Homework 3
H´
ector Guillermo Cu´
ellar R´
ıos
February 9, 2006
13.9 Prove the folowing.
(a) An accumulation point of a set
S
is either an interior point of
S
or a
boundary point of
S
.
Proof.
Suppose that an accumulation point
x
of a set
S
is neither an
interior point of
S
nor a boundary point of
S
, then
x
is an interior
point of
R
\
S
, thus there is a neighborhood
N
of
x
such that
N
⊆
R
\
S
it follow that
N
∩
S
=
∅
and
N
*
∩
S
=
∅
but this is contradicts the
definition of an accumulation point.
(b) A boundary point of a set
S
is either an accumulation point of
S
or
an isolated point of
S
.
Proof.
Lets consider several cases:
Case 1:
The boundary point
x
∈
S
. Now we consider two subcases:
Subcase a:
There is a deleted neighborhood
N
*
of
x
such that
N
*
∩
S
=
∅
then by definition 13.14
x
is an isolated point.
Subcase b:
For every deleted neighborhood
N
*
of
x, N
*
∩
S
=
∅
then
by definition 13.14
x
is an accumulation point.
Case 2:
The boundary point
x
∈
S
. By definition 13.3 we know that
for every neighborhood
N
of
x
,
N
∩
S
=
∅
. Since we know that
x
∈
S
it follows that for every deleted neighborhood
N
*
of
x
,
N
*
∩
S
=
∅
thus by definition 13.14
x
is an accumulation point of
S
.

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