary and choose consecutive
elements of the sequence so that xk+1 ∈
/ ∪kj=1 Bxj (ε). Let yn = xkn be some convergent
subsequence of xk with limn→∞ yn = y ∈ X. Since limn→∞ yn = y, there must exist an
n0 such that for all n ≥ n0 one gets yn ∈ By (ε/2). Let `, m ≥ n be two integers with
` < m. Then k` < km and therefore
d(xk` , xkm ) ≤ d(xk` , y) + d(y, xkm ) < ε/2 + ε/2 = ε
This implies that xkm ∈ Bxk` (ε) which is a contradiction since by definition xkm ∈
/
∪k<km Bxk (ε). Thus X must be totally bounded.
Lemma 6.16. If X is a second countable topological space then every open cover
F of X has a countable subcover. In particular, if (X, d) is a metric space with the
property that every subsequence xk ∈ X has a convergent subsequence, then every open
cover F of X has a countable subcover.
Proof. Let B = {U1 , U2 , U3 , ...} be countable basis for X and let F = {Vi | i ∈ I}
be an open cover of X. Since each Vi is an open set, there must exist a subset Ji ⊂ N
such that Vi = ∪j∈Ji Uj . Define F 00 = {Uj ∈ B | j ∈ Ji for some i ∈ I}. It is easy
to see that F 00 is an open cover of X for if x ∈ X is an arbitrary point then there is
some i ∈ I with x ∈ Vi and thus x ∈ Uj for some j ∈ Ji . On the other hand, F 00 is
a refinement of F by definition. For j ∈ N, let i(j) ∈ I be such that Uj ⊂ Vi(j) , then
F 0 = {Vi(j) | j ∈ N} is a countable subcover of F.
If (X, d) is a metric space in which every sequence has a convergent subsequence,
then according to lemma 6.15, X is totally bounded. According to lemma 6.13, X then
must be separable and so, by virtue of remark 6.14, X must also be second countable.
Thus the claim of the theorem them follows from the first paragraph of the proof. Theorem 6.17. A metric space (X, d) is compact if and only if every sequence
xi ∈ X has a convergent subsequence.
Proof. =⇒ Let X be a compact space and let xk ∈ X be an arbitrary sequence.
Suppose that there exists a point y ∈ X such that every neighborhood of y contains
infinitely many distinct points of the sequence xk . For such a point y we define a
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subsequence yn of xk as follows: Pick y1 ∈ By (1) ∩ {x1 , x2 , x3 , ...} arbitrarily and
inductively choose yn+1 from
yn ∈ [By (1/n) ∩ {x1 , x2 , x3 , ...}] − {y1 , ..., yn−1 },
It is easy to see that yn is a convergent subsequence of xk with limit y.
If this doesn’t occur, then every y ∈ Y possesses a neighborhood Uy that contains
only finitely many distinct points of the sequence xk . Then F = {Uy | y ∈ X} is an open
cover of X and so by compactness of X, it has a finite subcover F 0 = {Uy1 , ..., Uym }|
for some points y1 , ..., ym ∈ X. But then {x1 , x2 , x3 , ...} ⊂ Uy1 ∪ ... ∪ Uy1 showing that
(by the pigeon principle) some point y must equal xk for infinitely many distinct values
k1 , k2 , k3 , ... of k. But then the sequence xk1 , xk2 , xk3 , ..., being a constant sequence, is
a convergent subsequence of xk .
⇐= Suppose now that every sequence xk ∈ X has a convergent subsequence and
let F be an arbitrary open cover of X. Using lemma 6.16 we can find a countable
refinement G = {U1 , U2 , U3 , ...} of F. We will first show that G has a finite subcover.
To prove this, assume to the contrary that G has no finite subcovers. Form a sequence
xk ∈ X by choosing x1 ∈ X arbitrarily and by picking xk ∈ X − ∪k−1
i=1 Ui for k ≥ 2.
By the assumption, there must be a subsequence yn = xkn with converges to some
point y ∈ X. Since G is an open cover of X, there must be some index m ∈ N with
y ∈ Um . But then, by construction, yn ∈
/ Um for all n > m, a contradiction given that
limn→∞ yn = y. Therefore G must have a finite subcover G 0 = {Ui1 , ..., Uin } for some
indices i1 , ..., in ∈ N. Finally, for each of these indices, find an element Vij ∈ F with
Uij ⊂ Vij . Then F 0 = {Vi1 , ..., Vin } is a finite subcover of F. Since F was arbitrary, we
conclude that X must be compact.
Corollary 6.18. Every compact metric space X is totally bounded, separable and
second countable.
Proof. This follows from a combination of theorem 6.17, lemmas 6.13 and 6.15
and remark 6.14.
While theorem 6.17 provides a necessary and sufficient condition for a metric space
(X, d) to be compact, it is often tedious in practice to check that X has the property that every of its sequences has a convergent subsequence. We provide a second
characterization of compactness on metric spaces, one which involved easier to verify
conditions. To state that theorem, we first introduce two additional concepts.
Recall that a sequence xk ∈ Rn is said to be a Cauchy sequence (with respect to
the Euclidean topology) if for every ε > 0 there exists an integer n0 such that for
all m, n ≥ n0 the inequality d(xn , xm ) < ε holds. This definition involves only the
Euclidean metric on Rn and none of the other more advances structures on Rn . It is
thus not surprising that it transfers to any metric space.
Definition 6.19. Let (X, d) be a metric space. A sequence xk ∈ X is said to be a
Cauchy sequence if for every ε > 0 there exists an integer n0 such that for all m, n ≥ n0
6.2. COMPACTNESS FOR METRIC SPACES
87
we obtain d(xn , xm ) < ε. The metric space (X, d) is said to be a complete metric space
(or simply complete) if every Cauchy sequence is convergent.
Example 6.20. Consider Rn equipped with the Euclidean topology and let d be
the Euclidean metric. For a subset A ⊂ Rn , the metric space (A, d|A×A ) is complete if
and only if A is closed. In particular, (Rn , d) itself is complete.
To see this, suppose that A is closed in Rn and xk ∈ A be a Cauchy sequence.
Every Cauchy sequence is bounded and every bounded sequence in Rn has a convergent
subsequence yi = xki (these are standard results in analysis whose proofs we omit).
Let y = limi→∞ yk . Then for every ε > 0 the exists an n0 such that n, m ≥ n0 implies
that
d(yn , y) < ε/2
and
d(xn , xm ) < ε/2
But then d(xn , y) ≤ d(xn , yn ) + d(yn , y) < ε/2 + ε/2 = ε showing that xk converges to
y ∈ Rn . If we had y ∈ Rn − A then, since A is closed, Rn − A would be a neighborhood
of y which is impossible since xk ∈ A converges to y. Thus we must have y ∈ A so that
A is complete.
Conversely, suppose that A is complete. If A weren’t closed then Rn −A wouldn’t be
open and so we could find a point y ∈ Rn −A so that every neighborhood of y intersects
A. We can then define a sequence xk by choosing it arbitrarily from By (1/k). The
distance from xk to x` is bounded by 1/k + 1` showing that xk is Cauchy. But xk
converges to y which is not in A, a contradiction. Thus A must be closed.
The following theorem is the main result of this section.
Theorem 6.21. A metric space (X, d) is compact if and only if is complete and
totally bounded.
Proof. =⇒ Suppose that X is compact. To see that X is complete, let xk ∈ X
be any Cauchy sequence. By theorem 6.17 there is a convergent subsequence yn = xkn
of xk with limit y. But then xk itself has to converge to y. Namely, given any ε > 0
and by using the Cauchy condition, find an integer k0 such that n, m ≥ k0 implies that
d(xn , xm ) < ε/2. For that same ε > 0, using the convergence of yn to y, find an n0
such that n ≥ n0 implies that d(yn , y) < ε/2. Then for all k ≥ max{k0 , n0 } we obtain
d(xk , y) ≤ d(xk , yk ) + d(yk , y) < ε/2 + ε/2 = ε
To see that X is totally bounded, pick an arbitrary ε > 0 and consider the open
cover Fε of X defined for as
Fε = {Bx (ε) | x ∈ X}
Since X is compact, Fε must have a finite subcover, showing that X is totally bounded.
⇐= Assume that X is complete and totally bounded. We will show that X is
compact by relying on theorem 6.17. Specifically, we will show that every sequence
xk ∈ X has a convergent subsequence.
Since X is totally bounded, there is a finite cover of X with balls of radius 1, say
F1 = {Bx1,1 (1), ..., Bxn1 ,1 (1)}. Since the sequence xk has infinitely many terms, one of
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6. COMPACTNESS
the balls from F1 must contain infinitely many elements of the said sequence. Without
loss of generality, assume that Bx1,1 (1) is such a ball. Using total boundedness of X
once more, we find that there is a finite cover F2 = {Bx1,2 (1/2), ..., Bxn2 ,2 (1/2)} of X
by balls of radius 1/2. Since Bx1,1 (1) contains infinitely many elements of the sequence
xk , there must be a ball in F2 whose intersection with Bx1,1 (1) contains infinitely many
elements of the sequence xk . Again, and without loss of generality, we assume that
Bx1,2 (1/2) is that ball so that Bx1,1 (1) ∩ Bx1,2 (1/2) contains infinitely many elements
of the sequence xk . Proceeding inductively, we obtain a family of balls Bx1,k (1/k) of
shrinking radii such that for any n ∈ N the set ∩nj=1 Bx1,j (1/j) contains infinitely many
elements of the sequence xk .
To find a convergent subsequence of xk , pick indices k1 < k2 < k3 < ... such that
xk` ∈ ∩`j=1 Bx1,j (1/j) and set yn = xkn . Then yn is a Cauchy sequence since d(yn , ym ) <
max{1/n, 1/m}. Since X is assumed to be complete, yn must be a convergent sequence
and so we have generated a convergent subsequence of xk . This completes the proof. Corollary 6.22. Let A be a subspace of n-dimensional Euclidean space Rn . Then
A is compact if and only if it is closed and bounded.
Proof. If A is bounded then it is also totally bounded (exercise ??) and if in
addition it is also closed, then it is complete by example 6.20. Theorem 6.21 then
shows that A is compact.
Conversely, if A ⊂ Rn is compact, then by theorem 6.21 is must be complete and
hence closed by example 6.20. Theorem 6.21 also shows that A has to be totally
bounded and therefore bounded (exercise ??).
Example 6.23. The n-dimensional sphere S n is the subspace of Rn+1 consisting of
vectors of norm 1:
S n = {(x1 , ..., xn+1 ) ∈ Rn+1 | x21 + ... + x2n+1 = 1}
It is quite obviously bounded and it is also closed. An easy way to verify closedness
of S n is to consider the continuous function f : Rn+1 → R given by f (x1 , ..., xn+1 ) =
x21 + ... + x2n+1 . Then S n = f −1 (1) and {1} ⊂ R is a closed set. Point (a) of theorem
3.9 then guarantees that S n is also closed. According to corollary 6.22, S n is compact.
Corollary 6.24. A subset A of a discrete space X is compact if and only if its is
finite.
6.3. Properties of compact spaces
In this section we return to the investigation of properties of compact spaces already
initiated in section 6.1. Some of the properties enlisted in this section, rely on the results
about metric spaces obtained in the preceding section.
Theorem 6.25. Let X be a compact topological space and let f : X → R be
a continuous function to the Euclidean line. Then f attains both its maximum and
minimum value.
6.3. PROPERTIES OF COMPACT SPACES
89
Proof. By part (b) of theorem 6.9, the image f (X) is a compact subspace of R. By
corollary 6.22, f (X) must be closed and bounded and thus f (X) = [a1 , b1 ] ∪ ... ∪ [an , bn ]
for some ai , bi ∈ R with ai ≤ bi . This shows that both the supremum sup f (X) =
max{b1 , .., bn } and the infimum inf f (X) = min{a1 , ..., an } are contained in f (X) and
are therefore attained at some points xmax , xmin ∈ X.
Theorem 6.26. Let (X, d) be a metric space and let F be an open cover of X. If
X is compact then there exists a real number L > 0, called the Lebesgue number of F,
such that for every open ball Bx (L) of radius L (and with x ∈ X arbitrary), there exists
an element U ∈ F with Bx (L) ⊂ U .
Proof. Let x ∈ X be an arbitrary element. Since F is an open cover of X, there
must be some element Ux ∈ F such that x ∈ Ux . Let rx > 0 be chosen so that
Bx (rx ) ⊂ Ux . Consider then the open cover G of X given by G = {Bx (rx /2) | x ∈ X}.
Since X is compact, G has a finite subcover G 0 = {Bx1 (rx1 ), ..., Bxn (rxn )} for some
points x1 , ..., xn ∈ X. Let L > 0 be chosen that
nr
rx o
x1
, ..., n
0 < L < min
2
2
0
Let x ∈ X be any point. Since G is a cover, there is some index i ∈ {1, ..., n} such
that x ∈ Bxi (rxi /2). But then, if y ∈ Bx (L), we obtain
d(y, xi ) ≤ d(y, x) + d(x, xi ) < L + rxi /2 < rxi /2 + rxi /2 = rxi
showing that Bx (L) ⊂ Bxi (rxi ) ⊂ Uxi ∈ F. Since x was chosen arbitrarily, the claim
of the theorem is proved.
The following theorem has already been established for compact metric space (see
theorem 6.17). We show here that it remains valid in general compact spaces provided
they are first countable (which metric spaces always are, see example 2.42).
Proposition 6.27. Every sequence xk in a first countable, compact space (X, TX )
has a convergent subsequence.
Proof. Case 1. Suppose that we can find a point x ∈ X with the property that
every neighborhood U of x contains infinitely many elements of the sequence xk . Let
Bx = {U1 , U2 , U3 , ...} be a countable neighborhood basis at x. Define the subsequence
yn of xk by picking y1 from U1 ∩ {x1 , x2 , x3 , ...} at random and by picking yn from the
non-empty set
U1 ∩ ... ∩ Un ∩ ({x1 , x2 , x3 , ...} − {y1 , ..., yn−1 })
in an arbitrary fashion. Then, given any neighborhood V of x, we can find an integer n0
such that Un0 ⊂ V and consequently, yn ∈ V for all n ≥ n0 . In particular, limn→∞ yn =
x.
Case 2. If the above case does not occur, then for every point x ∈ X there has to
exist a neighborhood Ux containing only finitely many points of the sequence xk . By
compactness of X, we can reduce the open cover {Ux | x ∈ X} of X to a finite subcover
{Ua1 , ..., Uan } with a1 , ..., an ∈ X. But then, by the pigeonhole principle, one of the sets
90
6. COMPACTNESS
Uai must contain infinitely many elements of the sequence xk , a contradiction. Therefore case 2 cannot occur while in case 1 we were able to find a convergent subsequence
of xk .
The next set of theorems shows that the compactness of X “enhances”the separation axioms T0 − T5 that X possesses. By this we mean, for example, that if a compact
space X is T2 then it is automatically T3 , something that if of course false in general
for non-compact spaces. The chief reason for this is that closed subsets of X behave in
many ways like points, given that every open cover of a closed subset of X has a finite
subcover. How exactly this plays out can be seen in the proof of the next theorem.
Theorem 6.28. Let X be a compact topological space.
(a) If X is Hausdorff then X is also T3 and T4 .
(b) If X is T3 then it is also T4 .
The next lemma is used in the proof of theorem 6.28, it is the analogue for T3 spaces
of what lemma 4.25 is for T4 spaces.
Lemma 6.29. A topological space (X, TX ) is regular if and only if for every closed
set A ⊂ X and every point x ∈ X − A there exists an open set V ⊂ X such that
x ∈ V ⊂ V̄ ⊂ X − A
Proof. =⇒ Let X be a T3 space and let A ⊂ X be a closed set and x ∈ X − A
any point. By the T3 property there are disjoint open sets Ux , UA ⊂ X with x ∈ Ux ,
A ⊂ UA . From Ux ∩ UA = ∅ it follows that Ux ⊂ X − UA . Since X − UA is a closed
set, we find that Ūx ⊂ X − UA and since X − UA ⊂ X − A we get the desired relation
x ∈ Ux ⊂ Ūx ⊂ X − A, thus simply choose V = Ux .
⇐= Suppose that for every closed subset A ⊂ X and for every point x ∈ X − A
there exists an open set V with x ∈ V ⊂ V̄ ⊂ X − A. We define the open sets Ux and
UA as V and X − V̄ respectively. Then A ⊂ UA and Ux ∩ UA = ∅ and so X is a T3
space.
Proof. (Of theorem 6.28) (a) The proof of part (c) of theorem 6.9 shows that in a
compact Hausdorff space X, for every closed subset A and every point x ∈ X −A, there
exist disjoint open subsets Vx and UA with x ∈ Vx and A ⊂ UA . In the notation of that
proof, the set Vx is defined asVx,a1 ∩ ... ∩ Vx,an while UA is given by Ua1 ,x ∪ ... ∪ Uan ,x .
This already shows that X is T3 .
To verify the T4 property of X, let A, B ⊂ X be two disjoint closed subsets of
X. For a fixed point b ∈ B, let Vb,A and UA,b be disjoint open sets with b ∈ Vb,A and
UA,b . The set FB = {Vb,A | b ∈ B} is an open cover of B. Since B is a closed subset
of the compact space X, according to part (a) of theorem 6.9, B itself is compact
and so FB has a finite subcover FB0 = {Vb1 ,A , ..., Vbn ,A } for some points b1 , ..., bn ∈ B.
Set VB = ∪ni=1 Vbi ,A and UA = ∩ni=1 UA,bi . These are open sets containing B and A
respectively. Moreover, VB ∩ UA = ∅ for if we had a point y ∈ VB ∩ UA then y would
lie in some Vbj ,A and at the same time UA,bj , a contradiction since these two sets are
disjoint by construction. Therefore X is T4 .
6.4. THE ONE POINT COMPACTIFICATION
91
(b) Let A, B ⊂ X be two disjoint and closed subsets of X. According to lemma
6.29, for every point a ∈ A there exists a neighborhood Ua of a such that
a ∈ Ua ⊂ Ūa ⊂ X − B
Then {X − A, Ua |a ∈ A} is an open cover of X and has a finite subcover {X −
A, Ua1 , Ua2 , ..., Uan }. In a similar vein, every point b ∈ B has a neighborhood Vb with
b ∈ Vb ⊂ V̄b ⊂ X − A
and the open cover {X − B, Vb |b ∈ B} has a finite subcover {X − B, Vb1 , Vb2 , ..., Vbn }
(for convenience and without loss of generality, we assume that this cover has as many
elements as the cover {X − A, Ua1 , Ua2 , ..., Uan }). We now define the open sets
U10 = Ua1 − V̄b1
U20 = Ua2 − (V̄b1 ∪ V̄b2 )
..
.
V10 = Vb1 − Ūa1
V20 = Vb2 − (Ūa1 ∪ Ūa2 )
..
.
Un0 = Uan − (V̄b1 ∪ ... ∪ V̄bn )
Vn0 = Vbn − (Ūa1 ∪ ... ∪ Ūan )
Observe that Ui0 ∩ Vj0 = ∅ for all indices i, j ∈ {1, ..., n}. Since V̄b1 ∪ ... ∪ V̄bn ⊂ X − A
we see that UA = ∪ni=1 Ui0 is an open set containing A and likewise, VB = ∪nj=1 Vj0 is an
open set containing B. Moreover, the set UA and VB are disjoint for if they were not
then we would obtain Ui0 ∩ Vj0 6= ∅ for some indices i, j, an impossibility. Thus X is
T4 .
6.4. The one point compactification
Definition 6.30. A compactification of a topological space (X, TX ) is a compact
topological space (Y, TY ) that contains X as a dense subspace.
The goal of this section is to show that compactifications always exist for every
space X. We will describe one method for obtaining a compactification called the
one point compactification, though there are many others (exercise ??). Here are the
details.
If (X, TX ) is already compact then X is its own compactification. Thus, assume
that X is not compact. Let p be an abstract point not contained in X and define Y
as a set by Y = X ∪ {p}. Define TY to be the following collection of subsets of Y :


Either
p
∈
/
U
and
U
∈
T


X
or
U ⊂ Y (6.1) TY =

p ∈ U and X − U is a compact closed subset of X. 
Note that even when U ∈ TY is of the second kind (i.e. the kind with p ∈ U ), the set
U − {p} is still an open subset in X. We claim that TY is indeed a topology on Y :
1. ∅, X ∈ TY since p ∈
/ ∅, X and ∅, X ∈ TX .
2. Let Ui ∈ TY , i ∈ I and set U = ∪i∈I Ui . If for all i ∈ I we hve p ∈
/ Ui and
Ui ∈ TX then p ∈
/ U and U ∈ TX showing that in this case U ∈ TY . On the
92
6. COMPACTNESS
other hand, if p ∈ Uj for some j ∈ I, then X − U is a closed and compact
subset of X. But then p ∈ U and
X − U = X − ∪i∈I Ui = ∩i∈I (X − Ui ) ⊂ X − Uj
This shows that X − U is a closed subset of a closed compact subspace X − Uj
of X. Consequently, X − U is also closed and compact (see part (a) of theorem
6.9) and so U ∈ TY .
3. Let V1 , ..., Vn ∈ TY and set V = V1 ∩ ... ∩ Vn . If there is at least one index
j ∈ {1, ..., n} with p ∈
/ Vj and Vj ∈ TX then p ∈
/ V and V ∈ TX (see the first
line after (6.1)). On the other hand, if p ∈ Vi for all i = 1, ..., n then X − Vi is
closed and compact for all i. But then
X − V = X − ∩ni=1 Vi = ∪ni=1 (X − Vi )
A finite union of closed and compact subspaces of X is again a closed and
compact subspace of X. Therefore, V ∈ TY .
We remark that the definition of TY makes it immediate that the subspace topology
on X induced by TY is simply TX . Therefore we can view (X, TX ) as a subspace of
(Y, TY ).
We next turn to the verification of the density of X in Y . Since Y = X ∪ {p},
it is clear that X̄ must either be X or Y (since these are the only two subsets of Y
containing X). But X̄ = X could happen only if X were closed in Y . By definition of
TY , this would mean that {p} would have to be open in Y which in turn would mean
that Y − {p} = X would need to be a closed and compact subspace of X. But our
assumption was that X is not compact, and so {p} is not open in Y and consequently
X is not closed in Y . This excludes the possibility X̄ = X leaving only X̄ = Y .
It remains to be seen that (Y, TY ) is a compact space. Thus, let F be an open cover
of Y . There must be some element U0 ∈ F which contains p and so by the definition of
TY (6.1), X − U0 is a compact subspace of X. Thus, X − U0 is covered by finitely many
U1 , ..., Un ∈ F showing that Y is covered by the finite subcover {U0 , U1 , ..., Un } ⊂ F of
F. Ergo, Y is compact. We summarize our findings in the next theorem.
Theorem 6.31. Let (X, TX ) be a non-compact space and let Y = X ∪ {p} for
some abstract point p ∈
/ X. Then (Y, TY ), with TY defined as in (6.1), is a compact
space that contains X as a dense subspace. The space Y is referred to as the one point
compactification of X.
Example 6.32. Let (X, TX ) = (Rn , TEu ) be n-dimensional Euclidean space and let
p = ∞ be the symbol for an abstract point not in Rn . Consider also the n-dimensional
sphere S n defined as
S n = {(x1 , ..., xn+1 ) ∈ Rn+1 | x21 + ... + x2n+1 = 1}
equipped with the relative Euclidean topology on Rn+1 . We will show that the one point
compactification of Rn is homeomorphic to S n . Define the function f : Rn ∪ {∞} → S n
6.4. THE ONE POINT COMPACTIFICATION
93
as
 2xn
|x|2 − 1
2x1


, ..., 2
, 2
; x 6= ∞

2+1
|x|
|x|
+
1
|x|
+
1
f (x1 , ..., xn ) =



(0, 0, ..., 0, 1)
; x=∞
p
where |x| = x21 + ... + x2n . We first note that f is a bijection, this is easiest to see by
recognizing that its inverse function is given by
 y1
yn


, ...,
; y 6= (0, ..., 0, 1)

1 − yn+1
1 − yn+1
f −1 (y1 , ..., yn+1 ) =



∞
; y = (0, ..., 0, 1)
The function f −1 |S n −{(0,...,0,1)} : S n − {(0, ..., 0, 1)} → Rn is called the stereographic
projection from the north pole.
To see that f and f −1 are continuous we will show that they are open maps. If
U ∈ TRn ∪{∞} is an open set with ∞ ∈
/ U or if V ⊂ S n is an open set with (0, ..., 0, 1) ∈
/ V,
−1
n
n
n
then f (U ) and f (V ) are open since f |R : R → S − {(0, ..., 0, 1)} is quite obviously
a homeomorphism (since the coordinate functions of f −1 |S n −{(0,...,0,1)} and of f |Rn are
continuous functions).
Next, pick a set U ∈ TRn ∪{∞} with ∞ ∈ U . According to (6.1), Rn − U is a compact
subspace of Rn and so, according to corollary 6.22, it is also bounded. This allows us
to pick an R > 0 so that Rn − B0 (R) ⊂ U (where 0 ∈ Rn is the origin). But then
2
B(0,...,0,1) (ρ) ∩ S n ⊂ f (U ) with ρ = √
R2 + 1
showing that (0, ..., 0, 1) ∈ f (U ) has a neighborhood contained in f (U ). By continuity
of f |Rn , every point y ∈ f (U ) − {(0, ..., 0, 1)} also has a neighborhood contained in
f (U ) and so f (U ) is open and therefore f −1 is continuous.
A neighborhood basis for S n around (0, ..., 0, 1) is given by {B(0,...,0,1) (ρ)∩S n | ρ > 0}.
To show that f −1 is open, it suffices to show that f −1 (B(0,...,0,1) (ρ) ∩ S n ) is an open
subset of Rn . But this is a completely explicit computation and one finds that
ρ
p
f −1 (B(0,...,0,1) (ρ) ∩ S n ) = Rn − B0 (R) with R =
1 − 1 − ρ2
Note that Rn − B0 (R) inTRn ∪{∞} . For any open subset V ⊂ S n − {(0, ..., 0, 1)}, f −1 (V )
is obviously open and so f −1 is an open map and therefore f is continuous and thus a
homeomorphism.
Recall that compactness of S n was first demonstrated in example 6.23. The present
example is another, somewhat roundabout way, to verify the same fact.
Example 6.33. Consider the included point topology Tp on R. The space (R, Tp )
is not compact according to example 6.6. Thus, it makes sense to define Y = R ∪ {∞}
to be the one point compactification of (R, Tp ). To describe the elements of TY , note
that a subset A ⊂ R is compact with respect to Tp if and only if it is finite. For if
94
6. COMPACTNESS
A = {a1 , a2 , ...} is infinite and p ∈
/ A, then the set {{a1 }, {a2 }, ...} is an open cover
of A with no proper subcovers at all. On the other hand, if A is infinite and p ∈ A,
then the set {{a1 , p}, {a2 , p}, ..} is again an open cover of A with no proper subcovers.
A subset A ⊂ R is closed if it doesn’t contain p. Consequently, closed and compact
subsets of (R, Tp ) are precisely finite subsets of R that don’t contain p. From this we
obtain
TY = {U | (p ∈ U and ∞ ∈
/ U ) or (p, ∞ ∈ U and R − U is finite)}
We see that the one point compactification of (R, Tp ) contains features of the finite
complement topology. Observe that every sequence in Y converges to ∞.
6.5. Flavors of compactness
Definition 6.34. Let (X, TX ) be a topological space and F an open cover of X.
We shall say that F is locally finite if every point x ∈ X has a neighborhood Ux that
intersects only finitely many elements from F.
For later intentions, it is not sufficient to define local finiteness to be the property
that every point x ∈ X intersects only finitely many sets in F. A priori, it is not
clear that this is actually a weaker definition than definition 6.34, though example 6.36
below confirms that it is. Note that every finite cover F is locally finite.
Example 6.35. Consider the cover F = {ha + 1, a − 1i | a ∈ Z} of (R, TEu ). This is
locally finite since for every point x ∈ R, the interval hx − 1, x + 1i intersects at most
2 elements from F.
On the other hand, G = {ha, bi | a, b ∈ Q, a < b} is not locally finite since, given
any point x ∈ R and any neighborhood Ux of x, there are infinitely many rational
numbers contained in Ux . Let q1 , q2 , q3 , ... ∈ Ux ∩ Q be an infinite sequence chosen so
that qi < qi+1 , then hq1 , qi i intersects Ux for every i = 2, 3, 4, ....
Example 6.36. Consider the excluded point topology T p on R and let
F = {U ⊂ R | U = R or U = {x} for some x ∈ R − {p} }
Then F is an open cover of R with the property that every point x ∈ R is contained
in either exactly two elements of F (if x 6= p) or in only one such element (if x = p).
But p ∈ R has no neighborhood that intersects only finitely many elements from F.
We conclude that F is not a locally finite cover even though every point in R only
intersects finitely many sets in F.
Definition 6.37. Let (X, TX ) be a topological space.
(a) We shall say that X is locally compact if for every point x ∈ X and every
neighborhood U of x, there exists a neighborhood V of x such that x ∈ V ⊂
V̄ ⊂ U and with V̄ compact.
(b) We call X σ-compact if there exists a sequence A1 , A2 , A3 , ... of compact subsets
of X such that X = ∪∞
i=1 Ai .
6.5. FLAVORS OF COMPACTNESS
95
(c) The space X is said to be paracompact if every open cover F of X has a locally
finite refinement F 0 .
Remark 6.38. We comment briefly on each point from the previous definition
before moving on. Regarding part (a) from definition 6.37, some authors define local
compactness of X as meaning that every point x ∈ X has a neighborhood with compact
closure. While this is implied by our definition, it is not equivalent to it (see example
??).
If X is a compact space then we can take each Ai from part (b) of definition 6.37
to be X. This shows that a compact space is automatically σ-compact.
Similarly, if X is compact, then every of its open covers F has a finite subcover F 0 .
Since every finite cover is necessarily locally finite, we see that if X is compact then it
is also paracompact.
Example 6.39. Euclidean space (Rn , TEu ) is σ-compact since each of Ai = B0 (i),
n
i ∈ N is compact and clearly Rn = ∪∞
i=1 Ai . As we already saw, R is not compact.
Example 6.40. The exluded point topology T p on R is not locally compact.
Namely, the point p itself has only one neighborhood, R itself. But we saw in example 6.7 that (R, T p ) is not compact
Theorem 6.41. Let (X, TX ) be a locally compact, non-compact, Hausdorff space
and let (Y, TY ) be its one point compactification (with Y = X ∪ {∞}, see (6.1)). Then
Y is a compact Hausdorff space.
Proof. Let x, y ∈ Y be two distinct points. If x, y ∈ X then, since X is Hausdorff
and since TX ⊂ TY , x and y possess disjoint neighborhoods Ux and Uy .
On the other hand, if y = ∞ and thus x 6= ∞, then x lies in X and local compactness
of X guarantees the existence of a neighborhood Ux of x in X with Ū compact. But
then Uy = Y − Ū belongs to TY since Ū is compact and closed (the latter because
it is a compact subspace of a Hausdorff space, see part (c) of theorem 6.9). Clearly
Ux ∩ Uy = ∅ demonstrating that Y is Hausdorff.
Corollary 6.42. Let X be a locally compact, Hausdorff space. Then X is completely normal (part (b) of definition 4.2).
Proof. If X is compact, let (Y, TY ) = (X, TX ) and if X is not compact, let (Y, TY )
be the one point compactification of (X, TX ). Then Y is a compact, Hausdorff space
(either by assumption on X or by theorem 6.41) and so Y is a normal space according
to part (a) of theorem 6.28. The Hausdorff condition of Y implies that Y has the
T0 property while corollary 4.28 shows that Y is a T3 1 space. The conclusion of the
2
corollary now follows from part (a) of theorem 4.4 by which the subspace of a completely
regular space is again completely regular.
Example 6.43. Let X be a locally compact, Hausdorff space. Then for any two
distinct points x, y ∈ X there is a continuous function f : X → R such that f (x) = 0
and f (y) = 1.