Since (
x
n
) is bounded,
∃
M
∈
R
such that

x
n
 ≤
M
for all
n
∈
N
.
By definition, we have
that
s
n
≥
x
n
≥ 
M
. Therefore, since
n
∈
N
was arbitrary, (
s
n
)
n
∈
N
is bounded below by

M
and decreasing and we conclude that
S
:= inf
s
n
= lim
n
→∞
s
n
. We will inductively construct a
subsequence (
x
n
k
)
k
∈
N
which converges to
S
.
The first element
Simply take
x
n
1
=
x
1
.
The
k
th element (
k
>
1
)
Assume we constructed
x
n
k

1
.
We want to find
x
n
k
such that

x
n
k

S

<
1
/k
. Since,
S
= lim
m
→∞
s
m
,
∃
M
0
k
∈
N
such that

s
m

S

<
1
/
(2
k
) for any
m
≥
M
0
k
. In particular, we can find
M
k
> n
k

1
such that

s
M
k

S

<
1
/
(2
k
). Now since
s
M
k
= sup
n
≥
M
k
x
n
,
∃
n
k
≥
M
k
> n
k

1
such that
s
M
k
≤
x
n
k
< s
M
k
+ 1
/
(2
k
). Then,

x
n
k

S
 ≤ 
x
n
k

s
M
k

+

s
M
k

S

<
1
/
(2
k
) + 1
/
(2
k
) = 1
/k.
Since the sequence (
x
n
k
)
k
∈
N
satisfies

x
n
k

S

<
1
/k
for all
k
, by the Archimedean property,
∀
>
0
,
∃
K
( )
∈
N
such that 1
/K
( )
<
⇒ 
x
n
k

S

<
∀
k
≥
K
( )
and we conclude
x
n
k
→
S
as
k
→ ∞
.
11
11. Let
L
⊂
R
.
The set
L
is said to be open if for any
x
∈
L
there exists
>
0 such that
(
x

, x
+ )
⊂
L
. The set
L
is said to be closed if its complement
L
c
=
{
x
∈
R
:
x /
∈
L
}
is open.
(a) Prove that
L
is closed if and only if for any convergent sequence (
x
n
) with
x
n
∈
L
, the
limit
x
= lim
n
→∞
x
n
=
x
is also an element of
L
.
(b) Let (
x
n
) be a bounded sequence. A point
x
∈
R
is called an accumulation point of (
x
n
) if
there exists a subsequence (
x
n
k
) of (
x
n
) such that lim
k
→∞
x
n
k
=
x
. We denote by
L
the
set of all accumulation points of (
x
n
). By the BolzanoWeierstraß Theorem, the set
L
is
nonempty. Prove that
L
is a bounded closed set.
(c) Let (
x
n
) be a bounded sequence, let
L
be as in part (b) and let
S
be as in problem 1. Prove
that
S
= sup
L
.
Solution:
(a)(
⇒
) Let
L
be a closed set. Let (
x
n
)
n
∈
N
be converging sequence with
x
n
∈
L
and lim
n
→∞
x
n
=
x
.
Then, in particular,
∀
>
0,
∃
n
∈
N
such that

x
n

x

<
.
Therefore,
∀
>
0,
can find
x
n
∈
L
such that
x
n
∈
(
x

, x
+ ). Therefore,
∀
>
0, (
x

, x
+ )
6⊆
L
c
.
Equivalently,
@
>
0 such that (
x

, x
+ )
⊆
L
c
. Since
L
c
is open, we must conclude
that
x /
∈
L
c
and therefore that
x
∈
L
.
(
⇐
) Let
L
be a set that is not closed. Then
L
c
is not open, which implies that
∃
x
∈
L
c
such
that
∀
n
∈
N
, (
x

n

1
, x
+
n

1
)
6⊆
L
c
⇔
(
x

n

1
, x
+
n

1
)
∩
L
6
=
∅
. Hence,
∀
n
∈
N
,
∃
x
n
∈
L
such that
x
n
∈
(
x

n

1
, x
+
n

1
). We have constructed a sequence (
x
n
)
n
∈
N
in
L
with the property that
∀
n
∈
N
,

x
n

x

< n

1
, which implies that lim
n
→∞
x
n
=
x
where
x /
∈
L
. Therefore, it is not true that for any converging sequence (
x
n
)
n
∈
N
with
x
n
∈
L
, the limit
x
= lim
n
→∞
x
n
is also an element of
L
.
Remark:
A lot of people did well on the first implication. For the second, a lot of people have confused
“
L
is not closed” and “
L
is open”: they do not mean the same thing. Some sets are both open and
closed (clopen) and some sets are neither closed nor open.