Definition 4.35 A topological space is said to be compact if any open cover admits
a finite subcover. A subspace in a topological space is compact if it is compact for the induced (subspace) topology.
4.4 Compact Spaces 73
Equivalently, a subspace K of X is compact iff for any familyA of open sets in
X with K ⊂ ∪{A | A ∈ A}, there exist finitely many A1, . . . , An ∈ A such that
K ⊂ A1∪ · · · ∪ An.
Example 4.36 The spaceRnis not compact, for any n> 0; to show this it’s enough
to find an open cover without finite subcovers. Consider the open balls with centre 0 and radius a natural number
Rn= ∪+∞
m=1B(0, m).
If such cover had a finite subcover, there would exist m1, . . . , mk∈ N with
Rn= ∪k
i=1B(0, mi) = B(0, max(m1, . . . , mk)),
which is patently false.
Example 4.37 Any finite set is compact, irrespective of the topology we put on it: for
any open cover we can define a finite subcover by choosing one open set containing each point.
A discrete topological space is compact if and only if it is finite. In a discrete space X , in fact, the family of singletons{{x} | x ∈ X} is an open cover. This has a finite subcover iff X is finite.
Theorem 4.38 Let f : X → Y be a continuous map. If X is compact the range
f(X) is a compact subspace in Y .
Proof LetA be a family of open sets in Y covering f (X). Then the family { f−1(A) | A ∈ A} is an open cover for X, so there are finitely many A1, . . . , An ∈ A with
X= f−1(A1) ∪ · · · ∪ f−1(An). This implies that f (X) ⊂ A1∪ · · · ∪ An.
Theorem 4.39 The closed interval[0, 1] is compact in R.
Proof LetA be a family of open subsets in R that covers [0, 1]:
[0, 1] ⊂ ∪{A | A ∈ A}.
Write X ⊂ [0, +∞[ for the set of points t such that the closed interval [0, t] is contained in a finite union of open subsets inA.
The interval[0, 0] is contained in an open set of A, so 0 ∈ X. Let’s call b the least upper bound of X . If b > 1 there exists t ∈ X such that 1 ≤ t ≤ b, whence [0, 1] ⊂ [0, t] is contained in a finite union of open subsets in A. Now suppose
b ≤ 1, and let’s prove this leads to a contradiction. If b ≤ 1 there’s an A ∈ A such
that b∈ A, and A being open implies the existence of δ > 0 such that ]b − δ, b + δ[ ⊂ A.
On the other hand, by the properties of the supremum there’s a t ∈ X such that
[0, t] ⊂ A1∪ · · · ∪ An.
Therefore if 0≤ h < δ,
[0, b + h] = [0, t] ∪ [t, b + h] ⊂ [0, t] ∪ ]b − δ, b + δ[ ⊂ A ∪ A1∪ · · · ∪ An,
and then b+ h ∈ X for any real number 0 ≤ h < δ.
Example 4.40 The real lineR isn’t homeomorphic to the interval [0, 1], for [0, 1] is
compact, whereasR is not.
Proposition 4.41 1. Any closed subspace in a compact space is compact. 2. Finite unions of compact subspaces are compact.
Proof Let’s begin with (1) and take Y closed in the compact space X . We need to
prove that for any familyA of open subsets such that Y ⊂ ∪{A | A ∈ A}, there are finitely many sets A1, . . . , An ∈ A such that Y ⊂ A1∪ · · · ∪ An. The open family A ∪ {X − Y } is a cover of the compact space X, so we can pick A1, . . . , An ∈ A
such that X = (X − Y ) ∪ A1∪ · · · ∪ An. Immediately, then, Y ⊂ A1∪ · · · ∪ An.
Now let us see to (2). Let K1, . . . , Knbe compact subspaces in some X , andA
a family of open subsets in X covering K = K1∪ · · · ∪ Kn. For any h = 1, . . . , n
there is a finite number ofAh⊂ A that covers the compact set Kh, so K is contained
in the finite union∪{A | A ∈ A1∪ · · · ∪ An}.
Corollary 4.42 A subspace inR is compact if and only if it is closed and bounded.
Proof Note preliminarily that the definition of a compact set involves only the notions
of openness and cover, so it must be invariant under homeomorphism; that is, any space homeomorphic to a compact space is itself compact.
Let A ⊂ R be closed and bounded. Then A ⊂ [−a, a] for some a > 0. The interval[−a, a] is homeomorphic to [0, 1] and hence compact. Therefore A is closed in a compact set, whence compact.
Conversely, if A⊂ R is compact, the family of open intervals { ] − n, n[ | n ∈ N} covers A and so we can find a finite subcover:
A⊂ ] − n1, n1[ ∪ · · · ∪ ] − ns, ns[ .
Therefore A ⊂ ] − N, N[, where N is the largest among n1, . . . , ns. This implies
A is bounded. Now for any p ∈ A the map f (x) = 1/(x − p) is continuous and
defined onR − {p}. Its range f (A) is compact, hence bounded, so p ∈ ¯A and then
A is closed.
Corollary 4.43 Let X be a compact topological space. Any continuous map f: X →
R has a maximum and a minimum.
Proof The range f(X) is compact in R, so closed and bounded; but every closed
4.4 Compact Spaces 75 Theorem 4.44 Consider a closed map f: X → Y with Y compact. If f−1(y) is
compact for every y∈ Y , then X is compact.
Proof Without loss of generality we may assume f is onto (otherwise we can always
replace Y with f(X)). For any subset A ⊂ X define
A= {y ∈ Y | f−1(y) ⊂ A} .
As Y − A = f (X − A) and f is closed, it follows that Ais open provided A is open. Take an open coverA of X and call B the family of finite unions of elements of
A. The family B= {B| B ∈ B} is an open cover of Y : if y ∈ Y , the fibre f−1(y)
is compact, so there are A1, . . . , Am ∈ A such that f−1(y) ⊂ A1∪ · · · ∪ Am, and
then y∈ B, where B= A1∪ · · · ∪ Am.
But as Y is compact, we have a finite sequence B1, . . . , Bn ∈ B such that Y =
B1∪ · · · ∪ Bn. Then X = B1∪ · · · ∪ Bn, and since every Bi is a finite union inA,
we’ve found a finite subcover forA.
Proposition 4.45 LetB be a basis of the space X. If every cover of X made by
elements ofB has a finite subcover, X is compact.
Proof For any open U ⊂ X we set BU = {B ∈ B | B ⊂ U}. By assumption B is a
basis, so for any open set U we have U = ∪{B | B ∈ BU}.
LetA be an open cover of X and consider the open family C = ∪A∈ABA. It’s clear thatC covers X by basis elements, and because of the assumptions we’ve made C has a finite subcoverF. For any open set U ∈ F let’s choose an open set A(U) ∈ A such that U ∈ BA(U), i.e. U ⊂ A(U). Then {A(U) | U ∈ F} is a finite subcover of
A.
Proposition 4.46 Let K1 ⊃ K2 ⊃ · · · be a countable descending chain of non-
empty, closed and compact sets. Then
{Kn| n ∈ N} = ∅.
Proof Given n∈ N, the set K1− Knis open in K1. Now observe that the intersection
of the closed sets Knis empty precisely when the sets K1− Kncover K1. Exercises
4.24 Consider the metric spaceQ of rational numbers with the Euclidean distance.
Prove that
K =
x∈ Q | 0 ≤ x ≤√2 is closed and bounded, yet not compact.
4.25 (♥) A family of subsets A in a set X is said to enjoy the finite-intersection property if for every (non-empty) finite subfamilyF ⊂ A we have ∩{A | A ∈
Prove that a topological space is compact if and only if every closed family with the finite-intersection property has non-empty intersection. Convince yourself that Proposition4.46is a special case of this.
4.26 Let X be a compact space, U ⊂ X open and {Ci | i ∈ I } a closed family in X
with
i∈I
Ci ⊂ U .
Prove that there exist finitely many indices i1, . . . , in∈ I such that
Ci1∩ · · · ∩ Cin ⊂ U.
4.27 Prove that the space of Exercise3.3is compact and Hausdorff. Deduce that any set admits a topology that renders it a compact T2 topological space.
4.28 Denote byRutthe real line with the upper topology (Example3.3). Prove that
every compact, non-empty set inRuthas a maximum. Conclude that if X is compact,
every upper semi-continuous map f: X → R (Exercise3.28) has a maximum.
4.29 (♥) Let X be a compact space and f : R → X a continuous and closed map.
Show that there exists x ∈ X such that f−1(x) is infinite.
4.30 (K, ♥) Let X be compact and { fi: X → [0, +∞[ | i ∈ I } a non-empty family
of continuous maps. Prove that
f: X → [0, +∞[ , f(x) = inf{ fi(x) | i ∈ I },
has a maximum point in X . Find an example showing that there is no minimum, in general.