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every open cover of
A
in
X
contains a finite subcover of
A
(Exercise 5.5.1). Therefore,
when it comes to the compactness of a space, we do not have to distinguish between
whether we think of a space by itself or as a subspace of another space.
5.5.2. Theorem.
A closed subset
A
of a compact space
X
is compact.
Proof.
Let {
U
i
} be an open cover of
A
. Since
A
is closed,
X
-
A
is open and so
{
U
i
} » {
X
-
A
} is an open cover of
X
. Since
X
is compact, there is a finite subcover
and removing the set
X
-
A
, if it is in this subcover, will give us a finite subset of {
U
i
}
that covers
A
.
5.5.3. Theorem.
A compact subset
A
of a Hausdorff space
X
is closed.
Proof.
To show that
A
is closed, we need to show that
X
-
A
is open. The proof pro-
ceeds just like the proof for the case
X
=
R
n
in Theorem 4.2.4. Let
x
X
-
A
. Now
X
is a Hausdorff space. Therefore, for every
a
A
there is an open neighborhood
U
a
and
V
a
of
a
and
x
, respectively, such that
U
a
«
V
a
= f. The collection {
V
a
} is an open
cover of
A
. Since
A
is compact, there is a finite subcover {
V
a
i
}
1£i£k
. It follows that
k
I
U
=
a
i
i
=
1
is an open neighborhood of
x
contained in
X
-
A
.
An extremely important theorem with many consequences is the following.
5.5.4. Theorem.
(The Tychonoff Product Theorem) The product
X
1
¥
X
2
¥ ...¥
X
k
of nonempty topological spaces
X
i
is compact if and only if each
X
i
is compact.
Proof.
See [Eise74].
The Tychonoff product theorem is also true for infinite products.
The next theorem characterizes compact sets in Euclidean space and was already
stated and partially proved in Chapter 4 (Theorem 4.2.4). The proof relies on the
following:
5.5.5. Lemma.
(The Heine-Borel Theorem) A closed interval [a,b] in
R
is compact.
Proof.
Exercise 5.5.2. See [Eise74].
A subset
A
of
R
n
is compact if and only if it is closed and bounded.
5.5.6. Theorem.
Proof.
We already proved that compact implies closed and bounded in Theorem
4.2.4. We sketch a proof of the converse here.
Let
A
be a closed and bounded subset of
R
n
. First, consider the case n = 1. Now a
closed and bounded subset
A
in
R
is a closed subset of some interval [a,b]. But [a,b] is
compact by the Heine-Borel theorem, therefore, Theorem 5.5.2 implies that
A
is
