Question:
- Prove that the intersection of two compact sets is compact, using criterion (2).
- Prove that the intersection of two compact sets is compact, using criterion (1).
- Prove that the intersection of two compact sets is compact, using criterion (3).
Probably the most important new idea you'll encounter in real analysis is
the concept of compactness. It's the compactness of [a, b] that makes a
continuous function reach its maximum and that makes the Riemann in-
tegral exist. For subsets of R"h, there are three equivalent definitions of
compactness. The first, 9.2(1), promises convergent subsequences. The sec-
ond, 9.2(2) brings together two apparently unrelated adjectives, closed and
bounded. The third, 9.2(3), is the elegant, modern definition in terms of
open sets; it is very powerful, but it takes a while to get used to.
9.1. Definitions. Let S be a set in Rn. S is bounded if it is contained
in some ball B(0, R) about 0 (or equivalently in a ball about any point). A
collection of open sets {U} is an open cover of S if S is contained in U U(.
A finite subcover is finitely many of the Ua which still cover S. Following
Heine and Borel, S is compact if every open cover has a finite subcover.
9.2. Theorem. Compactness. The following are all equivalent conditions
on a set S in ]Rn.
(1) Every sequence in S has a subsequence converging to a point of S.
(2) S is closed and bounded.
(3) S is compact: every open cover has a finite subcover.
Criterion
(1)is the Bolzano-Weierstrass condition for compactness,
which you met for R in Theorem 8.3. The more modern Heine-Borel crite-
rion (3) will take some time to get used to. A nonclosed set such as (0, 1] is
not compact because the open cover {(1/n, oo)} has no finite subcover. An
unbounded set such as R is not compact because the open cover {(-n, n)}
has no finite subcover. This is the main idea of the first part of the proof.
Proof. We will prove that (3)->(2) -> (1) -> (3).
(3) ->(2). Suppose that S is not closed. Let a be an accumulation point
not in S. Then the open cover {{Ix - aI > 1/n}} has no finite subcover.
Suppose that S is not bounded. Then the open cover {{JxI < n}} has no
finite subcover.
(2) -> (1). Take any sequence of points in S C ]Rn. First look at just the
first of the n components of each point. Since S is bounded, the sequence
of first components is bounded. By Theorem 8.3, for some subsequence,
the first components converge. Similarly, for some further subsequence, the
second components also converge. Eventually, for some subsequence, all of
the components converge. Since S is closed, the limit is in S.
(1) =>. (3). Given an open cover {Ua}, first we find a countable subcover.
Indeed, every point x of S lies in a ball of rational radius about a rational
point, contained in some Ua. Each of these countably many balls lies in
some U,,. Let {Vi} be that countable subcover.
Suppose that {V} has no finite subcover. Choose xl in S but not in
V1. Choose X2 in S but not in V1 U V2. Continue, choosing xn in S but
not in U{V : 1 < i < n}, which is always possible because there is no finite
subcover. Note that for each i, only finitely many xn (for which n < i)
lie in V. By (1), the sequence xn has a subsequence converging to some
x in S, contained in some Vi. Hence infinitely xn are contained in Vi, a
contradiction.
9.3. Proposition. A nonempty compact set S of real numbers has a largest
element (called the maximum) and a smallest element (called the minimum).
Proof. We may assume that S has some positive numbers, by translating it
to the right if necessary. Since S is bounded, there is a largest integer part D
before the decimal place. Among the elements of S that start with D, there
is a largest first decimal place d1. Among the elements of S that start with
D.d1, there is a largest second decimal place d2. Keep going to construct
a = D.dld2d3.... By construction, a is in the closure of S.
Since S is closed, a lies in S and provides the desired maximum.
A minimum is provided by - max(-S).
