Given any fnite 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
²
6⊇
(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
)Forsome
ε>
0, and a fnite 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 fnite 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 fnite subset
{
U
m
1
m
n
}
oF
U
m
{
m
1
n
}
Then
U
m
1
∪···∪
U
m
n
−
1
)
6⊇
[0
,
∞
)
Theorem 2 (8.14)
Every closed subset
A
of a compact metric space
(
X, d
)
is compact.