Math 521, Lecture 2
Hints for Assignment # 3
Problem 1
Let
X
be a metric space, and let
E
⊂
X
be a subset. The
interior
of the set
E
is the
set
E
o
=
x
∈
E
there exists
r >
0 so that
B
(
x, r
)
⊂
E
. The
closure
of
E
is the set
E
=
E
∪
E
where
E
is the set of limit points of
E
.
(a)
Prove that
E
o
⊂
E
is always an open set, and that
E
⊃
E
is always a closed set.
Let
x
∈
E
o
.
By hypothesis, there exists
r >
0
so that
B
(
x, r
)
⊂
E
.
To show that
E
o
is open, it
suffices to show that
B
(
x, r
)
⊂
E
o
. Let
y
∈
B
(
x, r
)
. We showed in class that
B
(
x, r
)
is open. Thus
there exists
δ >
0
so that
B
(
y, δ
)
⊂
B
(
x, r
)
. But since
B
(
x, r
)
⊂
E
, this shows that
B
(
y, δ
)
⊂
E
, and
hence
y
∈
E
o
. Thus
B
(
x, r
)
⊂
E
o
.
To show that
E
is closed, it suffices to show that every limit point of
E
is contained in
E
. Let
x
be
a limit point of
E
.
Then for each
n
∈
N
there exists a point
y
n
∈
E
∪
E
such that
y
n
=
x
and
d
(
x, y
n
)
<
1
n
. It follows (why?) that
x
is either a limit point of
E
or a limit point of
E
. If
x
is a limit
point of
E
, then by definition it belongs to
E
which is a subset of
E
. If
x
is a limit point of
E
, then
by a problem on Assingment #2, it follows that
x
∈
E
. Thus in either case,
x
∈
E
.
(b)
Prove that
E
o
is the largest open set contained in
E
and that
E
is the smallest closed set
containing
E
. (Part of the problem is to figure out what “largest” and “smallest” should mean.)
By part (a),
E
o
is an open set contained in
E
. On the other hand, suppose that
V
is any open set
contained in
E
. Let
x
∈
V
. Then since
V
is open, there exists
r >
0
so that
B
(
x, r
)
⊂
V
. But since
V
⊂
E
, it follows that
B
(
x, r
)
⊂
E
. Hence
x
∈
E
o
, and consequently,
V
⊂
E
o
. Thus
E
o
is an open
set contained in
E
and it contains every open set contained in
E
. In this sense it is the largest open
set contained in
E
.
Also by part (a),
E
is a closed set containing
E
. Suppose
W
is a closed set containing
E
. Let
x
be
any limit point of
E
. Then
x
is also a limit point of
W
since
E
⊂
W
. But since
W
is closed, it follows
that
x
∈
W
. Thus
W
contains
E
. Since by hypothesis it contains
E
, it contains
E
∪
E
=
E
. Thus
we have shown that if
W
is any closed set containing
E
, then
W
contains
E
. It is in this sense that
E
is the smallest closed set containing
E
.

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