Graphics Reference
In-Depth Information
, there is an infinity of
n
N
For any given positive
, such that
1/
n
, and weknow this bcause(1/
n
) is a null sequence. In
other words, there is an infinity of points from the set
in
any neighbourhood of 0. See qn 3.61 for the term 'neighbourhood'. This
makes 0 a
cluster point
or an
accumulation point
for thest.
Thepoint
1/
n
n
N
is
not
a cluster point for the set because we can find a
positivevalueof
such that
1/
n
is only possiblefor the
singlenatural numbr
n
4.
48 Statea valueof
such that
1/
n
only holds for a
uniquenatural numbr
n
.
49 Statea positivevalueof
such that
1/
n
does not
hold for any natural number
n
. This means that the point
is
not
a
cluster point for the set
1/
n
n
N
.
1
2
50 Name three cluster points for the set
1
8
1
4
1
3
0
1
1
2
1
3
m
,
n
N
.
Name two points between 0 and 1 which are not cluster points for
thest.
51 Can a finite set of points on the number line have a cluster point?
52 Describe an infinite set of points that has no cluster point.
Aset
A
of real numbers has an upper bound
U
if
a
U
for all
a
A
,
and has a lower bound
L
if
L
a
for all
a
A
. A set is said to be
bounded
if it has an upper bound and a lower bound. The language
here is like that for the upper bound and lower bound of a sequence
introduced in qn 3.5.
53
The Bolzano—Weierstrass Theorem
Let
A
be an infinite set of real numbers with an upper bound
U
and a lower bound
L
.
(i) Construct a sequence of distinct points in
A
.
(ii) Is the sequence (
a
) bounded?
(iii) Why must the sequence (
a
) contain a convergent
subsequence?
(iv) If the convergent subsequence (
a
)
a
, must
a
A
? Must
L
a
U
?
(v) Whether
a
A
or not, show that
a
is a cluster point of
A
.
