MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTES
31
the sets in
C
, whose union contains
S
. That is,
C
k
∈ C
for all
k
, and
S
⊂ ∪
n
k
=1
C
k
.
Note that if
C
1
⊂
C
2
⊂ · · · ⊂
C
n
then
C
n
=
∪
n
k
=1
C
k
, so
S
⊂
C
n
.
Example.
Let
S
= [0
,
1] and let
C
=
{
(
−
1
50
,
1
50
)
,
(
1
2
,
3
2
)
,
(
1
4
,
3
4
)
,
(
1
8
,
3
8
)
,
(
1
16
,
3
16
)
,
(
1
32
,
3
32
)
,
(
1
64
,
3
64
)
,
(
1
128
,
3
128
)
,
· · · }
.
It is easy to see that in this last example,
C
is an infinite collection of open sets
whose union contains
S
. That is,
C
is an open cover of
S
. But in fact the first 7
sets listed in
C
have a union which already contains
S
.
These 7 sets comprise a
finite subcover. This example also illustrates the condition in (2) of the Heine-Borel
Theorem.
Theorem 3.30.
(The Heine-Borel theorem)
If
S
is a nonempty set of real numbers
then the following are equivalent:
(1)
S
is closed and bounded.
(2)
S
has the following property: whenever
C
is a collection of open sets whose
union contains
S
, then actually
S
is contained in the union of a finite
number of the sets in
C
.
(In English: every open cover of
S
has a finite
subcover.)
Proof.
(2)
⇒
(1)
Let
C
=
{
(
−
n, n
) :
n
∈
N
}
. By (2), this collection has a finite