[ Pobierz całość w formacie PDF ]
Obviously, then, the collection {Bx : x ∈ C} of open sets covers C. Hence, there must
be a finite sub-collection, say of sets Bx , . . . , Bx , that covers C. Since the union of
balls Bx , . . . , Bx must be bounded, this implies that C is bounded as well.
11.3 PROPOSITION. Every closed subset of a compact set is compact.
PROOF. Let D be compact. Let C ⊂ D be closed. Fix a collection of open sets that
covers C. Adding the open set E \ C to that collection, we obtain a collection of open
sets that covers D. Since D is compact, the latter collection has a finite sub-collection
that still covers D. Removing E \C from that sub-collection (if it were in), we obtain a
finite sub-collection of the original collection that covers C. Thus, C must be compact.
Compact Subspaces
Recall that every subset D of E can be regarded as a metric space by itself, with the
metric it inherits from E. Whether D is open or not as a subset of E, it is open
automatically when it is regarded as a metric space. The concept of compactness does
not suffer from such foolishness.
11.4 PROPOSITION. A set D is compact as a metric space if and only if it is compact
as a subset of E.
40
11
1
n
1
n
11. COMPACTNESS
PROOF. A subset of D is an open ball in the space D if and only if it has the form
B ∩ D for some open ball B of the space E. Since an open set is the union of all the
open balls it contains, it follows that A is an open subset of the space D if and only if
A = B∩D for some open subset B of the space E. Now, the definition of compactness
does the rest.
Cluster Points, Convergence, Completeness
This is to look into the connections between compactness and convergence.
11.5 DEFINITION. A point x in E is called a cluster point
of a subset A of E
provided that every open ball centered at x contains infinitely many points of A.
11.6 THEOREM. Every infinite subset of a compact set has at least one cluster point
in that compact set.
PROOF. We shall show that if C is compact, and A ⊂ C, and A has no cluster point
in C, then A is finite. Let A and C be such. Since no x in C is a cluster point of A, for
every x in C there is an open ball B(x, r) that contains only finitely many points of A.
Those open balls cover C obviously. Since C is compact, there must be a finte number
of them that cover C and, therefore, A. Since each one of those finitely many balls has
a finte number of points of A, the total number of points in A must be finite.
The following is the way compactness helps in discussing convergence. In particu-
lar, together with Proposition 10.4, it shows that every Cauchy sequence in a compact
set is convergent.
11.7 THEOREM. Every sequence in a compact set has a subsequence that converges
to some point of that set.
PROOF. Let C be compact. Let (xn) ⊂ C. If the set A = {x1, x2, . . .} is finite,
then at least one point of A, say x, appears infinitely often in the sequence, and hence
(x, x, . . .) is a subsequence, which obviously converges to x ∈ A ⊂ C. Now suppose
that A is infinite. By the preceding theorem, then A has a cluster point x in C. Since
each one of the balls B(x, 1/n), n = 1, 2, . . ., has infinitely many points in C, we
may pick k1 so that xk is in B(x, 1), pick k2 k1 so that xk is in B(x, 1/2), pick
k3 k2 so that xk is in B(x, 1/3), and so on. Obviously, (xk ) converges to x.
41
2
1
2
3
n
Other terms in common use include limit point, adherence point, point of accumulation, etc.
2
METRIC SPACES
PROOF. Let C be compact. The preceding theorem implies that every convergent
sequence in C converges to a point of C. Thus, C is closed by Theorem 9.5.
11.9 COROLLARY. Every compact metric space is complete. Every Cauchy sequence
in a compact metric space is convergent.
PROOF. The second statement is immediate from Theorem 11.7 and Proposition 10.4.
The first follows from the second by the definition of completeness.
Compactness in Euclidean Spaces
We have seen that, for an arbitrary metric space, every compact set is bounded and
closed (Proposition 11.2 and Corollary 11.8). In the case of Euclidean spaces, the
converse is true as well. This is called the Heine-Borel Theorem.
11.10 THEOREM. A subset of a Euclidean space is compact if and only if it is bounded
and closed.
We start by listing an auxiliary result that is trivial at least for R, R2, R3. We omit
its proof.
11.11 LEMMA. Let B be a bounded subset of a Euclidean space E. Then, for every
0 there is a finite collection of closed balls of radius that covers B.
Here is the proof of Theorem 11.10.
PROOF. As mentioned above, 11.2 and 11.8 prove the necessity part. We now prove
the sufficiency of the condition.
[ Pobierz całość w formacie PDF ]