Math 350
Advanced Calculus I
Fall 2009
Homework Solutions
Section 13
TOPOLOGY OF THE REALS
13.3
Find the interior of each set.
13.4
Find the boundary of each set.
(a)
Ω
1
n

n
∈
N
æ
Since int
S
⊆
S
, then the only possible interior points of the set are the elements of the
set itself. Suppose
x
0
= 1
/n
is an arbitrary element of the set
S
. Since every neighborhood
N
(1
/n
;
ε
) of
x
0
contains elements of
R
\
S
, then
x
0
is not an interior point. Therefore none of
the elements of
S
are interior points so
int
S
=
∅
.
For each element
x
0
= 1
/n
∈
S
, every neighborhood of
x
0
contains elements of
S
,
namely
x
0
itself, and elements of
R
\
S
. Therefore each point
x
0
∈
S
is a boundary point.
We want to show that 0 is also a boundary point of
S
. To show this, we need to show
that every neighborhood
N
(0;
ε
) of 0 contains points in
S
and points in
S
. Suppose
ε >
0
is arbitrary. Then
N
(0;
ε
) contains 0, which is not in
S
. Since
ε >
0, then the Archimedean
Property implies there exists
n
∈
N
such that 1
/n < ε
. Then 1
/n
∈
N
(0;
ε
). Since 1
/n
∈
S
,
then
N
(0;
ε
) contains a point in
S
. Therefore 0 is also a boundary point.
Finally, for any
x
0
/
∈
S
di±erent from 0, there exists a neighborhood which does not
contain elements of
S
. Choose
ε >
0 so that
N
(
x
0
;
ε
) misses all elements of
S
. Therefore
bd
S
=
S
∪{
0
}
=
{
1
,
1
2
,
1
3
, . . . ,
0
}
.
(c)
{
r
∈
Q

0
< r <
√
2
}
We want to show that
bd
S
=
£
0
,
√
2
§
=
©
x
∈
R

0
≤
x
≤
√
2
™
.
Suppose
x
0
is an arbitrary real number in
°
0
,
√
2
¢
. We want to show that
x
0
is a boundary
point. To show this, we need to show every neighborhood
N
(
x
0
;
ε
) of
x
0
contains points in
S
and points not in
S
. Suppose
ε >
0 is arbitrary. Then by the Density Theorems,
N
(
x
0
;
ε
)
contains rationals in
°
0
,
√
2
¢
and irrationals so contains points in
S
and points in
R
\
S
.
Similarly, each neighborhood of 0 or
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 Spring '08
 QIAN
 Calculus, Topology, Interior, Closed set, General topology, Boundary, boundary point

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