Given any finite subset
{
U
m
1
, . . . , U
m
n
}
of
U
, let
m
= max
{
m
1
, . . . , m
n
}
Then
∪
n
i
=1
U
m
i
=
U
m
=
1
m
,
2
⊇
(0
,
1]
so (0
,
1] is not compact.
Note that this argument does not work for [0
,
1]. Given an open cover
{
U
λ
:
λ
∈
Λ
}
, there must be some
λ
∈
Λ such that 0
∈
U
λ
, and therefore
U
λ
⊇
[0
, ε
) for some
ε >
0, and a finite number of the
U
m
’s we used
to cover (0
,
1] would cover the interval (
ε,
1]. This is not a proof that [0
,
1] is compact, since we need to
show that
every
open cover has a finite subcover, but it is suggestive, and we will soon see that [0
,
1] is
indeed compact.
Example:
[0
,
∞
) is closed but not compact. To see that [0
,
∞
) is not compact, let
U
=
{
U
m
= (
−
1
, m
) :
m
∈
N
}
Given any finite subset
{
U
m
1
, . . . , U
m
n
}
of
U
, let
m
= max
{
m
1
, . . . , m
n
}
Then
U
m
1
∪ · · · ∪
U
m
n
= (
−
1
, m
)
⊇
[0
,
∞
)
Theorem 2 (8.14)